LINEAR ALGEBRA
4th STEPHEN H.FRIEDBERG, ARNOLD J.INSEL, LAWRENCE E.SPENCE
Exercises Of Chapter 1-4
http : //math.pusan.ac.kr/caf e home/chuh/
Contents
§1. Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.
1.2.
1.3.
1.4.
1.5.
1.6.
1.7.
Introduction . . . . . . . . . . . . . . . . . . . . . . . .
Vector Spaces . . . . . . . . . . . . . . . . . . . . . . .
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . .
Linear Combinations and Systems of Linear Equations
Linear Dependence and Linear Independence . . . . . .
Bases and Dimension . . . . . . . . . . . . . . . . . . .
Maximal Linearly Independent Subsets . . . . . . . . .
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1
1
3
9
19
24
32
50
§2. Linear Transformations and Matrices . . . . . . . . . . . . . . . . . . . . . . . . 55
2.1.
2.2.
2.3.
2.4.
2.5.
2.6.
2.7.
Linear Transformations, Null Spaces, and Ranges . . . . . . . . .
The matrix representation of a linear transformation . . . . . . .
Composition of Linear Transformations and Matrix Multiplication
Invertibility and Isomorphisms . . . . . . . . . . . . . . . . . . . .
The change of Coordinate Matrix . . . . . . . . . . . . . . . . . .
Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Homogeneous Linear Differential Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
77
87
101
116
123
140
§3. Elementary Matrix Operations and Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
3.1.
3.2.
3.3.
3.4.
Elementary Matrix Operations and Elementary Matrices
The Rank of a Matrix and Matrix Inverse . . . . . . . .
Systems of Linear equations - Theoretical aspects . . . .
Systems of Linear equations - Computational aspects . .
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154
162
172
178
§4. Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
4.1.
4.2.
4.3.
4.4.
4.5.
Determinants of Order 2 . . . . . . . . . . . . .
Determinants of Order n . . . . . . . . . . . . .
Properties of Determinants . . . . . . . . . . . .
Summary-Important Facts about Determinants
A characterization of the Determinants . . . . .
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188
192
198
213
215
1.1.
§1.
Vector Spaces
1.1. Introduction
1. Only the pairs in (b) and (c) are parallel
(a) x = (3, 1, 2) and y = (6, 4, 2)
@ 0 6= t ∈ R s.t. y = tx
(b) (9, −3, −21) = 3(−3, 1, 7)
(c) (5, −6, 7) = −1(−5, 6, −7)
(d) x = (2, 0, −5) and y = (5, 0, −2)
@ 0 6= t ∈ R s.t. y = tx
2. (a) x = (3, −2, 4) + t(−8, 9, −3)
(b) x = (2, 4, 0) + t(−5, −10, 0)
(c) x = (3, 7, 2) + t(0, 0, −10)
(d) x = (−2, −1, 5) + t(5, 10, 2)
3. (a) x = (2, −5, −1) + s(−2, 9, 7) + t(−5, 12, 2)
(b) x = (−8, 2, 0) + s(9, 1, 0) + t(14, −7, 0)
(c) x = (3, −6, 7) + s(−5, 6, −11) + t(2, −3, −9)
(d) x = (1, 1, 1) + s(4, 4, 4) + t(−7, 3, 1)
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1.1.
4. x = (a1 , a2 , · · · , an ) ∈ Rn , i = 1, 2, · · · , n
0 = (0, 0, · · · , 0) ∈ Rn s.t. x + 0 = x, ∀x ∈ Rn
5. x = (a1 , a2 ) ⇒ tx = t(a1 , a2 ) = (ta1 , ta2 )
6. A + B = (a + c, b + d), M = ( a+c
, b+d
)
2
2
7.
C = (v − u) + (w − u) + u = v + w − u
−−→ 1 −−→ −→ 1 −→
OD + 2 DB = OA + 2 AC
i.e. w + 12 (v − w) = 12 (v + w) = u + 21 (v + w − u − u)
k
j
w
t
h
t
d
d
hOˆS‰P
iOŠS‹P
i
u
v
v
XT^
v
XT]
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1.2.
1.2. Vector Spaces
1.
(a) T
(b) F (If ∃00 s.t. x + 00 = x, ∀x ∈ V, then 00 = 0 + 00 = 00 + 0 = 0, ∴ 00 = 0)
(c) F (If x = 0, a 6= b, then a · 0 = b · 0 but a 6= b)
(d) F (If a = 0, x 6= y, then a · x = 0 = a · y but x 6= y)
(e) T
(f) F (An m × n matrix has m rows and n cilumns)
(g) F
(h) F (If f (x) = ax + b, g(x) = −ax + b, then degf =degg=1, deg(f + g)=0)
(i) T (p.10 Example4)
(j) T
(k) T (p.9 Example3)

0
0
2. 
0
0
0
0
0
0

0
0

0
0
3. M13 = 3, M21 = 4, M22 = 5
4.
(a)
µ
6 3 2
−4 3 9
¶
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1.2.


1 −1
(b) 3 −5
µ3 8
¶
8 20 −12
(c)
4 0 28 
30 −20
(d) −15 10 
−5 40
4
(e) 2x + x3 + 2x2 − 2x + 10
(f) −x3 + 7x2 + 16
(g) 10x7 − 30x4 + 40x2 − 15x
(h) 3x5 − 6x3 + 12x + 6
5.






8 3 1
9 1 4
17 4 5
U = 3 0 0  , D = 3 0 0  , T =  6 0 0 
3 0 0
1 1 0
4 1 0
(∗) The total number of crossings
Brook trout
Rainbow trout
Brown trout
Fall Spring Winter
17
4
5
6
0
0
4
1
0
6.




3 1 1 4
4 2 1 3
M = 5 1 1 4 , 2M − A = 4 0 1 3
3 1 2 6
5 2 1 9
35 suites were sold during the June sale
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1.2.
7. We are going to show that f (t) = g(t) and (f + g)(t) = h(t), ∀t ∈ S = {0, 1}
(i) f (0) = 1 = g(0), f (1) = 3 = g(1)
(ii) (f + g)(0) = 2 = h(0), (f + g)(1) = 6 = h(1)
∴ In F(S, R), f = g and f + g = h
8. VS 7, 8
9. (a) Exercise 1(b)
(b) If ∃y 0 s.t. x + y 0 = 0, then x + y = 0x + y 0
By the theorem 1.1, y = y 0
(c) a · 0 + a · 0 = a(0 + 0) = a · 0 = a · 0 + 0
By the theorem 1.1, a · 0 = 0
10. V = D(R), ∀s ∈ R
(1) ∀f, g ∈ V, f + g = g + f
(f + g)(s) = f (s) + g(s) = g(s) + f (s) = (g + f )(s)
∴ f +g =g+f
(2) ∀f, g, h ∈ V, (f + g) + h = f + (g + h)
((f + g) + h)(s) = (f + g)(s) + h(s) = f (s) + g(s) + h(s)
= f (s) + (g + h)(s) = (f + (g + h))(s)
∴ (f + g) + h = f + (g + h)
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1.2.
(3) ∃0 ∈ V s.t. f + 0 = f, ∀f ∈ V
(f + f 0 )(s) = f (s) + f 0 (s) = 0(s)
∴ f 0 (s) = 0(s) − f (s) = (0 − f )(s) = (−f )(s)
∴ f 0 = −f
(5) ∀f ∈ V, 1 · f = f
(1 · f )(s) = 1(f (s)) = f (s)
∴ 1·f =f
(6) ∀a, b ∈ F, (ab)f = a(bf )
((ab)f )(s) = (ab)f (s) = a(bf (s)) = a(bf )(s)
∴ (ab)f = a(bf )
(7) ∀a ∈ F, a(f + g) = af + ag
a(f + g)(s) = a(f (s) + g(s)) = af (s) + ag(s) = (af + ag)(s)
∴ a(f + g) = af + ag
(8) ∀a, b ∈ F, (a + b)f = af + bf
(a + b)f (s) = af (s) + bf (s) = (af + bf )(s)
∴ (a + b)f = af + bf
11. V = {0}, ∀a, b ∈ F
12.
(1) ∀f, g ∈ V, t ∈ R
(f + g)(−t) = f (−t) + g(−t) = f (t) + g(t) = (f + g)(t) ∴ f + g ∈ V
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1.2.
(g + f )(−t) = (g + f )(t) ∴ g + f ∈ V
∴ f +g =g+f
(2) ∀f, g, h ∈ V, (f + g) + h = f + (g + h)
(3) ∃0 ∈ V s.t. f + 0 = f, ∀f ∈ V
(4) ∃f 0 ∈ V s.t. f + f 0 = 0, ∀f ∈ V, f 0 = −f
(5) ∀f ∈ V, 1f = f
(6) ∀a, b ∈ F, (ab)f = a(bf )
((ab)f (−t) = (ab)f (t), ∴ abf ∈ V )
(7) ∀a ∈ F, ∀f, g ∈ V, a(f + g) = af + ag
(8) ∀a, b ∈ F, ∀f ∈ V, (a + b)f = af + bf
13. No, (VS 4) fails
(VS 3) ∃ (0, 1) ∈ V s.t. (a1 , a2 ) + (0, 1) = (a1 , a2 ), ∀(a1 , a2 ) ∈ V
(VS 4) If a2 = 0, then @(b1 , b2 ) ∈ V s.t. (a1 , 0) + (b1 , b2 ) = (0, 1)
14. Yes (∵ R ⊆ C)
15. No (∵ C
R)
α ∈ F = C, αx ∈
/ V = Rn , ∀x ∈ V
16. Yes (∵ Q ⊆ R)
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1.2.
17. No, (VS 5) fails
(VS 5) If a2 6= 0, then 1(a1 , a2 ) = (a1 , 0) 6= (a1 , a2 )
18. No, (VS 1) fails
(VS 5) If a1 6= b1 , then it fails to hold (VS 1)
19. No, (VS 8) fails
(VS 8) If c1 + c2 6= 0, c1 6= 0, c2 6= 0, then it fails to hold (VS 8)
20. (VS 1) ∀{an }, {bn } ∈ V
{an } + {bn } = {a1 + b1 , a2 + b2 , · · · } = {b1 + a1 , b2 + a2 , · · · } = {bn } + {an }
(VS 2) ({an } + {bn }) + {cn } = {an } + ({bn } + {cn })
(VS 3) ∃{0} s.t. {an } + {0} = {an }
(VS 4) ∃{−an } s.t. {an } + {−an } = {an } − {an } = {0}, ∀{an } ∈ V
(VS 5) ∃{1} s.t. {1}{an } = {an }, ∀{an } ∈ V
(VS 6) ∀α, β ∈ F, (αβ){an } = α(β{an })
(VS 7) ∀α ∈ F, ∀{an }, {bn } ∈ V, α({an } + {bn }) = α{an } + α{bn }
(VS 8) ∀α, β ∈ F, (α + β){an } = α{an } + β{an }
22. 2mn
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1.3.
1.3. Subspaces
1. (a) F (p.1 Definition of subspace)
(b) F (0 ∈
/ ∅)
(c) T (V and {∅} are subspaces of V )
(d) F (p.19 Theorem 1.4)
(e) F
(f) F (p.18 Example 4)
(g) F ((0, 0, 0) ∈ W , but (0, 0, 0) ∈
/ R2 )
2. (b), (c), (e), (f), (g) are not square matrices
(a) -5, (d) 12, (h) -6
3. ∀A, B ∈ Mm×n (F ), a, b ∈ F (1 ≤ i ≤ m, 1 ≤ j ≤ n)
(aA + bB)tij = (aA + bB)ji = (aA)ji + (bB)ji
= a(A)ji + b(B)ji = aAtij + bBijt = (aAt + bB t )ij
∴ (aA + bB)t = aAt + bB t
4. (At )tij = (At )ji = Aij
5. (A + At )t = At + (At )t = At + A = A + At
∴ A + At is symmetric
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1.3.
6. tr(aA + bB) =
n
P
(aA + bB)ii =
i=1


7. A = 

n
P
(aA)ii +
i=1
n
P
(bB)ii = atr(A) + btr(B)
i=1

a11



a22 O
O a33
⇒ At = A
a44
∴ A is symmetric
8. (a) Yes
(b) No ((0, 0, 0) ∈
/ W2 )
(c) Yes
(d) Yes
(e) No ((0, 0, 0) ∈
/ W5 )
(f) No x + y ∈
/ W6 ), ∀x, y ∈ W6
9. (1) W1 ∩ W3 = {0} is a subspace of R3
(2) W1 ∩ W4 = W1 is a subspace of R3
(3) W3 ∩ W4 = {(a1 , a2 , a3 ) ∈ R3 | 3a1 = 11a2 , 3a3 = 23a2 } is a subspace of R3
10. (i) W1 is a subspace of F n
(ii) W2 is not a subspace of F n
(∵ (0, 0, 0) ∈
/ W2 )
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1.3.
11. No (The given set is not closed under addition)
(Example) ∀f, g ∈ W
Let f (x) = an xn + an−1 xn−1 + · · · + a0 , degf = n
g(x) = bn xn + bn−1 xn−1 + · · · + b0 , degg = n
If bn = −an , then deg(f + g) = n − 1
∴ f +g ∈
/W
∴ W is not a subspace of V
15. Yes
17. (⇒) Theorem 1.3
(⇐) W is a subspace of V
(i) 0 ∈ W (a = 0 ∈ F : field )
(ii) ax ∈ W
(iii) x + y ∈ W
18. (⇒) Theorem 1.3
(⇐) W is a subspace of V
(i) 0 ∈ W
(ii) ax ∈ W (y = 0 ∈ W, by (i) )
(iii) x + y ∈ W (a = 1 ∈ F : field)
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1.3.
19. (⇐) If W1 ⊆ W2 , then W1 ∪ W2 = W2 is a subspace of V
If W2 ⊆ W1 , then W1 ∪ W2 = W1 is a subspace of V
∴ W1 ∪ W2 is a subspace of V
(⇒) If ∃ a ∈
/ W1 , a ∈ W2 ∃ b ∈
/ W2 , b ∈ W1 ,
then ab ∈ W1 ∪ W2
But if ab ∈ W1 , then a = (ab)b−1 inW1
if ab ∈ W2 , then b = (ba)a−1 inW2
It’s a contrdiction
∴ W1 ⊆ W2 or W2 ⊆ W1
20. Induction on n
In case of n = 2, it’s clear
Assume that this holds for n = k − 1 (k > 2)
By the induction hypothesis,
k−1
P
ai wi + ak wk ∈ W, wk ∈ W, ak ∈ F
i=1
∴ ∀wi ∈ W,
n
P
ai wi ∈ W , where ∀ai ∈ F, i = 1, 2, · · · , n
i=1
21. (i) {0} → 0, ∴ {0} ∈ W
(ii) lim{an } = a, lim{bn } = b
lim({an } + {bn }) = lim{an } + lim{bn } = a + b
∴ {an } + {bn } ∈ W
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1.3.
(iii) lim c{an } = c lim{an }
∴ c{an } ∈ W
∴ W is a subspace of V
22.
Let W1 = {g ∈ F (F1 , F2 ) | g(−t) = g(t), for each t ∈ F1 }
W2 = {g ∈ F (F1 , F2 ) | g(−t) = −g(t), for each t ∈ F1 }
(1) ∀g1 , g2 ∈ W1 , ∀c ∈ F
(i) 0 ∈ W1
(ii) (g1 + g2 )(−t) = g1 (−t) + g2 (−t) = g1 (t) + g2 (t) = (g1 + g2 )(t)
∴ g1 + g2 ∈ W1
(iii) cg1 (−t) = c(g1 (t)) = cg1 (t)
∴ cg ∈ W1
∴ W1 is a subspace of V
(2) ∀g1 , g2 ∈ W2 , ∀c ∈ F
(i) 0 ∈ W2
(ii) (g1 + g2 )(−t) = g1 (−t) + g2 (−t) = −g1 (t) − g2 (t) = −(g1 + g2 )(t)
∴ g1 + g2 ∈ W2
(iii) cg1 (−t) = c(−g1 (t)) = −cg1 (t)
∴ cg ∈ W2
∴ W2 is a subspace of V
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1.3.
23. (a) W1 + W2 = {x + y | x ∈ W1 and y ∈2 }
(i) 0 = 0 + 0 ∈ W1 + W2
(ii) ∀x1 + y1 , x2 + y2 ∈ W1 + W2
(x1 + y1 ) + (x2 + y2 ) = (x1 + x2 ) + (y1 + y2 ) ∈ W1 + W2
(iii) ∀x1 + y1 ∈ W1 + W2 , ∀c ∈ F
c(x1 + y1 ) = cx1 + cy1 ∈ W1 + W2
(b) ∀ W as a subspace of V s.t. W1 ⊆ W, W2 ⊆ W
∀x ∈ W1 ⊆ W, ∀y ∈ W2 ⊆ W ⇒ x + y ∈ W1 + W2 ⊆ W
24. V = F n
(i) W1 ∩ W2 = {(0, 0, · · · , 0)}
(ii) W1 + W2 ⊆ V is clear
∀v = (a1 , a2 , · · · , an ) = (a1 , a2 , · · · , an−1 , 0) + (0, 0, · · · , 0, an ) ∈ W1 + W2
∴ V = W1 + W2
25. (a) W1 ∩ W2 = {0}
(b) V = {f (x) ∈ P (F ) | f (x) = a0 + a1 x + · · · }
(∵) Since W1 ⊆ V and W2 ⊆ V , W1 + W2 ⊆ V is clear
∀f ∈ V, f = a0 + a2 x + · · · + a1 x + a3 x3 + · · · ∈ W1 + W2
∴ V = P (F ) = W1 ⊕ W2
26. (a) W1 ∩ W2 = {A ∈ Mm×n | Aij = 0 ∀i, j} = {0}
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1.3.
(b) W1+ W2 ⊆ V is clear 
a11 a12 · · · a1n
 a21 a22
a2n 


∀A =  ..
 ∈ Mm×n ,
...
 .

a
a
· · · amn
 
 m1 m2
0
0
a11 a12 · · · a1n


 0 a22
a
a
2n 
 21 0

+
A =  ..

.
.
.
..
..  
 ..
 .
am1 · · ·
0
0 · · · amn
∴ V = P (F ) = W1 ⊕ W2
···
..
.
am(n−1)

0


..  ∈ W1 + W2
.
0
27. (a) W1 ∩ W2 = {A ∈ Mm×n | Aij = 0 ∀i, j} = {0}
(b) W1+ W2 ⊆ V is clear 
a11 a12 · · · a1n
 0 a22
a2n 


∀A =  ..
..  ∈ Mm×n ,
.
.
 .
.
. 
0
0 · · · amn

 

a11 0 · · ·
0
0 a12 · · ·
a1n
 0 a22

0 
a2n 

 0 0

A =  ..
+


.
.
..  ∈ W1 + W2
.
.
.
.
.
.
 .
.
.
.  .
. 
0
0
amn
0 0 · · · a(n−1) n
∴ V = P (F ) = W1 ⊕ W2
28. (a) W1 ∩ W2 = {0}
(b) W1 + W2 ⊆ V is clear
t
t
) + ( A+A
) ∈ W1 + W2
∀A ∈ V, A = ( A−A
2
2
∴ V ∈ W1 + W2
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1.3.
(cf) char(F ) 6= 2
{ 12 (A − At )}t = 21 (At − A) = −{ 12 (A − At )} ∴ 21 (A − At ) ∈ W1
{ 12 (A − At )}t = 21 (A + At ) ∴ 12 (A + At ) ∈ W2
29. W1 : strictly lower triangular matrices
W2 : all symmetric matrices
(a) W1 ∩ W2 
= {0}



a11 a12 · · · a1n
a11 a12 · · · a1n
 a21 a22
 a12 a22
a2n 
a2n 




0
(b) Let A =  ..
 ∈ V and A =  ..
 ∈ W2
.
.
.
.
 .

 .

.
.
an1 an2 · · · ann
a1n a2n · · · ann
0
0
then ∀A = (A − A ) + A ∈ W1 + W2
∴ V ⊆ W1 + W2
∴ V = W1 + W2
30.
(⇒) If x = x1 + x2 = x3 + x4 , x1 , x3 ∈ W1 and x2 , x4 ∈ W2
then x1 − x3 = x4 − x2 ∈ W1 ∩ W2 = {0}
∴ x1 = x3 , x 2 = x4
(⇐) (i) V = W1 + W2 is clear
(ii) If w ∈ W1 ∩ W2 , then w = w + 0 = 0 + 0
∴ w=0
∴ W1 ∩ W2 = {0}
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1.3.
31. (a) (⇐) v ∈ W ⇒ v + W = W
∴ v + W is a subspace of V
(⇒) v + w1 , v + w2 ∈ W
v + w1 + v + w2 = v + (w1 + w2 + v) ∈ v + W
∴ w1 + w2 + v ∈ W
∴ v∈W
(b) v1 − v2 ∈ W ⇔ v1 − v2 + W = W ⇔ v1 + W = v2 + W
(c) Since v1 − v10 ∈ W and v2 − v20 ∈ W,
(v1 − v10 ) + (v2 − v20 ) = (v1 + v2 ) − (v10 + v20 ) ∈ W
∴ (v1 + v2 ) + W = (v10 + v20 ) + W
∴ (v1 + W ) + (v2 + W ) = (v10 + W ) + (v20 + W )
Since a(v1 − v10 ) ∈ W, a(v1 + W ) = a(v10 + W )
(d) S = V /W = {v + W | v ∈ V } is a vector space
(VS 1) (v1 +W )+(v2 +W ) = (v1 +v2 )+W = (v2 +v1 )+W = (v2 +W )+(v1 +W )
(VS 2) {(v1 + W ) + (v2 + W )} + (v3 + W ) = (v1 + v2 ) + W + v3 + W
= (v1 + v2 + v3 ) + W = v1 + (v2 + v3 ) + W
= {v1 + W } + {(v2 + v3 ) + W }
= (v1 + W ) + {(v2 + W ) + (v3 + W )}
(VS 3) ∃0 + W s.t. (v + W ) + (0 + W ) = (v + 0) + W = v + W
(VS 4) ∀v + W, (1 + W )(v + W ) = v + W
(VS 5) ∃ − v + W s.t. (v + W ) + (−v + W ) = 0 + W
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1.3.
(VS 6) ∀a, b ∈ F, (ab)(v +W ) = abv +W = a(bv)+W = a(bv +W ) = a(b(v +W ))
(VS 7) ∀a ∈ F, a(v1 + W + v2 + W ) = av1 + av2 + W = av1 + W + av2 + W =
a(v1 + W ) + a(v2 + W )
(VS 8) (a + b)(v + W ) = (a + b)v + W = av + bv + W = a(v + W ) + b(v + W )
18
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1.4.
1.4. Linear Combinations and Systems of Linear Equations
1. (a) T (0v = 0, ∀v ∈ V )
(b) F (p.30 span(∅) = {0})
(c) T (p.30 Theorem 1.5)
(d) F (p.27)
(e) T (p.27)
(f) F (p.33 2(b))
2. (a) {r(1, 1, 0, 0) + s(−3, 0, −2, 1) + (5, 0, 4, 0) | r, s ∈ R}
(b) (−2, −4, −3)
(c) There are no solutions
(d) {r(−8, 3, 1, 0) + (−16, 9, 0, 2) | r ∈ R}
(e) {r(0, −3, 1, 0, 0) + s(−3, −2, 0, 1, 0) + (−4, 3, 0, 0, 5) | r, s ∈ R}
(f) (3, 4, −2)
3. (a) yes (−2, 0, 3) = 4(1, 3, 0) + (−3)(2, 4, −1)
(b) Yes (1, 2, −3) = 5(−3, 2, 1) + 8(2, −1, −1)
(c) No
(d) Yes (2, −1, 0) = 45 (1, 2, −3) + 65 (1, −3, 2)
(e) No
(f) Yes (−2, 2, 2) = 4(1, 2, −1) + 2(−3, −3, 3)
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1.4.
4. (a) Yes (x3 − 3x + 5) = 3(x3 + 2x2 − x + 1) + (−2)(x3 + 3x2 − 1)
(b) No
(c) Yes 4, −3
(d) Yes −2, 5
(e) No
(f) No
5. (a) Yes (2, −1, 1) = 1(1, 0, 2) + (−1)(−1, 1, 1) ∈ span(S)
(b) No
(c) No
(d) Yes 2, −1
(e) Yes −1, 3, 1
(f) No
(g) Yes 3, 4, −2
(h) No
6. Let span{(1, 1, 0), (1, 0, 1), (0, 1, 1)} = W
∀v = (a1 , a2 , a3 ) ∈ F 3 ,
v = r(1, 1, 0)+s(1, 0, 1)+t(0, 1, 1) s.t. r = 12 (a1 +a2 −a3 ), s = 21 (a1 −a2 +a3 ), t =
1
(−a1
2
+ a2 + a3 ) ∈ W
∴ V ⊆ W Since W ⊆ F 3 is clear, W = F 3
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1.4.
7. ∀v = (a1 , a2 , · · · , an ) ∈ F n
v = a1 e1 + a1 e2 + · · · + an en , ∀ai ∈ F, i = 0, 1, · · · , n
8. ∀f (x) = a0 + a1 x + · · · + an xn ∈ Pn (F )
f (x) = a0 · 1 + a1 · x + · · · + an · xn , ∀ai F, i = 0, 1, · · · , n
µ
¶
a11 a12
9. ∀A =
∈ M2×2 (F )
µ a21 ¶a22
µ
¶
µ
¶
µ
¶
1 0
0 1
0 0
0 0
A = a11
+ a12
+ a21
+ a22
0 0
0 0
1 0
0 1
¶
a c
∈ M2×2 (F )
10. ∀A =
¶
¶
µ
µ c¶ b µ
0 1
0 0
1 0
∈ span{M1 , M2 , M3 }
+c
+b
A=a
1 0
0 1
0 0
µ
11. If x 6= 0, then span({x}) = {ax | a ∈ F } is the line through the origin
in R3 Otherwise span({x}) = {ax | a ∈ F } is the origin
12. (⇐) By the theorem 1.5, span(W ) is a subspace of V
∴ W is a subspace of V
(⇒) ∀v ∈ W, v = 1 · v ∈ span(W )
∴ W ⊆ span(W )
∀v ∈ span(W ), v = a1 v1 + · · · + an vn
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1.4.
Since ∀ai vi ∈ W, v =
n
P
∈W
i=1
∴ span(W ) ⊆ W
13. (i) S1 ⊆ S2 ⊆ span(S2 ) ⇒ span(S1 ) ⊆ span(S2 )
(ii) By (i) span(S1 ) = V ⊆ span(S2 ) ⊆ V
∴ span(S2 ) = V
14. S1 , S2 ⊆ V, S1 = {x1 , x2 , · · · , xm }, S2 = {xm+1 , xm+2 , · · · , xn }
(i) Since S1 , S2 ⊆ S1 ∪ S2 , span(S1 ), span(S2 ) ⊆ span(S1 ∪ S2 )
n
m
P
P
ai xi ∈ span(S1 ) + span(S2 ), ∀ai ∈ F
a i xi +
If v =
i=m+1
i=1
then v ∈ span(S1 ∪ S2 )
n
P
ai xi ∈ span(S1 ∪ S2 , ∀ai ∈ F
(ii) If v =
i=1
then v =
m
P
i=1
ai xi +
n
P
ai xi ∈ span(S1 ) + span(S2 )
i=m+1
15. Since S1 ∩ S2 ⊆ S1 ⊆ span(S1 ) and S1 ∩ S2 ⊆ S2 ⊆ span(S2 ),
span(S1 ∩ S2 ) ⊆ span(S1 ) ∩ span(S2 )
16. ∀v ∈ span(S), suppose v = a1 v1 + · · · + an vn = b1 v1 + · · · + bn vn
then (a1 − b1 )v1 + · · · + (an − bn )vn = 0
∴ (a1 − b1 ) = · · · = (an − bn ) = 0
∴ a1 = b1 , · · · , an = bn
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1.4.
17. W must be a finite set
(i) F is an infinite field
If ∃0 6= w ∈ W , then {aw | a ∈ F } ⇒ W = {0}
(ii) F is a finite field
If β = {w1 , · · · , wn }, then | W |=| F ||β|
∴ dim W < ∞
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1.5.
1.5. Linear Dependence and Linear Independence
1. (a) If S is a linearly dependent set, then each vector in S is a linear combination of other vector in S.
Ans : F
(Example) V = R3 , S={e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1), e4 = (1, 1, 0)}
{e1 , e2 , e3 } : linearly independent
{e1 , e2 , e4 } : linearly dependent
(b) Any set containing the zero vector is linearly dependent.
Ans : T (∵) ∀a ∈ F, 0 = a · 0, a 6= 0
(c) The empty set is linearly dependent.
Ans : F (∵) linearly dependent set must be non-empty.
(d) Subsets of linearly dependent sets are linearly dependent.
Ans : F (∵) theorem 1.6
(e) Subsets of linearly independent sets are linearly independent.
Ans : T (∵) the corollarly from theorem 1.6
(f) If a1 x1 + a2 x2 + · · · + an xn = 0 and x1 , x2 , · · · , xn are linearly independent,
then all the scalars ai are zero.
Ans : T (∵) from the definition.
2. (a), (d), (e), (g), (h), (j) : linearly independent
(b), (c), (f), (i) : linearly dependent
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1.5.
3. In M2×3 (F ), prove that the set

 
 
 
 

0 0
0 0
1 0
0 1 
 1 1
0 0 , 1 1 , 0 0 , 1 0 , 0 1


0 0
0 0
1 1
1 0
0 1
is linearly dependent.
(∵) a + d = 0, b + d = 0, c + d = 0,
a + e = 0, b + e = 0, c + e = 0
⇒ a = −d, b = −d, c = −d, d, e = d
∴ the given set is linearly dependent.
4. In F n , let ej denote the vector whose jth coordinate is 1 and whose other
coordinates are 0. Prove that {e1 , e2 , · · · , en } is linearly independent.
(∵) (a1 , a2 , · · · , an ) = a1 e1 + a2 e2 + · · · + an en = (0, · · · , 0)
∴ a1 = a2 = · · · = an = 0
∴ {e1 , e2 , · · · , en } is linearly independent.
5. Show that the set {1, x, x2 , · · · , xn } is linearly independent in Pn (F ).
(∵) If a0 + a1 x1 + a2 x2 + · · · + an xn = 0 · 1 + 0 · x1 + 0 · x2 + · · · + 0 · xn ,
then ∀ai = 0, 1 ≤ i ≤ n.
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1.5.
6. In Mn×n (F ), let E ij denote the matrix whose only nonzero entry is 1 in
the ith row and ith column.
Prove that {E ij : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is linearly independent.
(∵) ∀A ∈ Mn×n (F ),
If A = a11 E 11 + a12 E 12 + · · · + ann E nn = 0
then,

a11 a12 · · ·
 a21 a22 · · ·

 ..
..
 .
.
an1 an2 · · ·
∴ ∀aij = 0,
 
a1n
0
0
a2n 
 
..  =  ..
.  .
ann
0 ···
0 ···
..
.
0 0 ···

0
0

.. 
.
0
where 1 ≤ i ≤ m, 1 ≤ j ≤ n
7. Recall from Example 3 in section 1.3 that the set of diagonal matrices in
M2×2 (F ) is a subspace.
Find a linearly independent set that generates this subspace.
¶ µ
¶
µ
0 0
1 0
,
(∵)
0
0
µ
¶
µ0 1 ¶
µ
¶
a 0
1 0
0 0
∀
=a
+b
, a, b ∈ F
0 b
0 0
0 1
8. Let S = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} be a subset of the vector space F 3 .
(a) Prove that if F = R, then S is linearly independent.
(b) Prove that if F has characteristic 2, then S is linearly independent.
(∵) (a) a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (0, 0, 0),
26
∀a, b, c ∈ F
PNU-MATH
1.5.
then a + b = 0, a + c = 0, b + c = 0
∴a=b=c=0
(b) If a = b = c = 1, then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (2, 2, 2) = (0, 0, 0)
∴ {(1, 1, 0), (1, 0, 1), (0, 1, 1)} is linearly dependent
9. Let u and v be distinct vectors in a vector space V .
Show that {u, v} is linearly dependent if and only if u or v is a multiple of the
other.
(∵) u 6= v, {u, v} : linearly independent
(⇒) If au+bv= 0,
(1)a 6= 0 ⇒ u =
−a
v
b
(2)a = 0 ⇒ b 6= 0 ⇒ v =
−a
u
b
(⇐)u = kv ⇒ 1u − kv = 0
10. Give an example of three linearly dependent vectors in R3 such that none of
the three is a multiple of another.
(Example) {(−1, 0, −1), (1, −1, 0), (0, 1, 1)}
11. Let S = {u1 , u2 , · · · , un } be a linearly independent subset of a vector space
V over the field Z2 . How many vectors are there in span(S)?
| span(S) |=| F ||n| , when the set is linearly independent.
* what if the given set is not linearly independent?
27
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1.5.
then the number of vectors in the set is smaller than | F ||n| .
(Example) ai = {0, 1},
v1 = (−1, 0, 1), v2 = (1, −1, 0), v3 = (0, 1, 1)
span(S)={(0, 0, 0), (0, 1, 1), (1, 1, 0), (1, 0, 1)}
12. Prove Theorem 1.6 and its corollary.
Let S1 = {u1 , · · · , un }, S2 = {u1 , · · · , un , v}
(i) a1 u1 + a2 u2 + · · · + an un + an+1 v = 0, not all ai 6= 0 (n ≥ 1) then v =
ai
b1 u1 + · · · + bn un , where bi = − an+1
∈F
∴ S2 is linearly independent
(ii) a1 u1 + · · · + an un + 0v = 0 ⇒ a1 = · · · = an = 0
∴ S1 is linearly independent
13. Let V be a vector space over a field of characteristic not equal to two.
(a) Let u and v be distinct vectors in V .
Prove that {u, v} is linearly independent if and only if {u + v, u − v} is linearly
independent.
(∵)(⇒) Suppose a(u + v) + b(u − v) = 0, a, b ∈ F
(a + b)u + (a − b)v = 0
∴a=b=0
(⇐) {u + v, u − v} is linearly independent.
)(u + v) + ( a−b
)(u − v) = 0
au + bv = ( a+b
2
2
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1.5.
) = 0 and ( a−b
)=0
( a+b
2
2
∴a=b=0
(b)
(⇒) a(u + v) + b(u + w) + c(v + w) = 0
⇒ (a + b)u + (a + c)v + (b + c)w = 0
∴ a+b=a+c=b+c=0
∴ a=b=c=0
(⇐) au + bv + cw = ( a+b
)(u + v) + ( a+c
)(u + w) + ( b+c
)(v + w)
2
2
2
( a+b
) = ( a+c
) = ( b+c
)=0
2
2
2
∴ a=b=c=0
14. (⇐) By the exercise 1(b), if S = {0}, then S is linearly dependent
If v = a1 u1 + · · · + an un , then a1 u1 + · · · + an un − 1 · v = 0
∴ S = {u1 , · · · , un , v} is linearly dependent
(⇒) If S 6= {0}, ∃ai 6= 0 s.t. a1 u1 + · · · + an un + an+1 v = 0
Let an+1 6= 0
ai
∈F
v = b1 u1 + · · · bn vn ∈ span({u1 , · · · , un }), where bi = − an+1
15. (⇐) By the theorem 1.6
(⇒)
If u1 = 0, then it’s clear
So we may assume u1 6= 0
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1.5.
Let k ≥ 0 be the first integer s.t. u1 , · · · , uk linearly independent and {u1 , · · · , uk , uk+1 }
linearly dependent
So a1 u1 + · · · + ak uk + ak+1 uk+1 = 0 for some scalar a1 , a2 , · · · , ak+1 (not all zero)
If ak+1 = 0, then a1 u1 + · · · + ak uk + ak+1 uk+1 = a1 u1 + · · · + ak uk = 0
∴ a1 = · · · = ak = ak+1 = 0
It’s a contradiction
ai
Thus uk+1 = b1 u1 + · · · + bk uk ∈ span(u1 , · · · , uk ), where bi = − ak+1
16. (⇒) By the corollary of theorem 1.6
(⇐) S ⊆ S is linearly independent
17. Let M (1) = (a11 , 0, · · · , 0)t , M (2) = (a12 , a22 , 0, · · · , 0)t , · · · , M (n) = (a1n , a2n , · · · , ann )t
M ∈ span{M (1) , M (2) , · · · , M (n) | aii 6= 0}
Suppose k1 M (1) + · · · + kn M (n) = 0
k1 = k2 = · · · = kn = 0, ∀aii 6= 0
∴ S is linearly independent
18. f0 (x) = a0
f1 (x) = a0 + a1 x
..
.
fn (x) = a0 + a1 x + · · · + an xn
⇒ k0 f0 (x) + k1 f1 (x) + · · · + kn fn (x) = 0
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1.5.
⇒ kn an = 0
(kn−1 + kn )an−1 = 0
..
.
(k0 + · · · + kn )a0 = 0
∴ ∀ki = 0
19.
(a1 A1 + · · · + ak Ak )t = (0)t ⇒ a1 At1 + · · · + ak Atk = 0
∴ a1 = · · · = ak = 0
∴ {At1 , · · · , Atk } is linearly independent
20.
aert + best = 0, r 6= s
⇒ a + be(s−r)t = 0
Since e(s−r)t 6= 0, a = b = 0
∴ {ert , est } is linearly independent
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1.6.
1.6. Bases and Dimension
1. (a) F (∵)∅ is a basis for the zero vector space.
* span{∅} = {0} and ∅ is linearly independent.
(b) T (∵) Theorem 1.9 ; If a vector space V is generated by a finite set S, then
some subset of S is a basis for V .
(c) F (Counterexample) {1, x, x2 , · · · } is a basis for P (F )
(d) F (∵) Corollary 2 (c) from Theorem 1.10
Every linearly independent subset of V can be extended to a basis for V .
(e) T (∵) Corollary 1 from Theorem 1.10
Let V be a vector space having a finite basis. Then every basis for V contains
the same number of vectors.
(f) F (∵) {1, x, x2 , · · · , xn } is a basis for Pn (F )
(g) F (∵) The dimension of Mm×n (F ) is m × n
(h) T (∵) Replacement theorem.
(i) F
(∵) Theorem 1.8. Let V be a vector space and β = {u1 , u2 , · · · , un } be a subset
of V . Then β is a basis for V if and only if each v ∈ V can be uniquely expressed
as a linear combination of vectors of β.
(Example) V = R2 , S = {v1 = (1, 0), v2 = (0, 1), v3 = (1, 1)}
(a, b) = av1 + bv2 + 0v3 = 0v1 + (b − a)v2 + av3
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1.6.
(j) T
Theorem 1.11. Let W be a subspace of a finite-dimensional vector space V . Then
W is finite-dimensional and dim(W)≤ dim(V). Moreover, if dim(W)=dim(V),
then V = W .
(k) T
(i) The vector space {0} has dimension zero and 0 is unique element in V .
So V has exactly one subspace with dimension 0.
(ii) From the theorem 1.11, W ≤ V as a subspace
If dim(W )=dim(V ), then V = W
So V has exactly one subspace with dimension n.
(l) T
(⇒) If S is linearly independent,
Let W be a space spanned by S.
Then S is a basis for W (the corollary 2 from 1.10)
And dimW = n
∴W =V
∴ S is a basis for V
(⇐) If S is a generating set for V that contains n vectors,
then by the corollary 2 from 1.10, S is linearly independent.
2. (a) {(1, 0, −1), (2, 5, 1), (0, −4, 3)} : a basis for R3
(∵) 0 = a · (1, 0, −1) + b · (2, 5, 1) + c · (0, −4, 3) = (a + 2b, 5b − 4c, −a + b + 3c)
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1.6.
∴a=b=c=0
(*)
¯
¯
¯1 0 −1¯
¯
¯
¯2 5
¯ = 27 6= 0
1
¯
¯
¯0 −4 3 ¯
∴ {(1, 0, −1), (2, 5, 1), (0, −4, 3)} is linearly independent.
(b) {(2, −4, 1), (0, 3, −1), (6, 0, −1)} : a basis for R3
(∵) 0 = a · (2, −4, 1) + b · (0, 3, −1) + c · (6, 0, −1)
= (2a + 6c, −4a, a − b − c)
∴a=b=c=0
(c) {(1, 2, −1), (1, 0, 2), (2, 1, 1)} : a basis for R3
(d) {(−1, 3, 1), (2, −4, −3), (−3, 8, 2)} : a basis for R3
(e) {(1, −3, −2), (−3, 1, 3), (−2, −10, −2)}
(∵) 0 = a · (1, −3, −2) + b · (−3, 1, 3) + c · (−2, −10, −2)
= (a − 3b − 2c, −3a + b − 10c, −2a + 3b − 2c)
∴ a = 2b, c =
−1
b
2
∴ ∃(4, 2, 1) 6= (0, 0, 0)
(*)
¯
¯ ¯
¯
¯ 1 −3 −2¯ ¯1 −3 −2¯
¯
¯ ¯
¯
¯−3 1
¯ = ¯0 −8 −3¯ = 0
3
¯
¯ ¯
¯
¯−2 10 −2¯ ¯0 −16 −6¯
∴ {(1, −3, −2), (−3, 1, 3), (−2, −10, −2)} is linearly dependent.
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1.6.
3. (a) {−1 − x + 2x2 , 2 + x − 2x2 , 1 − 2x + 4x2 }
(∵) 0 = a · (−1 − x + 2x2 ) + b · (2 + x − 2x2 ) + c · (1 − 2x + 4x2 )
= (1 + 2a − 2b + 4c)x2 ) + (−a + b − 2c)x + (−a + 2b + c)
∴ a = −5c, b = −3c
∴ ∃(−5, 3, 1) 6= (0, 0, 0)
(*)
¯
¯ ¯
¯
¯−1 −1 2 ¯ ¯1 0 0¯
¯
¯ ¯
¯
¯2
1 −2¯¯ = ¯¯2 −1 2¯¯ = 0
¯
¯ 1 −2 4 ¯ ¯1 −3 6¯
∴ {−1 − x + 2x2 , 2 + x − 2x2 , 1 − 2x + 4x2 } is linearly dependent.
(b) {1 + 2x + x2 , 3 + x2 , x + x2 } : a basis for P 2 (R)
(c) {1 − 2x − 2x2 , −2 + 3x − x2 , 1 − x + 6x2 } : a basis for P 2 (R)
(d) {−1 + 2x + 4x2 , 3 − 4x − 10x2 , −2 − 5x − 6x2 }: a basis for P 2 (R)
(e) {1 + 2x − x2 , 4 − 2x + x2 , −1 + 18x − 9x2 }
(∵) 0 = a · (−1 − x + 2x2 ) + b · (4 − 2x + x2 ) + c · (−1 + 18x − 9x2 )
= (−a + b − 9c)x2 ) + (2a − 2b + 18c)x + (a + 4b − c)
∴ a = −11c, b = −2c
∴ ∃(−11, −2, 1) 6= (0, 0, 0)
4. No.
|{x3 − 2x2 + 1, 4x2 − x + 3, 3x − 2}| = 3 and dim(P3 (R)) = 4
The generating set for V contains at least 4 vectors.
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5. No.
Any n + 1 or more vectors in V are linearly dependent.
Since dim(R3 ) = 3, every linearly independent set contains at most 3 vectors.
6.
½µ
¶ µ
1 0
0
,
0 0
0
½µ
¶ µ
1 0
1
,
0 0
0
½µ
¶ µ
1 1
0
,
1 1
1
¶ µ
1
0
,
0
1
¶ µ
1
1
,
0
1
¶ µ
1
0
,
1
1
¶ µ
0
0
,
0
0
¶ µ
1
1
,
0
1
¶ µ
0
0
,
1
0
0
1
1
1
0
1
¶¾
¶¾
¶¾
7.
Select any nonzero vector in the given set, say u1 , to be a vector in the basis.
Since u3 = −4u1 , the set {u1 , u3 } is linearly dependent.
Hence we don’t include u3
On the other hand, the set {u1 , u2 } is linearly independent. Thus we include u2 .
And the set {u1 , u2 , u4 } is linearly dependent. So we exclude u4 .
∴ {u1 , u2 , u5 } is a basis for R3 .
(*)
¯
¯ ¯
¯
¯
¯
¯2 −3 1 ¯ ¯0 −11 5 ¯
¯−11 5 ¯
¯
¯ ¯
¯
¯=0
¯1 4 −2 ¯ = ¯1 4
−2 ¯¯ = − ¯¯
¯ ¯
¯
33 −15¯
¯1 37 −17¯ ¯0 33 −15¯
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1.6.
∴ {u1 , u2 , u4 } is linearly dependent.
8. {u1 , u3 , u4 , u8 } is a basis for W .
9. a, b, c, d ∈ F , (a1 , a2 , a3 , a4 ) = au1 + bu2 + cu3 + du4
= (a, a + b, a + b + c, a + b + c + d)
∴ a = a1 , b = a2 − a1 , c = a3 − a2 , d = a4 − a3
∴ (a1 , a2 , a3 , a4 ) = a1 u1 + (a2 − a1 )u2 + (a3 − a2 )u3 + (a4 − a3 )u4
10. (a)
f0 (x) = 13 (x2 − 1), f1 (x) = − 12 (x2 + x − 2), f2 (x) = 16 (x2 + 3x + 2)
2
P
∴ g(x) =
bi fi (x) = −4x2 − x + 8
i=0
(b)
f0 (x) =
1
(x2
35
2
P
∴ g(x) =
1
− 4x + 3), f1 (x) = − 10
(x2 + x − 12), f2 (x) =
1
(x2
14
+ 3x − 4)
bi fi (x) = −3x + 12
i=0
(c)
1
f0 (x) = − 15
(x3 − 3x2 − x + 3), f1 (x) = − 18 (x3 − 2x2 − 5x + 6), f2 (x) =
1
1
− 12
(x3 − 7x2 − 6), f3 (x) = 40
(x3 + 2x2 − x − 2)
3
P
∴ g(x) =
bi fi (x) = −x3 + 2x2 + 4x − 5
i=0
(d)
1
f0 (x) = − 12
(x3 + x2 − 2x), f1 (x) = 16 (x3 + 2x2 − 3x), f2 (x) = − 61 (x3 + 4x2 + x −
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1.6.
1
(x3 + 5x2 + 6x)
6), f3 (x) = 12
3
P
∴ g(x) =
bi fi (x) = −3x3 − 6x2 + 4x + 15
i=0
11. (i)
We need to show that {u + v, au} is linearly independent
k1 , k2 are scalars,
If k1 (u + v) + k2 (au) = (k1 + ak2 )u + (k1 v) = 0
Since {u, v} is a basis for V ,
∴ k1 + ak2 = 0, k1 = 0 (a is a nonzero scalar)
∴ k1 = k2 = 0
(ii)
If k1 (au) + k2 (bv) = (ak1 )u + (bk2 )v = 0
Since a 6= 0, b 6= 0 and {u, v} is a basis for V .
∴ k1 = k2 = 0
∴ {au, bv} is a basis for V .
12. k1 (u + v + W ) + k2 (v + w) + k3 (w) = 0, k1 , k2 , k3 ∈ F
⇒ k1 u + (k1 + k2 )v + (k1 + k2 + k3 )w = 0
⇒ k1 = k2 = k3 = 0
13. {a(1, 1, 1)|a ∈ R} is a solution set
∴ {(1, 1, 1)} is a basis for the given system
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1.6.
14. {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1)} : a basis for W1
{(1, 0, 0, 0, −1), (0, 1, 1, 1, 0)} : a basis for W2
∴ dim(W1 )=4, dim(W2 )=2
(*) The dimension of the solution space AX = 0 is equal to n−rankA
(n is the number of rows of A)
(i) a1 − a3 − a4 = 0 i.e. a1 + 0a2 − a3 − a4 + 0a5 = 0—(*)
 
a1
a2 
 

A = (1, 0, −1, −1, 0)1×5 , X = 
a3 
a4 
a5
(*) is equal to AX = 0
The dimension of the solution space AX = 0 is equal to n−rankA = 5 − 1 = 4
(ii) a2 = a3 , a2 = a4 , a1 + a5 = 0


0 1 −1 0 0
A = 0 1 0 −1 0
1 0 0
0 1 3×5
∴ rank(A)=3
The dimension of the solution space AX = 0 is equal to n−rankA = 5 − 3 = 2
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1.6.
15.
(i) {(
n
P
Eij , where i 6= j) and (∗∗)} : a basis for W .
i,j=1
(∗∗) = {(1, −1, 0, · · · , 0), (0, 1, −1, · · · , 0), · · · , (0, · · · , 1, −1, 0), (0, · · · , 1, −1)}
that is, ith component of the element in (∗∗) is −1
(ii) Dim(W )=n2 − n + (n − 1) = n2 − 1
(∵) n2 : the dim of Mn×n (F ), n : the number of vectors consist of diagonal,
(n − 1) : the number of vectors consist of (∗∗)
* When charF =2, if At = −A, then aij = −aji and aii = −aii
∴ aii = 0
(Example)
W = {(a1 , a2 , a3 , a4 , a5 )|
n
P
a1 = 0} ≤ R5 is a subspace of V , dim(W ) = 4
i=1
β = {v1 = (1, −1, 0, 0, 0), v2 = (1, 0, −1, 0, 0), v3 = (1, 0, 0, −1, 0), v4 = (1, 0, 0, 0, −1)}
(or β = {(1, −1, 0, 0, 0), (0, 1, −1, 0, 0), (0, 0, 1, −1, 0), (0, 0, 0, 1, −1)})
From a1 + a2 + a3 + a4 + a5 = 0, a5 = −(a1 + a2 + a3 + a4 )
(a1 , a2 , a3 , a4 , a5 ) = (a1 , a2 , a3 , a4 , −(a1 + a2 + a3 + a4 )
= (a1 , 0, 0, 0, −a1 ) + (0, a2 , 0, 0, −a2 ) + (0, 0, a3 , 0, −a3 ) + (0, 0, 0, a4 , −a4 )
16.
∀A ∈ W : the set of all upper triangular n × n matrices
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1.6.


a11 a12 · · · a1n
 0 a22 · · · a2n 


 ..
.. 
.
.
 .
.
. 
0
0 · · · ann
(i) {Eij |Eij = 0, i > j} is a basis for W
(ii) dim(W ) = 1 + 2 + · · · + n = 12 n(n + 1)
17.
∀A ∈ W : the set of all skew-symmetric n × n matrices


a1n
a2n 

a3n 
 , aij = −aji
.. 
. 
0
0 a12 a13 · · ·
 a21 0 a23 · · ·

 a31 a32 0 · · ·

 ..
...
 .
an1 an2 an3 · · ·
(i) A basis for W

0
1



1
0


–—–



 
0
0
  0
0
,
  1
0
——-
1
0
0


,···



,

–—–
0
1





1 


0
(ii) dim(W ) = 1 + 2 + · · · + (n − 1) = 12 n(n − 1)
(iii) In case of charF =2
if At = −A, then aij 6= −aji
(∵) if aii = −aii ; aii = 0 in char(F )=2
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1.6.
18. V consists of all sequences {an } in F that have only a finite number of
nonzero terms an
Let {ei } = {0, · · · , 0, 1, 0 · · · , 0} i.e. the ith term is 1
then {ei | i = 1, · · · , n} is a basis for V
19. (⇐) Suppose each v ∈ V can be uniquely expressed as a linear combination of vectors of β, then clearly β spans V .
If 0 = a1 u1 + a2 u2 + · · · + an un
And we also have 0 = 0 · u1 + 0 · u2 + · · · + 0 · un
By hypothesis, the representation of zero as a linear combination of the ui is
unique. Hence each ai = 0, and the ui are linearly independent.
20. (a) Prove that there is a subset of S that is a basis for V . (Be careful
not to assume that S is finite.)
If V = (0) ⇒ ∅(⊆ S) a basis. So we may assume that V 6= (0)
Let C = {B|B ⊆ S, B is linearly independent}
then ∀B ∈ C, |B| ≤ n(= dim V )
(i) Choose B 0 ∈ C with maximal element
i.e. B 0 ∈ C and ∀B ∈ C, |B| ≤ |B 0 |
(ii)
Claim S ⊆ span(B 0 )
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1.6.
(∵) If not, S * span(B 0 ), then ∃v ∈ S, v ∈
/ span(B 0 )
By the theorem 1.7(p.39), B 0 ∪ {v} is linearly independent
Since B 0 ( B 0 ∪ {v} ⊆ S, this contradicts to the maximality of B 0
And V = span(S) ⊆ span(span(B 0 )) = span(B 0 )
∴ V = span(B 0 )
Therefore B 0 is a basis for V
(b)
Let Q ⊆ V s.t. span(Q) = V and | Q |< n
From (a), we can find a subset Q0 of Q is a basis for V .
This contradicts to the following fact :
If β = {v1 , v2 , · · · , vn } is a basis for V , then any set of n + 1 vectors in V is
linearly dependent
∴ Any spanning set S for V must contain at least n vectors.
21. (i) Suppose dim V = ∞
⇒ V has an infinite set of linearly independent
(ii) By the Replacement theorem
22. dim(W1 ∩ W2 ) = dim(W1 ) ⇔ W1 ∩ W2 = W1 ⇔ W1 ⊆ W2
23. (a) dim(W1 ) = dim(W2 ) if and only if v ∈ span{v1 , v2 , · · · , vk } = W1
(b) If dim(W1 ) 6= dim(W2 ), then dim(W1 ) < dim(W2 )
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1.6.
(∵) By (a), v ∈ W1 and v ∈
/ W2
By the exercise 20, dim(W1 ) ≤ k and dim(W2 ) ≤ k + 1
∴ dim(W1 ) ≤ dim(W2 )
∴ dim(W1 ) < dim(W2 )
24. f (x) = kn xn + · · · + k1 x + k0 , kn 6= 0, ∀ki ∈ R
Let a0 f (x) + a1 f 0 (x) + · · · + an f (n) (x) = 0, for some scalars a0 , · · · , an ∈ R
then (a0 kn )xn + (a0 kn−1 + a1 kn )xn−1 + · · · + (a0 k1 + 2!a1 k2 + 3!a2 k3 + · · · +
n
P
m!am km ) = 0
n!an−1 kn )x + (
m=0
By equating the coefficient of xk on both sides of this equation for k = 0, 1, 2, · · · , n,
we obtain a0 = a1 = · · · = an = 0 (since charR=0)
It follows from (b) of corollary 2 (p.48) that {f (x), f 0 (x), · · · , f (n) (x)} is a basis
for Pn (R).
∴ ∀g(x) ∈ Pn (R), ∃ c0 , · · · , cn ∈ R s.t. g(x) = c0 f (x) + c1 f 0 (x) + · · · + cn f (n) (x)
25. Z = {(v, w) | v ∈ V and w ∈ W } = V × W
dim(Z) = dim(Z) × dim(W ) = mn
β = {v1 , · · · , vm } a basis for V
γ = {w1 , · · · , wn } a basis for W
then α = {(v1 , 0), · · · , (vm , 0), · · · , (0, w1 ), · · · , (0, wn )} a basis for V × W
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26. W = {f ∈ Pn (R) | f (a) = 0} is a subspace of V = Pn (R)
i.e. ∀f ∈ W forms (x − a)(an−1 xn−1 + · · · + a1 x + a0 )
∴ dim(W ) = n
27.
dim(W1 ∩Pn (F )) =
n
2
n+1
2
if n : even
,
if n : odd
dim(W2 ∩Pn (F )) =
n
2
+ 1 if n : even
if n : odd
n+1
2
28.
dim V = 2n
β = {v1 , v2 , · · · , vn } a basis for V over C, dim V = n
β 0 = {v1 , v2 , · · · , vn , v1 i, v2 i, · · · , vn i} a basis for V over R
a1 v1 + · · · + an vn + b1 vv i + · · · + bn vn i = 0
⇒ (a1 + ib1 )v1 + · · · + (an + ibn )vn = 0
29.
Let β = {u1 , · · · , uk , v1 , · · · , vm , w1 , · · · , wp } (i) β spans W1 + W2
p
k
m
k
P
P
P
P
∀v ∈ W1 + W2 , v = ( ai ui +
bi vi ) + ( ci ui +
di wi ) ∈ span(β), w1 ∈
i=1
i=1
i=1
i=1
W1 , w2 ∈ W2 (ii) β is linearly independent
Suppose that a1 u1 +· · ·+ak uk +b1 v1 +· · ·+bm vm +c1 w1 +· · ·+cp wp = 0−−−−(∗)
, where a1 , · · · , ak , b1 , · · · , bm , c1 , · · · , cp ∈ F
p
k
m
P
P
P
Let v =
ai ui +
bi vi = − ci wi ∈ W1 ∩ W2
i=1
i=1
i=1
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1.6.
Since {u1 , · · · , uk } is a basis for W1 ∩ W2 , ∃di ∈ F s.t. v =
∴
k
P
di ui +
i=1
p
P
k
P
d i ui
i=1
ci w i = 0 ⇒ c1 = c2 = · · · = cp = 0
i=1
By (∗), a1 = · · · = ak = b1 = · · · = bm = 0
∴ β is linearly independent
∴ β is a basis for W1 + W2
∴ dim(W1 + W2 ) = dim(W1 ) + dim(W2 ) − dim(W1 ∩ W2 )
(b) dim(W1 + W2 ) = 0 ⇔ W1 ∩ W2 = ∅
30.
(i) W1 and W2 are subspaces of V
µ
¶
µ
¶
a b
d e
∀A =
,B =
∈ W1 , α ∈ F
c µ
a
f d¶
αa+d αb+e
αA + B =
∈ W1
αc+f
αa+d
µ
¶
µ
¶
0 a
0 c
∀A =
,B =
∈ W2 , α ∈ F
−a µb
−c¶ d
0
αa+c
αA + B =
∈ W2
−αa-c b+d
(ii) dim(W1 ) = 3, dim(W2 ) = 2, dim(W1 ∩ W2 ) = 1 and dim(W1 + W2 ) = 4
31. (a)
(∵) W1 ∩ W2 is a subspace of W1 ,
By the theorem 1.11, dim(W1 ∩ W2 ) ≤ dim(W1 ) ≤ n
∴ dim(W1 ∩ W2 ) ≤ n
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1.6.
(b)
dim(W1 + W2 ) = dim(W1 ) + dim(W2 ) − dim(W1 ∩ W2 ) ≤ m + n − dim(W1 + W2 )
∴ dim(W1 + W2 ) ≤ m + n
32.
(a) dim(W1 ∩ W2 ) = dim(W2 )
W1 = {(a1 , a2 , 0)|a1 , a2 ∈ F }, dim(W1 )=2
W2 = {(a1 , 0, 0)|a1 ∈ F }, dim(W2 )=1
(b) dim(W1 ∩ W2 )=0
W1 = {(a1 , 0, 0)|a1 ∈ F }, dim(W1 )=1
W2 = {(0, a2 , a3 )|a2 , a3 ∈ F }, dim(W2 )=2
(c) W1 = {(a1 , 0, a3 )|a1 , a3 ∈ F }, dim(W1 )=2
W2 = {(a1 , a2 , 0)|a1 , a2 ∈ F }, dim(W2 )=2
33. V = W1 ⊕ W2 ⇔ β1 ∩ β2 = ∅, β1 ∪ β2 : a basis for V
(⇒)
(i) Suppose that a1 u1 + a2 u2 + · · · + b1 w1 + b2 w2 + · · · = 0 ai , bj ∈ F
Then a1 u1 + a2 u2 + · · · = −(b1 w1 + b2 w2 + · · · ) ∈ W1 ∩ W2 = {0}
∴ ∀ai = bj = 0
∴ β1 ∪ β2 is linearly independent
Let v = u + w ∈ W1 + W2
Since β1 , β2 spans W1 , W2 , respectively
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1.6.
∃ ai , bj ∈ F s.t. v = a1 u1 + a2 u2 + · · · + b1 w1 + b2 w2 + · · ·
∴ β1 ∪ β2 spans V
(ii) If ∃0 6= u ∈ β1 ∩ β2 , then W1 ∩ W2 6= {0}
∴ β1 ∩ β2 = ∅
(⇐)
(i) Since β1 ∪ β2 spans V, v ∈ span(β1 ∪ β2 )
∴ v = a1 u1 + +a2 u2 + · · · + b1 w1 + b2 w2 + · · · ∈ W1 + W2
∴ V = W1 + W2
(ii) Since β1 ∩ β2 = ∅, W1 ∩ W2 = {0}
∴ V = W1 ⊕ W2
34. (a)
Let β1 = {u1 , u2 , · · · , uk } be a basis for W1
and we can extend it to a basis for V , say β
Let β = {u1 , u2 , · · · , uk , uk+1 , · · · , un } and β2 = {uk+1 , uk+2 , · · · , un }
By the exercise 33, V = W1 ⊕ W2
(b)
V = W1 ⊕ W2 , W1 = {(a, 0)|a ∈ R}, W2 = {(0, b)|b ∈ R}
V = W1 ⊕ W20 , W1 = {(a, 0)|a ∈ R}, W20 = {(−a, b)|a, b ∈ R}
35.
(a) β 0 = {uk+1 + W, uk+2 + W, · · · , un + W } is a basis for V /W . (p.23)
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1.6.
(i) ∀α ∈ V /W = {v + W |v ∈ V }
α = (a1 u1 + a2 u2 + · · · + an un ) + W
= (a1 u1 + W ) + (a2 u2 + W ) + · · · + (an un + W )
= a1 (u1 + W ) + · · · + ak (uk + W ) + ak+1 (uk+1 + W ) + · · · + an (un + W )
= ak+1 (uk+1 + W ) + ak+2 (uk+2 + W ) + · · · + an (un + W )
∴ β 0 spans V /W
(ii) Suppose ak+1 (uk+1 + W ) + ak+2 (uk+2 + W ) + · · · + an (un + W ) = W, ai ∈
F, i = k + 1, · · · , n
then ak+1 uk+1 + ak+2 uk+2 + · · · + an un = 0
Since {uk+1 , · · · , un } is linearly independent, ak+1 = · · · = an = 0
∴ β 0 is a basis for V /W .
(b) dim(V ) = k + n, dim(W ) = k, dim(V /W ) = n − k
∴ dim(/W V ) = dim(V ) − dim(W )
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1.7.
1.7. Maximal Linearly Independent Subsets
1. Label the following statement as true or false.
(a) Every family of sets contains a maximal element. (F)
(∵) Let F be the family of all finite subsets of an infinite set S.
then F has no maximal element.
(b) Every chain contains a maximal element. (F)
(∵) If A : a partial ordered set and every chain (6= ∅) of A has an upper bound,
then A has a maximal element.
(ex) Z, Q, R
(c) If a family of sets has a maximal element, then that maximal element is
unique. (F)
(∵) Let S = {a = x3 −2x2 −5x−3, b = 3x3 −5x2 −4x−9, c = 2x3 −2x2 +12x−6}
then {a, b}, {a, c}, {b, c} are maximal linearly independent subsets of S
So maximal element need not be unique.
(d) If a chain of sets has a maximal element, then that maximal element is unique.
(T)
(e) A basis for a vector space is a maximal linearly independent subset of that
vector space. (T)
(f) A maximal linearly independent subset of a vector space is a basis for that
vector space. (T)
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1.7.
2. Show that the set of convergent sequences is an infinite-dimensional subspace
of the vector space of all sequences of real numbers. ( See Exercise 21 in Section
1.3.)
(i) By the exercise 21 of section 1.3, W is a suspace of V
(ii) {1, 1, · · · } is a basis for W
3. Let V be the set of real numbers regarded as a vector space over the field
of rational numbers. Prove that V is infinite-dimensional.
Let V be a finite-dimensional vector space
β = {v1 , · · · , vn }
Since π ∈ V, π = α1 v1 + · · · + αn vn , ∀αi ∈ Q, i = 1, · · · , n
But π is a transcendental number in Q
It’s a contrdiction
Thus V is infinite dimensional
4. Let W be a subspace of a (not necessarily finite-dimensional) vector space
V . Prove that any basis for W is a subset of a basis for V .
(∵) Let βW be a basis for W , then βW ⊆ V
(i) By theorem 1.13, ∃ β : a maximal linearly independent subset of V that contains βW .
(ii) We are going to show that β is a basis for V which contains βW .
Since β is linearly independent, so it suffices to show that β spans V .
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1.7.
If v ∈ V and v is not contained in β (i.e. v is not in span(β) ),
then β ∪ {v} is linearly independent.
This contradicts to the maximality of β.
∴ ∀v ∈ V, v ∈ span(β) ∈ V
∴ span(β) = V .
∴ β is a basis for V which contains βW .
5.
(∵) (⇒) Let β = {v1 , v2 , · · · } is a basis for V .
If v ∈ V , then v ∈ spanV
Thus v is a linear combination of the vectors of β (that is v can be expressed by
some finite vectors of β and scalars in F )
Suppose that v = a1 v1 + a2 v2 + · · · + an vn and v = b1 v1 + b2 v2 + · · · + bn vn
are two such representation of v. Then
0 = (a1 − b1 )v1 + (a2 − b2 )v2 + · · · + (an − bn )un
Since β is linearly independent, it follows that ai = bi , where 1 ≤ i ≤ n
(⇐) Suppose each v ∈ V can be uniquely expressed as a linear combination of
vectors of β, then clearly β spans V .
If 0 = a1 u1 + a2 u2 + · · · + an un
And we also have 0 = 0 · u1 + 0 · u2 + · · · + 0 · un
By hypothesis, the representation of zero as a linear combination of the ui is
unique. Hence each ai = 0, and the ui are linearly independent.
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1.7.
6.
(i) Since S1 ∈ F , F 6= ∅
Let C be a chain in F and U = ∪{A | A ∈ C}
Clearly S1 ⊆ U ⊆ S2
Let a1 u1 + · · · + an un = 0 s.t. ∀ui ∈ U, ∀ai ∈ F, i = 1, · · · , n
Since ui ∈ A, ∃Ai ⊆ C s.t. ui ∈ Ai
Since C is a chain, ∃Ak a.t. ∀Ai ⊆ Ak
Thus ∀ui ∈ Ak ∈ F
∴ a1 = · · · = an = 0
So U is n upper bound of C
By the maximal principle, F has a maximal element β
i.e. β is a maximal linearly independent subset of V
(ii) ∀v ∈ S2 , if v ∈ β, then v ∈ span(β)
If v ∈
/ β, β ∪ {v} is linearly independent
This contradicts to the maximality of β
∴ S2 ⊆ span(β)
Then V = span(S2 ) ⊆ span(β) ⊆ V
∴ span(β) = V
Therefore β is a basis for V s.t. S1 ⊆ β ⊆ S2
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1.7.
7. Let S = {A ⊆ β | A ∩ S = ∅, A ∪ S is linearly independent }
S 6= ∅ (∵ A = ∅)
Let C be a nonempty chain in S and B = ∪{A | A ∈ C}
If ∃x ∈ B ∩ S, then x ∈ B and x ∈ S
Since A ⊆ B, x ∈ A for some A ∈ C
thus x ∈ A ∩ S = ∅
It’s a contradiction, so B ∩ S = ∅
If a1 u1 + · · · + am um + b1 v1 + · · · + bn vn = 0, ∀ai , bj ∈ F, ui ∈ B, vj ∈ S
Since B = ∪{A | A ∈ C}, ∃A1 , · · · , Am ∈ C s.t. ∀ui ∈ Ai
So we may assume that u1 , · · · , um ∈ Am
then u1 , · · · , um , v1 , · · · , un ∈ Am ∪ S is linearly independent
∴ ∀ai = 0, ∀bj = 0
∴ B ∪ S is linearly independent
i.e. B is an upper bound of C
By the maximal principle, S has a maximal element, say S1
Clearly S1 ⊆ β, S1 ∩ S = ∅, S1 ∪ S2 is linearly independent
So we only need to show that either { S1 ∪ S is a maximal linearly independent
subset of V or V } or {β ⊆ span(S1 ∪ S)}
∀v ∈ β, If v ∈ S1 ∪ S, then v ∈ span(S1 ∪ S) If v ∈
/ S1 ∪ S, then S1 ∪ S ∪ {v} is
linearly independent
It’s a contradiction, ∴ β ⊆ span(S1 ∪ S)
Therefore S1 ∪ S is a basis for V
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2.1.
§2.
Linear Transformations and Matrices
2.1. Linear Transformations, Null Spaces, and Ranges
1. (a) T
(b) F (∵) If ∀x, y ∈ V and c ∈ F , T(x + y) =T(x)+T(y) and T(cx) = cT(x),
then T is a linear transformation
(c) F (∵) T is linear and one-to-one if and only if
(d) T (∵) T(0V ) =T(0V + 0V ) =T(0V )+T(0V ) ∴ T(0V ) = 0W
(e) F (∵) p.70 Theorem 2.3
(f) F (∵) T is linear and one to one, then
(g) T (∵) p.73 Corollary to Theorem 2.6
(h) F (∵) p.72 Theorem 2.6
2.
(1) T ((a1 , a2 , a3 ) + (b1 , b2 , b3 )) = T (a1 + b1 , a2 + b2 , a3 + b3 )
= ((a1 + b1 ) − (a2 + b2 ), 2(a3 + b3 ))
= (a1 − a2 ), 2a3 ) + (b1 − b2 ), 2b3 )
= T (a1 , a2 , a3 ) + T (b1 , b2 , b3 )
T (c(a1 , a2 , a3 )) = T (ca1 , ca2 , ca3 ) = (ca1 −ca2 , 2ca3 ) = c(a1 −a2 , 2a3 ) = cT (a1 , a2 , a3 )
Thus, T is linear.
(2) N(T)=span{(1, 1, 0)}
R(T)=span{(1, 0), (0, 1)}
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2.1.
(3) Dim(V)=nullity(T) + rank(T)= 1 + 2
(4) T is not one-to-one (∵N(T)6= {0} = {(0, 0, 0)})
T is onto (∵ 2=rank(T)=dim(W)=2)
3.
(1) T is linear.
(2) N(T)=span{(0, 0)}
R(T)=span{(1, 0, 0), (0, 0, 1)}
(3) Dim(V)=nullity(T) + rank(T)= 0 + 2
(4) T is one-to-one (∵N(T)= {0})
T is not onto (∵ 2=rank(T)6= dim(W)=3)
4.
(1) T is linear.
¶ µ
¶
0 1
1 0
,
,
(2) N(T)=span
0 ¶
0 µ 0 0 ¶ µ
µ
¶ µ
¶
1 0 0
0 1 0
0 0 1
0 0 0
R(T)=span
,
,
,
0 0 0
0 0 0
0 0 0
1 1 1
(3) Dim(V)=nullity(T) + rank(T)= 4 + 2
µ
(4) T is not one-to-one (∵N(T)6= {0})
T is not onto (∵ 2=rank(T)6= dim(W)=4)
5. (1) T is linear.
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2.1.
(2) N(T)=span{0}
R(T)=span{1, x, x2 }
(3) Dim(V)=nullity(T) + rank(T)= 0 + 3
(4) T is one-to-one (∵N(T)= {0})
T is not onto (∵ rank(T)6= dim(W))
6. (1) T is linear.
(2) N(T)={A | T (A) = tr(A) = 0}=span{Eij , Eij0 },
where Eij denote the matrix whose ij-entry is 1 and zero elsewhere
and Eij0 denote the matrix of a11 component is 1, ajj component is −1 and others
are all zero. (2 ≤ j ≤ n)
(*) β = {Eij | i 6= j} ∪ {E11 − Eii | 2 ≤ j ≤ n}
R(T)={T (A) | A ∈ M atn×n }=span{1}
(3) n2 =Dim(V)=nullity(T) + rank(T)= (n2 − 1) + 1
* dim N(T)=n2 − 1 (p.56 exercise 15, sec 1.6)
(4) T is not one-to-one but onto (∵ rank(T)= dim(W))
7.
(1) If T is linear, then T (0) = 0
T (0) = T (0 + 0) = T (0) + T (0)
∴ T (0) = 0
(2)
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2.1.
If T is linear, T (cx + y) = T (cx) + T (y) = cT (x) + T (y)
Suppose T (cx + y) = cT (x) + T (y), then
c = 1, T (x + y) = T (x) + T (y)
j = 0, T (cx) = cT (x)
(3) 0 = T (0) = T (x − x) = T (x) + T (−x)
⇒ T (−x) = −T (x) ∀x
So T (x − y) = T (x) + T (−y) = T (x) − T (y)
(4)
(⇒) If T is linear, then T (
n
P
a i xi ) =
i=1
n
P
T (ai xi ) =
i=1
n
P
ai T (xi )
i=1
(⇐) clear
8. (a)
(i) Tθ ((a1 , a2 ) + (b1 , b2 )) = Tθ (a1 + b1 , a2 + b2 )
= ((a1 + b1 ) cos θ − (a2 + b2 ) sin θ, (a1 + b1 ) sin θ + (a2 + b2 ) cos θ)
= (a1 cos θ − a2 sin θ, a1 sin θ + a2 cos θ) + (b1 cos θ − b2 sin θ, b1 sin θ + b2 cos θ)
= Tθ (a1 , a2 ) + Tθ (b1 , b2 )
(ii) Tθ c((a1 , a2 )) = Tθ (ca1 , ca2 ) = (ca1 cos θ − ca2 sin θ, ca1 sin θ + ca2 cos θ))
= c(a1 cos θ − a2 sin θ, a1 sin θ + a2 cos θ) = cTθ (a1 , a2 )
(b)
(i) T ((a1 , a2 ) + (b1 , b2 )) = T (a1 + b1 , a2 + b2 ) = (a1 + b1 , −a2 − b2 ) = (a1 − a2 ) +
(b1 − b2 ) = T (a1 + a2 ) + T (b1 + b2 )
(ii) T (c(a1 , a2 )) = (ca1 , −ca2 ) = c(a1 , −a2 ) = cT (a1 , a2 )
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2.1.
9. (a)
T ((a1 + a2 ) + (b1 + b2 )) = (1, a2 + b2 )
T (a1 + a2 ) + T (b1 , b2 ) = (2, a2 + b2 )
(b)
T (c(a1 , a2 )) = (ca1 , c2 a22 ) cT (a1 , a2 ) = (ca1 , ca21 )
(c)
T ((a1 + a2 ) + (b1 + b2 )) = (sin(a1 + b1 ), 0)
T (a1 + a2 ) + T (b1 , b2 ) = (sin a1 + sin b1 , 0)
(d)
T ((a1 + a2 ) + (b1 + b2 )) = (|a1 + b1 |, a2 + b2 )
T (a1 + a2 ) + T (b1 , b2 ) = (|a1 | + |b1 |, a2 + b2 )
(e)
T ((a1 + a2 ) + (b1 + b2 )) = (a1 + b1 + 1, a2 + b2 )
T (a1 + a2 ) + T (b1 , b2 ) = (a1 + b1 + 2, a2 + b2 )
10.
(a) T (2, 3) = −1T (1, 0) + 3T (1, 1) = (5, 11)
(b) T is one-to-one
T (x, y) = xT (1, 0)+yT (0, 1) = xT (1, 0)+y(T (1, 1)−T (0, 1)) = x(1, 4)+y((2, 5)−
(1, 4)) == (x + y, 4x + y) = (0, 0)
∴ (x, y) = (0, 0)
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2.1.
11. T (8, 11) = 2T (1, 1) + 3T (2, 3) = (5, −3, 16)
12. No (T (2, 0, 6) 6= −2T (1, 0, 3))
13.
S={v1 , v2 , · · · , vk } s.t T (vi ) = wi , i = 1, 2, · · · , k
k
P
Suppose
ai vi = 0 for some scalars ai ∈ F ,
i=1
then we need to show that all ai = 0
k
k
k
P
P
P
ai T (vi ) = ai
wi = T (0) = 0
T ( ai vi ) =
i=1
i=1
i=1
Since wi ’s are linearly independent, so ai = 0
∴ S is linearly independent.
14. (a)
(⇒) {v1 , v2 , · · · , vn } is linearly independent subset of V
P
If
ai T (vi ) = 0, then
P
P
T ( ai vi ) = 0 and
ai vi ∈ N (T ) = {0}
∴ ∀ai = 0
∴ {T (v1 ), · · · , T (vn )} is linearly independent subset of W
P
(⇐) Let w1 =
ai T (vi ), ∀ai , bi ∈ F
P
If w1 = w2 , then (ai − bi )T (vi ) = 0
Since T (vi )0 s are linearly independent, ∀ai = bi
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2.1.
∴
P
ai vi =
P
b i vi
∴ T is one-to-one
(b)
(⇒) By (a) , T (S) is linearly independent
(⇐) Let T (S) = {T (v1 ), · · · , T (vn )} ” linearly independent
P
P
If
ai vi = 0, then
ai T (vi ) = T (0) = 0
∴ ∀ai = 0
∴ S = {v1 , · · · , vn } is linearly independent
(c)
By (b), T (β) is linearly independent
By the theorem 2.2, R(T ) = span(T (β))
Since T is onto, span(T (β)) = W
∴ T (β) is a basis for W
15. (a)
Rx
Rx
Rx
(i) T (f (x)+g(x)) = 0 (f (t)+g(t))dt = 0 f (t)dt+ 0 g(t)dt = T (f (x))+T (g(x))
Rx
Rx
(ii) T (cf (x)) = 0 cf (t)dt = c 0 f (t)dt = cT (f (x))
Rx
Rx
(b) If T (f (x)) = T (g(x)), then 0 f (t)dt = 0 g(t)dt
Rx
i.e. 0 (f (t) − g(t))dt = 0
∴ f (x) = g(x)
(c) {x, x2 , · · · } is a basis for R(T )
Since spanR(T ) 6= P (R), T is not onto
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2.1.
16. (a)
Let f (x) = ax + b, g(x) = ax + d, b 6= d (i.e. f (x) 6= g(x))
But T (f (x)) = T (g(x))
(b) β = {1, x, x2 , · · · } is a standard basis for P (R)
It suffices to show that β ⊆ R(T ) = span(T (β))
1
∀xn ∈ β, xn = T ( n+1
xn ) ∈ R(T )
17. (a) By the theorem 2.3, dim R(T ) ≤ dim V
By the assumption, dim V < W
∴ dim R(T ) < W
∴ T can’t be onto
(b) If T is one-to-one i.e. nullity(T ) = 0, then dim V = rank(T ) > dim W
It’s a contradiction to dim R(T ) ≤ dim W
18. Let ∀(a, b) ∈ R2 , T (a, b) = (0, a)
Then N (T ) = {(0, b)|b ∈ R}, R(T ) = {(0, a)|a ∈ R}
∴ N (T ) = R(T )
19. ∀(a, b) ∈ R2 , Let T (a, b) = (0, a) and U (a, b) = (0, 2a)
Then T 6= U, N (T ) = {(0, b)|b ∈ R}, N (U ) = {(0, b)|b ∈ R}
R(T ) = {(0, a)|a ∈ R}, R(U ) = {(0, 2a)|a ∈ R}
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2.1.
20.
(a) Let w1 , w2 ∈ T (v1 ), a ∈ F
then ∃v1 , v2 ∈ V1 s.t. wi = T (vi ), i = 1, 2
So aw1 + w2 = aT (v1 ) + T (v2 ) = T (av1 + v2 ) ∈ T (v1 )
∴ T (v1 ) is a subspace of W
(b) Let K = {x ∈ V |T (x) ∈ W1 }, a ∈ F
Let x1 , x2 ıK s.t. T (x1 ) = w1 , T (x2 ) = w2 ∈ W1
T (ax1 + x2 ) = aT (x1 ) + T (x2 ) = aw1 + w2 ∈ W1
ax1 + x2 ∈ K, a ∈ F
∴ {x ∈ V |T (x) ∈ W1 } is a subspace of V
21. (a)
(i) T (c(a1 , a2 , · · · )+(b1 , b2 )) = T (ca1 +b1 , ca2 +b2 , · · · ) = (ca2 +b2 , ca3 +b3 , · · · ) =
c(a2 , a3 · · · ) + (b2 , b3 , · · · ) = cT (a1 , a2 , · · · ) + T (b1 , b2 , · · · )
(ii) U (c(a1 , a2 , · · · ) + (b1 , b2 )) = U (ca1 + b1 , ca2 + b2 , · · · ) = (0, ca1 + b1 , ca2 +
b2 , · · · ) = (0, ca1 , ca2 , · · · ) + (0, b1 , b2 , · · · ) = cU (a1 , a2 , · · · ) + U (b1 , b2 , · · · )
(b)
(i) T is not one-to-one
0 6= (1, 0, 0, · · · ) ∈ N (T )
(ii) T is onto
R(T ) = {(a2 , a3 , · · · )|∀ai ∈ F } = span{e1 , e2 , · · · } = V
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2.1.
(c)
(i) If U (a1 , a2 , · · · ) = (0, a1 , a2 , · · · ) = (0, 0, · · · ), then ∀ai = 0
∴ N (U ) = {0}
(ii) U is not onto
R(U ) = {(0, a1 , a2 , · · · )|∀ai ∈ F } 6= V
22.
(i) T : R3 → R is linear
Let β = {e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)} a standard basis for R3
Put T (e1 ) = a, T (e2 ) = b, T (e3 ) = c
then ∀(x, y, z) ∈ R3 ,
T (x, y, z) = T (xe1 + ye2 + ze3 ) = xT (e1 ) + yT (e2 ) + zT (e3 ) = ax + bycz
(ii) T : F n → F is linear
Let β = {e1 , e2 , · · · , en } a standard basis for F n
Put T (e1 ) = a1 , T (e2 ) = a2 , · · · , T (en ) = an
then T (x1 , x2 , · · · , xn ) = a1 x1 + · · · + an xn
(iii) T : F n → F m
For ∀ j (i ≤ j ≤ m), T (ej ) =
T (x1 , x2 , · · · , xn ) = T (
n
P
m
P
i=1
xj e j ) =
aij wi
n
P
xj T (ej ) =
j=1
j=1
n
P
j=1
xj (
m P
n
n
n
n
P
P
P
P
( )aij xj wi = ( a1j xj ,
a2j xj , · · · ,
amj xj )
i=1 j=1
j=1
j=1
m
P
i=1
aij wi ) =
n P
m
P
aij xj wi =
j=1 i=1
j=1
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2.1.
23. T : R3 → R linear transformation
By the exercise 22, ∃a, b, c ∈ R s.t. T (x, y, z) = ax + by + cz
N (T ) = {(x, y, z) ∈ R3 | T (x, y, z) = 0}
= {(x, y, z) ∈ R3 | ax + by + cz)0}
= R3 (where a = b = c = 0)
a plane through 0 (where a2 + b2 + c2 6= 0)
24.
(a) The projection on the y-axis along the x-axis
¶
µ
0 0
2
T (a, b) = (0, b), ∀(a, b) ∈ R , A =
0 1
(b)
T (a, b) = (a − b, a − b), ∀(a, b) ∈ R2
(a, b) 7→ (0, 0) if a = b
(a − b, a − b) otherwise
25.
(a) Let W1 = {(a, b, 0)|a, b ∈ R}, W2 = {(0, 0, c)|c ∈ R}
Then R3 = W1 ⊕ W2
∀x = (a, b, c) ∈ R3 s.t. x1 = (a, b, 0) ∈ W1 and x2 = (0, 0, c) ∈ W2
T (x) = x1


1 0 0
∴ T is the projection on the W1 along the W2 , A = 0 1 0
0 0 0
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2.1.
(b) T (a, b, c) = (0, 0, c)
(c) Let W1 = {(a, b, 0)|a, b ∈ R}, W2 = L = {(a, 0, a)|a ∈ R}
then R3 = W1 + W2
For all x = (a, b, c) ∈ R3 s.t. x = x1 + x2 , x1 = (a − c, b, 0) ∈ W1 , x2 = (c, 0, c) ∈
W2
T (a, b, c) = (a − c, b, 0) ∈ W1


1 0 −1
∴ T : R3 R3 is the projection on W1 along the W2 , A = 0 1 0 
0 0 0
26.
∀x ∈ V s.t x = x1 + x2 , x1 ∈ W1 , x2 ∈ W2 and T(x) = x1
(1) T is linear
∀x, y ∈ V s.t. x = x1 + x2 , y = y1 + y2 , x1 , y1 ∈ W1 , x2 , y2 ∈ W2
T (x + y) = T ((x1 + y1 ) + (x2 + y2 )) = x1 + y1 =T(x)+T(y)
T (cx) = T (cx1 + cx2 ) = cx1 = T (x)
(2) W1 = {x ∈ V |T(x) = x}
(⊆) If x1 ∈ W , then T (x1 ) = x1
∴ x1 ∈ {x ∈ V | T (x) = x}
(⊇) If x ∈ V s.t T(x) = x and we have T (x) = x1
(∴) x = x1 ∈ W1
(b)
(1)W1 = R(T )
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2.1.
(⊇) ∀ T (x) ∈ R(T ), T (x) = x1 ∈ W1 ∴ R(T ) ⊆ W1
(⊆) ∀ x1 ∈ W1 , x1 = T (x1 ) ∈ R(T ) ∴ W1 ⊆ R(T )
(2)W2 = N (T )
∀x2 ∈ W2 , T (x2 ) = 0 ∴ x2 ∈ N (T )
∴ W2 ⊆ N (T )
(⊆) ∀x ∈ N (T ), T (x) = x1 = 0 ∴ x = x1 + x2 = 0 + x2 = x2 ∈ W2
(⊇) ∴ N (T ) ⊆ W2
(c) Describe if W1 = V
T (x) = x, ∀x ∈ W1 = V
∴ T is the identity transformation(IV )
(d) Describe if W1 = 0
T (x) = x ⇔ x = 0 ( or (0) = W1 = R(T ) ) ∴ T is the zero transformation(T0 )
27.
Claim : ∀W ≤ V, ∃W 0 ≤ V as subspace, T = W +W 0 and T : V → V , T (V ) = W
(a) From the exercise 34 in Section 1.6(p.58),
If W is any subspace of a finite dimensional vector space,
∃ W 0 s.t V = W ⊕ W 0
so ∀x ∈ V, ∃!x1 ∈ W x2 ∈ W 0 s.tx = x1 + x2
Define T : V → V T (x) = x1 is a desired linear transformation
Then T is a projection on W along W 0
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2.1.
* Say dim V = n ≥ 2
{v1 , v2 , · · · , vm } : a basis for W
{v1 , v2 , · · · , vm , vm+1 , · · · , vn } : a basis for V
Let W 0 = span{vm+1 , · · · , vn } and W 00 = span{vm+1 − v1 , vm+2 − v3 , · · · , vn − v5 }
then V = W ⊕ W 0 = W ⊕ W 00
Clearly W 0 6= W 00
(∵) If W 0 = W 00 , then vm+1 , vm+1 − v1 ∈ W 00
∴ v1 ∈ W ∩ W 0 = (0)
It’s a contradiction
(b) Example
(example 1) the projection on W along W10
{(a, b)) = (0, b − 31 a) + (a, 13 a)}
(example 2) the projection on W along W20
(a, b) = (0, b) + (a, 0)
28.
(1) {0} is T -invariant
∀x ∈ {0}, T (x) = 0 ∈ {0} (∵ T is linear)
(2) V is T -invariant (T(V)⊆V)
∀x ∈ V , T (x) ∈ V (∵ T : V → V )
(3) R(T ) is T -invariant (T(T(V))⊆T(V))
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∀, T (x) ∈ R(T ),
T (T (x)) ∈ T (V ) ⊆ R(T )
(4) N (T ) is T-invariant
∀x ∈ N (T ), T(x) = 0 ∈ N (T ) (∵ N (T ) ≤ V as subspace, so N (T ) has zero)
29.
If T (W ) ⊆ W , ∀x, y ∈ W, and c ∈ F
TW (x + y) = T (x + y) = T (x) + T (y) = TW (x) + TW (y) (W is T-invariant and T
is linear)
TW (cx) = T (cx) = cT (x) = cTW (x)
∴ TW is linear
30. ∀x ∈ V, T (x) = x1 s.t x = x1 + x2 , x1 ∈ W, x2 ∈ W 0
(1) W is T -invariant
∀x1 ∈ W , T(x1 ) = x1 ∈ W ∴ T(W ) ⊆ W
(2) TW = IW
TW : W → W, ∀x1 ∈ W, TW (x1 ) = x1
IW : W → W, ∀x1 ∈ W, IW (x1 ) = x1
∴ ∀x1 ∈ W, TW (x1 ) = IW (x1 )
∴ TW = I W
31. V = R(T ) ⊕ W , W is T-invariant
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2.1.
(a)
T (W ) ⊆ T (V ) ∩ W = (0) ⇒ T (W ) = 0
∴ T (W ) ⊆ N (T ) (b)
Since V = R(T ) ⊕ W ,
so dim V = dim R(T ) + dim W ≤ rank(T ) + nullity(T ) = dim V
Since dim V < ∞, dim W = nullity(T )
(c) (Example 1) Exercise 21, left shift
(Example 2) β = {v1 , v2 , · · · } for V
T : V → V, T (vi ) = 0 if i is odd
i
2
if i is even
Then R(T ) = V, N (T ) = span({v1 , v3 , v5 , · · · }), W = (0)
∴ V = R(T ) ⊕ W
(Example 3) dim V = ℵ0 ,
{v1 , v2 , v3 , v4 , v5 , v6 , · · · } : a basis for V
{v1 , v2 , v3 , v5 , v6 , v7 , v9 , v10 , · · · } : a basis for R(T )
{v3 , v4 , v7 , v8 , v11 , v12 , v15 , v16 , · · · } : a basis for N (T )
W = span{v4 , v8 , v12 , · · · }
i.e. V = R(T ) ⊕ W, W
N (T )
32.
(1) N (TW ) = N (T ) ∩ W
(⊆) N (TW ) = {x1 ∈ W | TW (x1 ) = 0}
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2.1.
∀x1 ∈ N (TW ), x1 ∈ W
TW (x1 ) = T (x1 ) = 0 i.e. x1 ∈ N (T )
∴ N (TW ) ⊆ N (T ) ∩ W
(⊇) If x1 ∈ N (T ) ∩ W , then x1 ∈ W and TW (x1 ) = T (x1 ) = 0 (∵ x1 ∈ N (T ))
∴ N (T ) ∩ W ⊆ N (TW )
(2) R(TW ) = T (W ), R(TW ) = {TW (x1 ) | x1 ∈ W }
(⊆) ∀x1 ∈ W, TW (x1 ) = T (x1 ) ∈ T (W )
(⊇) ∀x1 ∈ W, T (x1 ) = TW (x1 ) ∈ R(TW )
33. Prove theorem 2.2 in case β is infinite
Claim : R(T ) = span(T (β)) = span{T (v1 ), T (v2 ), · · · }
(⊇) Clearly T (vi ) ∈ R(T ) for each i.
Since R(T ) ≤ V as subspace,
R(T ) ⊇ span{T (v1 ), T (v2 ), · · · } = span(T (β))
(⊆) Suppose that w ∈ R(T ), then w =T(v) for some v ∈ V
Since β is a basis for V ,
each v ∈ V can be uniquely expressed as a linear combination of vectors of β
It means there exist a finite number of vectors v1 , v2 , · · · , vn in β and scalars
n
P
a1 , a2 , · · · , an in F s.t v =
ai vi
i=1
Since T is linear, w = T (v) =
n
P
ai T (vi ) ∈ spanT (β)
i=1
∴ R(T ) ⊆ span(T (β))
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2.1.
34. Generalization of theorem 2.6
Claim : ∀f : β → W, ∃! T (x) = f (x), ∀xi ∈ β
Let f : β → W s.tf (xi ) = wi where β = {xi |i ∈ I}
Note that ∀v ∈ V, v = ai vi in a unique way where ai ∈ F, xi ∈ β
P
Define T : V → W by T (v) =
ai f (xi )
i
(i) T is well-defined and unique
(ii) T is linear
(iii) T (x) = f (x), ∀x ∈ β
(*) Well-definess
(Example 1) V = R2 , T : R2 → R by v = a1 v1 + a2 v2 7→ (a1 + a2 )
β = {v1 = (1, 1), v2 = (1, −1)}
∀v = (a, b) = ( a+b
, a+b
) + ( a−b
, − a−b
)=
2
2
2
2
a+b
v1
2
+
a−b
v2
2
Then actually T : (a, b) 7→ a
(Example 2) V = R2 , T : R2 → R by v = a1 v1 + a2 v2 + a3 v3 7→ (a1 + a2 + a3 )
β = {v1 = (1, 1), v2 = (1, −1), v3 = (0, 1)}
∀v = (a, b) =
a+b
v1 + a−b
v2 + 0 · v3
2
2
7→ a and v = (a, b) = av1 + 0 · v2 + (b − a)v3 7→ b
∴ T is not well-defined
35.
(a) Suppose V = R(T ) + N (T )
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2.1.
Claim : R(T ) ∩ N (T ) = {0}
(⊇) Since R(T ), N (T ) ≤ V as subspaces,
So R(T ) and N (T ) have {0}
∴ R(T ) ∩ N (T ) ⊇ {0}
(⊆) If v ∈ R(T ) ∩ N (T ),
then ∃x ∈ V s.t T (x) = v and T (v) = 0
v = T (x) = T 2 (x) = T (v) = 0
(∵ T is a projection on R(T ) along N (T ), then T 2 = T )
∴ R(T ) ∩ N (T ) = {0}
(b) Suppose R(T ) ∩ N (T ) = {0}
We are going to show that V = R(T ) + N (T )
Let dim V = n, rank(T ) = m, null(T ) = k and n = m + k,
and let β1 = {w1 , w2 , cdots, wm } a basis for R(T )
β2 = {u1 , u2 , cdots, uk } a basis for N (T )
Claim : β = β1 +β2 = {w1 , w2 , cdots, wm , u1 , u2 , cdots, uk } is linearly independent
If a1 w1 + · · · + am wm + b1 u1 + · · · + bk uk = 0, (∀ai , bj ∈ F )
then a1 w1 + · · · + am wm = −(b1 u1 + · · · + bk uk ) ∈ R(T ) ∩ N (T ) = {0}
∴ a1 = · · · = am = b1 · · · = bk = 0
Since dim V = n, β is a basis for V ,
∀v ∈ V , v = c1 w1 + · · · + cm wm + d1 u1 + · · · + dk uk = 0, (∀ci , dj ∈ F )
let w = c1 w1 + · · · + cm wm and u = d1 u1 + · · · + dk uk
then v = w + u ∈ R(T ) + N (T )
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2.1.
∴ V = R(T ) ⊕ N (T )
36.
(a) We are going to show that if V is infinite-dimensional, then V doesn’t hold
the result of Exercise 35(a)
From Exercise 21, ∀v = (a1 , a2 , · · · ) ∈ V , T : V → V is left shift
then N (T ) = {(a1 , 0, 0, · · · ) | a1 ∈ F }
R(T ) = {(a2 , a3 , a4 , · · · ) | ∀ai ∈ F, i = 2, 3, · · · }
∴ ∀v ∈ V, v = (a1 , a2 , a3 , · · · ) = (a1 , a2 , a3 , · · · ) + (0, 0, 0, · · · ) ∈ R(T ) + N (T )
∴ V = R(T ) + N (T )
But R(T ) ∩ N (T ) = {(a, 0, 0, · · · ) | a ∈ F } 6= {0}
∴ V 6= R(T ) ⊕ N (T )
(b) Find T1 : V → V s.t R(T1 ) ∩ N (T1 ) = {0} but V 6= R(T1 ) ⊕ N (T1 )
∀v ∈ V, T1 (v) = T (a1 , a2 , · · · ) = (0, a1 , a2 , · · · )
then N (T1 ) = {0}
R(T1 ) = {(0, a1 , a2 , · · · ) | ai ∈ F }
∴ R(T1 ) ∩ N (T1 ) = {0}, but
for 0 6= a ∈ F, ∃v = (a, 0, 0, · · · ) is not in R(1 ) + N (T1 )
∴ R(T1 ) + N (T1 )
V
∴ V 6= R(T1 ) ⊕ N (T1 )
37. We are going to show that T (αx) = αT (x), α ∈ Q
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2.1.
∀x, y ∈ V, T (x + y) = T (x) + T (y)
Let α = ab , a, b ∈ Z, a 6= 0
T (αx) = T ( ab x) = T ( ab a · xa ) =
b
a
· aT ( xa ) =
b
a
· T (a xa ) = ab T (x)
∴ T is linear
38.
∀x, y ∈ C s.t x = a + bi, y = c + di, ∀a, b, c, d ∈ R, α ∈ C
T(x + y)=T((a + c) + (b + d)i) = (a + c) − (b + d)i = (a − bi) + (b − di)=T(x)+T(y)
In case α = i, T(αx)=T(−b + ai) = −b − ai
but αT(x) = i(a − bi) = b − ai
∴ T(αx) 6= αT(x)
∴ T is not linear
39.
Claim : ∃ T:R → R is an additive function that is not linear
let V be the set of real numbers regarded as a vector space over the field of rational numbers.
Since every vector space has a basis(p.61),so V has a basis β
For fixed x, y ∈ β s.t x 6= y
Define f : β → V by f (x) = y, f (y) = x and f (z) = z otherwise
By Exercise 34, ∃! T:V→V : linear transformation over Q s.t T(u)=f(u), ∀u ∈ β
Then T is additive(from exercise 37) but T is not linear over R
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2.1.
(∵) In case α = xy ,
T(αx)=T( xy x)=T(y) = f (y) = x
αT(x)= xy T(x) = xy f (x) = xy y =
y2
x
∴T(αx) 6= αT(x) (∵ x 6= y)
∴ T is not linear
40. η: V→V/W, η(v) = v + W (v + W = 0 in V/W ⇔ v ∈ W )
(a) Prove η is linear and N(η)=W
∀v1 , v2 ∈ V , c∈ F
(1) η(cv1 + v2 ) = (cv1 + v2 ) + W = cv1 + W + v2 + W = c(v1 + W ) + (v2 + W ) =
cη(v1 ) + η(v2 )
(2) ∀v + W ∈ V /W, ∃v ∈ V
(3) N(η)=W
(∵) (⊇) If w ∈ W , then η(w) = w + W = 0 + W
∴ w ∈N(η)
(⊆) If v ∈N(η), then η(v) = 0 + W = w + W for some w ∈ W
∴ v ∈W
(b) Suppose V is finite-dimensional (kerη = W, R(η) = V /W )
dimN(η)=dim(W), Rank(η)=dim(V/W)=dimV-dimW (p.58 Sec1.6 Exercise35)
dimV=dimN(η)+Rank(η)=dim(W)+dim(V/W)
(c) same
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2.2.
2.2. The matrix representation of a linear transformation
1. β = {v1 , v2 , · · · , vn }, γ = {w1 , w2 , · · · , wm } bases for V and W, respectively
(a) T (p.82 Theorem 2.7(a))
(b) T (p.73 The corollary to Theorem 2.6 and p.80)
(p.80) Let T:V→W is linear. Then for each j, 1 ≤ j ≤ n, there exist unique
scalars aij , bij ∈ F, 1 ≤ i ≤ m s.t
m
m
P
P
T(vj ) =
aij wi , U(vj ) =
bij wi for 1 ≤ j ≤ n
i=1
i=1
Suppose [T ]γβ = (aij )m×n ,[U ]γβ = (bij )m×n
m
m
P
P
aij wi = bij wi =U(vj ) for all aij , bij ∈ F, ∀vj ∈ β.
If [T ]γβ = [U ]γβ , then T(vj ) =
i=1
i=1
Hence T=U
(c) F ([T ]γβ is an n × m matrix )
(d) T (p.83 Theorem 2.8 (a))
(e) T (0 ∈ L(V, W )) and Theorem 2.7 (a))
(f) F (p.104 L(V, W ) ∼
= Mm×n (F ), L(W, V ) ∼
= Mn×m (F ))
(cf ) (L(V, W ) ∼
= L(W, V )) but (L(V, W ) 6= L(W, V ))
2. Compute [T ]γβ
(a) T(1,0)=(2,3,1)=2w
1 + 3w2 + 1w3 , T(0,1)=(−1, 4, 0) = −1w1 + 4w2 + 0w3


2 −1
[T ]γβ = 3 4 
1 0
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2.2.
(b) T(1,0,0)=(2,1), T(0,1,0)=(3,0), T(0, 0, 1) = (−1, 1)
µ
¶
2 3 −1
γ
[T ]β =
1 0 1
(c) [T ]γβ = (1, 0, −3)


0 2 1
(d) [T ]γβ = −1 4 5
 1 0 1 
1 0 ··· 0
1 0 · · · 0


(e) [T ]γβ =  .. ..
.
 . . · · · .. 
1 0 ··· 0


0 0 ··· 0 1
0 0 · · · 1 0


 .. ..

γ
.
.
(f) [T ]β =  . . · · · . 


0 1 · · · 0 0
1 0 ··· 0 0
γ
(g) [T ]β = (1, 0, · · · , 0, 1)
3.
(a) T(1,0)=(1,1,2)=− 31 (1, 1, 0) + 0(0, 1, 1) + 23 (2, 2, 3)
T(0,1)=(−1,
1, 0) + 1(0, 1, 1) + 0(2, 2, 3)
 1 0, 1)=−1(1,

− 3 −1
1
[T ]γβ =  0
2
0
3
(b) T(1,2)=(−1, 1, 4)=− 73 (1, 1, 0) + 2(0, 1, 1) + 32 (2, 2, 3)
T(2,3)=(−1, 2, 7)=− 11
(1, 1, 0) + 3(0, 1, 1) + 34 (2, 2, 3)
3
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2.2.
 7

− 3 − 11
3
3 
[T ]γα =  2
2
3
4
3
µ
¶
0
1 0
4. β = {e1 =
, e2 =
0 0
µ
¶
µ0
1 0
0
T
= 1 + 0x + 0x2 , T
0
0
µ
¶
µ0
0 0
0
T
= 0 + 0x + 0x2 , T
1 0
0

1 1 0 0
[T ]γβ = 0 0 0 2
0 1 0 0
µ
µ
¶
µ
¶
¶
0 0
0 0
1
}, γ = {1, x, x2 }
, e4 =
,e =
0 1
1 0
0¶ 3
1
= 1 + 0x + 1x2 ,
0¶
0
= 0 + 2x + 0x2
1
5.α = {e1 , e2 , e3 , e4 }, β = {1, x, x2 } , γ = {1}
(a)
µ
1
T
µ0
0
T
µ0
0
T
µ1
0
T
0
¶ µ
¶
0
1 0
=
= 1e1 + 0e2 + 0e3 + 0e4 ,
0¶ µ0 0¶
1
0 0
=
= 0e1 + 0e2 + 1e3 + 0e4 ,
0¶ µ1 0¶
0
0 1
=
= 0e1 + 1e2 + 0e3 + 0e4 ,
0¶ µ0 0¶
0
0 0
=
= 0e1 + 0e2 + 0e3 + 1e4
1
0 1


1 0 0 0
0 0 1 0 

[T ]γ = 
0 1 0 0 
0 0 0 1
µ
¶
0 2
(b) T(1)=
= 0e1 + 2e2 + 0e3 + 0e4 ,
0
0
µ
¶
1 2
T(x)=
= 1e1 + 2e2 + 0e3 + 0e4 ,
0 0
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2.2.
µ
0
T(x )=
0

0

2
[T ]γβ = 
0
2
2
¶
2
= 0e1 + 2e2 + 0e3 + 2e4 ,
2

1 0
2 2

0 0
2 2
(c) T(e1 )=1, T(e2 )=0, T(e3 )=0, T(e4 )=1
[T ]γα = (1, 0, 0, 1)
(d) T(1)=1, T(x)=2, T(x2 )=4, T(f (x))=f (2)
[T ]γβ = (1, 2, 4)
(e) A=1e1 + (−2)e2 + 0e3 + 4e4
[A]α = (1, −2, 0, 4)T
(f) f (x) = 3 · 1 + (−6)x + 1x2
[f (x)]β = (3, −6, 1)T
(g)γ = {1}
[a]γ = a
6. Theorem 2.7 (b)
L(V, W ) is a vector space over F
(a) T0 ∈ L(V, W ), T0 : the zero transformation
(b) aT + U ∈ L(V, W ), ∀T, U ∈ L(V, W ), ∀a ∈ F
7. Theorem 2.8 (b)
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2.2.
Let β = {v1 , v2 , · · · , vn } and γ = {w1 , w2 , · · · , wm } bases for V and W, respectively
and let [T ]γβ = (aij )m×n , aij ∈ F
m
m
m
P
P
P
then (aT )vj = aT (vj ) = a aij wi =
a(aij wi ) = (aaij )wi , ∀i
So
[aT ]γβ
i=1
i=1
= (aaij )m×n = a(aij )m×n =
a[T ]γβ
i=1
8. β = {v1 , · · · , vn } : a basis for V
n
n
P
P
∀x, y ∈ V s.t x =
ai vi , y =
bi vi ∈ V, ai , bi , c ∈ F
i=1
i=1
T(x) = [x]β = (a1 , · · · , an )t
n
P
and cx + y = (cai + bi )vi , then
i=1
T(cx + y)=[cx + y]β =(ac1 + b1 , · · · , can + bn )t = (ca1 , · · · , can )t + (b1 , · · · , bn )t =
c(a1 , · · · , an )t + (b1 , · · · , bn )t = c[x]β + [y]β = cT(x)+T(y)
∴ T is linear
(Indeed T is an isomorphism)
9.
∀x = a + bi, y = c + di ∈ C, α, a, b, c, d ∈ R
(a) T(αx+y)=T((αa+c)+(αb+d)i)=(αa+c)−(αb+d)i = (αa−αbi)+(c−di) =
α(a − bi) + (c − d)i = αT(a + bi)+T(c + di) = αT(x)+T(y)
∴ T is linear
(b) T(1)=1=1·1 + 0 · i and T(i) = −i = 0 · 1 + (−1) · i
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2.2.
µ
A=[T]β =
¶
1 0
0 −1
10. Compute [T]β
T(v1 ) = v1 + v0 = v1 = 1 · v1 + 0 · v2 + · · · + 0 · vn
T(v2 ) = v2 + v1 = v1 + v2 = 1 · v1 + 1 · v2 + · · · + 0 · vn
T(v3 ) = v3 + v2 = v2 + v3 = 0 · v1 + 1 · v2 + 1 · v3 + · · · + 0 · vn
..
.
T(vn ) = v
= 0 · v1 + 0 · v2 + · · · + 1 · vn−1 + 1 · vn
n + vn−1 = vn−1 + vn
1 1 0 ··· 0 0
0 1 1 · · · 0 0


 .. .. ..
.. .. 

. .
∴ [T]β =  . . .

0 0 0 · · · 1 0


0 0 0 · · · 1 1
0 0 0 ··· 0 1
11.
We choose a basis {w1 , · · · , wk } of W and extend it to a basis of V ({w1 , · · · , wk , v1 · · · , vs })
Let dimV=n and n=k+s
TW (w1 )=T(w1 )=a11 w1 + · · · + a1k wk + 0 · v1 + · · · + 0 · vs
TW (w2 )=T(w2 )=a21 w1 + · · · + a2k wk + 0 · v1 + · · · + 0 · vs
..
.
TW (wk )=T(wk )=ak1 w1 + · · · + akk wk + 0 · v1 + · · · + 0 · vs
T(v1 )=b11 w1 + · · · + b1k wk + c11 v1 + · · · + c1s vs
T(v2 )=b21 w1 + · · · + b2k wk + c21 v1 + · · · + c2s vs
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2.2.
..
.
T(vs )=bs1 w1 + · · · + bsk wk + cs1 v1 + · · · + css vs
The matrix of T is the transpose of the matrix of coefficients in the above system
of equations

a11 a12
a21 a22
 .
..
 .
.
 .

a
a
∴ [T]β =  k1 k2
0
 0

0
 0
 .
..
 ..
.
0
0
···
···
···
···
···
···
a1k b11
a2k b21
..
..
.
.
akk bk1
0 c11
0 c21
..
..
.
.
0
b12 · · ·
b22 · · ·
..
.
bk2 · · ·
c12 · · ·
c22 · · ·
..
.
cs1 cs2 · · ·

b1s
b2s 
.. 

. 

bks 

c1s 

c2s 
.. 
. 
css
12. T : V → V : the projection on W along W 0
Let β1 = {v1 , · · · , vk }, γ1 = {vk+1 , · · · , vn } bases for W and W 0 , respectively
Since V = W ⊕ W 0 , β = β1 ∪ γ1 is an ordered basis for V
Claim : [T]β is a diagonal matrix
Since β = {v1 , · · · , vk , vk+1 , · · · , vn }, ∀x ∈ V, x =
So T(x) =
n
P
n
P
ai vi , ai ∈ F
i=1
ai Tvi
i=1
T(v1 ) = v1 = 1 · v1 + 0 · v2 + · · · + 0 · vn (∵ v1 ∈ W )
..
.
T(vk ) = vk = 0 · v1 + 0 · v2 + · · · + 1 · vk + · · · + 0 · vn (∵ vk ∈ W )
T(vk+1 ) = 0 = 0 · v1 + 0 · v2 + · · · + 0 · vn (∵ vk+1 ∈ W 0 )
..
.
T(vn ) = 0 = 0 · v1 + 0 · v2 + · · · + 0 · vn (∵ vn ∈ W 0 )
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2.2.

1 0 ··· 0 0
0 1 · · · 0 0
. .
.. ..
. .
. .
. .

∴ [T]β = 0 0 · · · 1 0

0 0 · · · 0 0
. .
.. ..
 .. ..
. .
0 0 ··· 0 0
∴ [T ] is a diagonal matrix
···
···
···
···
···

0
0
.. 

.

0
← k − th

0
.. 
.
0 n×n
13. β = {v1 , v2 , · · · } : a basis for V
Claim: If aT + bU = 0, then a = b = 0, a, b ∈ F
Let aT (v) + bU (v) = (aT + bU )(v) = 0(v) = 0
⇒ aT (v) = −bU (v) ∈ R(T ) ∩ R(U ) = 0
⇒ (aT )(v) = 0, (bU )(v) = 0, ∀v ∈ V
⇒ aT = 0, bU = 0
⇒∗ a = 0, b = 0 (∵ T, U 6= 0)
d (∗ )
a ∈ F, T ∈ L(V, W )
If aT = 0 ⇒either a = 0 or T = 0
(∵) Assume a 6= 0 ⇒ ∃a−1 ∈ F
⇒ T = 1 · T = (a−1 a)T = a−1 (aT ) = a−1 0 = 0
or
Assume that T 6= 0 ⇒
∃v ∈ F s.t. T (v) 6= 0
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⇒ a(T (v)) = (aT )(v) = T0 (v) = 0
⇒ a=0
(∵) aw = 0 ⇒ a = 0 or w = 0 c
14.
Claim: If (a1 T1 +a2 T2 +· · ·+an Tn )(x) = 0(x), then a1 = a2 = · · · = an = 0, ai ∈ F
Let f (x) = k0 + k1 x + k2 x2 + · · · + kn xn ∈P(R)
(a1 T1 +a2 T2 +· · ·+an Tn )(f (x))= a1 T1 (f (x))+a2 T2 (f (x))+· · ·+an Tn (f (x))=a1 f 0 (x)+
a2 f 00 (x) + · · · + an f (n) (x) = 0
By Exercise 24. Sec1.6(p.56), we obtain a1 = a2 = · · · = an = 0
∴ {T1 ,T2 , · · · ,Tn } is a linearly independent subset of L(V ), ∀n ∈ Z + .
15. S 0 = {T∈ L(V, W ) | T(x) = 0, ∀x ∈ S}
(a) S 0 is a subspace of L(V, W )
(1) ∀x ∈ V, T0 ∈ L(V, W ) s.t T(x)=0
∴ ∀x ∈ S, T0 (x) = 0
∴ T0 ∈ S 0
(2) If T1 , T2 ∈ S 0 , α ∈ F, ∀x ∈ S
(αT1 +T2 )(x) = αT1 (x)+T2 (x) = α · 0 + 0 = 0
∴ ∀x ∈ S, (αT1 +T2 )(x) = 0
∴ αT1 +T2 ∈ S 0
(b) S10 = {T∈ L(V, W ) | T(x) = 0, ∀x ∈ S1 }
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and S20 = {T∈ L(V, W ) | T(x) = 0, ∀x ∈ S2 }
If T∈ S20 , then ∀x ∈ S2 , T(x) = 0
Since S1 ⊆ S2 , ∀x ∈ S1 , T(x) = 0
∴ T∈ S10
∴ S20 ⊆ S10
(c) (V1 + V2 )0 = V10 ∩ V20
(⊆) Since V1 , V2 ⊆ V1 + V2 and (b)
(V1 + V2 )0 ⊆ V10 and (V1 + V2 )0 ⊆ V20
∴ (V1 + V2 )0 ⊆ V10 ∩ V20
(⊇) ∀ T∈ V10 ∩ V20
Claim: T∈ (V1 +V2 )0 (i.e. ∀x ∈ V1 +V2 , T(x) = 0 s.t x = x1 +x2 , x1 ∈ V1 , x2 ∈ V2 )
T(x) =T(x1 + x2 )=T(x1 )+T(x2 ) = 0 + 0 = 0 (∵ T is linear and T∈ V10 ∩ V20 ,)
∴ T∈ (V1 + V2 )0
16.
Let β = {v1 , · · · , vn } and γ = {w1 , · · · , wn } : bases for V and W, respectively
Claim : [T]β is a diagonal matrix
Define T(vi ) = wi for i = 1, 2, · · · , n,
then [T]γβ = In
∴ [T]γβ is a diagonal matrix
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2.3. Composition of Linear Transformations and Matrix
Multiplication
1.
(a) F ([U T ]γα = [U ]γβ [T ]βα ) (p.88 Theorem 2.11)
(b) T (p.91 Theorem 2.14)
(c) F ([U (w)]γ = [U ]γβ [w]β for all w ∈ W )
(d) T (Since IV (vj ) = vj , 1 ≤ i, j ≤ n [IV ]α =In
(e) F
In case T:V→V, [T 2 ]α = [T · T ]α = [T ]α [T ]α = ([T ]α )2
(f) F
µ1
1
2
¶
2
If I 6=A= 23
1 , then A =I
−
µ 2 ¶2
0 1
(cf) A=
1 0
(g) F
(T:F n → F m ⇔T=LA for some A ∈ Mm×n (F )
(h) F
µ
¶
0 0
If A=
, then A2 = 0 even though A6= 0
1 0
The cancelation property for multiplication in fields is not valid for matrices.
(i) T (p.93 Theorem 2.15(c))
(j) T (p.89 Definition)
2.
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(a)
µ
20 −9 18
(1) A(2B+3C)=
5 10 8
¡
¢
(2) (AB)D= 29 −26
¡
¢
(3) A(BD)= 29 −26
(b)
(1)
(2)
(3)
(4)
(5)
¶
µ
¶
2 −3 4
A =
µ5 1 2 ¶
23 19 0
At B=
26 −1 10
12
t  
BC = 16
¡ 19
¢
CB= 27 7 9
¡
¢
CA= 20 26
t
3.
(a)
(1) [U ]γβ
U (1) = U (1 + 0 · x + 0 · x2 ) = (1, 0, 1)
U (x) = U (0 + 1 · x + 0 · x2 ) = (1, 0, −1)
2
U (x2 ) = U(0 + 0 · x +
1 · x ) = (0, 1, 0)
1 1 0
∴ [U ]γβ = 0 0 1
1 −1 0
(2) [T ]β
T (1) = T (0 · (3 + x) + 2 · 1) = 2 = 2 · 1 + 0 · x + 0 · x2
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2.3.
T (x) = T (1 · (3 + x) + 2 · x) = 3x + 3 = 3 · 1 + 3 · x + 0 · x2
T (x2 ) = T (2x · (3 + x) + 2 · x2 ) = 4x2 + 6x = 0 · 1 + 6 · x + 4 · x2
(3) [U T ]γβ = [U ]γβ [T ]β
(U T )(1) = U (T (1)) = U (2 · 1 + 0 · x + 0 · x2 ) = (2, 0, 2)
(U T )(x) = U (T (x)) = U (3 · 1 + 3 · x + 0 · x2 ) = (6, 0, 0)
2
(U T )(x2 ) =U (T (x2 )) =
 U (0 · 1 + 6 · x + 4 · x ) = (6, 4, −6)
2 6 6
∴ [U T ]γβ = 0 0 4 
2 0 −6


 
2 6 6
1 1 0
2 3 0
And [U ]γβ [T ]β = 0 0 1 0 3 6 = 0 0 4 
2 0 −6
1 −1 0
0 0 4
γ
γ
∴ [U T ]β = [U ]β [T ]β
(b)


3
(1) [h(x)]β = −2
1¡
¢
(2) [U (h(x))]γ = 1 1 5
 
1 1 0
3
¡
¢
γ



= 1 1 5
0
0
1
−2
(3) [U (h(x))]γ = [U ]β [h(x)]β =
1 −1 0
1
4.

1
0
(a) [T (A)]α = [T ]α [A]α = 
0
0

0
0
1
0
0
1
0
0
   
0
1
1
 4  −1
0
  =  
0 −1  4 
1
6
6
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2.3.


 
0  
−6
4


2    2 
−6 =  
0
0
3
2
6
 
1

¡
¢
¡ ¢
3

(c) [T (A)]γ = [T ]γα [A]α = 1 0 0 1 
2 = 5
4 
¡
¢ 6
¡ ¢
(d) [T (f (x))]γ = [T ]γβ [A]β = 1 2 4 −1 = 12
2
0

2
(b) [T (f (x))]α = [T ]αβ []β = 
0
0
1
2
0
0
5. Theorem 2.12
(a) (D+E)A=DA+EA
m
m
m
P
P
P
(Dik + Eik )Akj =
(Dik Akj + Eik Akj ) =
[(D + E)A]ij = (D + E)ik Akj =
m
P
k=1
Dik Akj +
k=1
m
P
k=1
k=1
Eik Akj = (DA)ij + (EA)ij = [DA + EA]ij
k=1
(b) a(AB) = (aA)B = A(aB) for any scalar a
n
n
P
P
a(Aik Bkj ) =
(aAik )Bkj = [(aA)B]ij
We have [a(AB)]ij =
and [(aA)B]ij =
n
P
k=1
(aAik )Bkj =
k=1
n
P
k=1
Aik (aBkj ) = [A(aB)]ij
k=1
∴ a(AB) = (aA)B = A(aB)
(c) Aij =
n
P
k=1
Aik δkj =
n
P
Aik (In )kj = (A · In )ij
k=1
(d) Let dimV=n and β = {v1 , · · · , vn }
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IV : V → V s.t IV (vi )
= vi = 0 · v1 + · · · + 0 · vi−1 + 1 · vi + 0 · vi+1 + · · · + vn
1 ··· 0


∴ [IV ]β =  . . .  = In
0 ··· 1
Corollary.
k
k
P
P
(1) A( ai Bi ) =
ABi
A(
k
P
i=1
ai Bi ) = A(a1 B1 + · · · + ak Bk ) = A(a1 B1 + · · · + A(ak Bk ) = a1 AB1 + · · · +
i=1
ak ABk =
(2) (
k=1
k
P
k
P
ai ABi
i=1
ai Ci )A = (a1 C1 + · · · + ak Ck )A = (a1 C1 )A + · · · + (ak Ck )A =
i=1
k
P
ai Ci A
i=1
6. Theorem 2.13
(b) 
 
B1j
B11 · · ·
 B2j   B21 · · ·

 
vj =  ..  =  ..
 .   .
Bnj
∴ vj = Bej
Bn1 · · ·
B1j · · ·
B2j · · ·
..
.

B1n
B2n 
¡
..  0 · · ·
. 
Bnj · · ·
Bnn
0 1(j − th) 0 · · ·
7. Theorem 2.15
(c) ∀x ∈ F n
(1) LA+B (x) = (A + B)x = Ax + Bx = LA (x) + LB (x) = (LA + LB )(x)
∴ LA+B = LA + LB
(2) LaA (x) = (aA)x = a(Ax) = (aLA )(x)
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¢t
0
2.3.
∴ LaA = aLA
(f) LIn = IF n
LIn (x) = In x = x and IF n (x) = x ∀x ∈ F n
∴ LIn = IF n
8.
(a)
(1) (T (U1 + U2 ))(x) = T ((U1 + U2 )(x)) = T (U1 (x) + U2 (x)) = T (U1 (x)) +
T (U2 (x)) = (T U1 )(x) + (T U2 )(x) = (T U1 + T U2 )(x)
∴ T (U1 + U2 ) = T U1 + T U2
(2) (U1 + U2 )T (x) = U1 (T (x)) + U2 (T (x)) = (U1 T )(x) + (U2 T )(x) = (U1 T +
U2 T )(x)
∴ (U1 + U2 )T = U1 T + U2 T
(b)
T (U1 U2 )(x) = T (U1 (U2 (x))) = (T U1 )(U2 (x)) = (T U1 )U2 (x)
∴ T (U1 U2 ) = (T U1 )U2 (c)
T I(x) = T (I(x)) = T (x) ∴ T I = T
IT (x) = I(T (x)) = T (x) ∴ IT = T
(d)
(1) a(U1 U2 )(x) = aU1 (U2 (x)) = (aU1 )(U2 (x)) = ((aU1 )U2 )(x)
∴ a(U1 U2 ) = (aU1 )U2
(2) U1 (aU2 )(x) = U1 (aU2 (x)) = aU1 (U2 (x)) = a(U1 U2 )(x)
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∴ U1 (aU2 ) = a(U1 )U2 )
More general result
Let V,U,W be vector spaces over K. Suppose the following mappings are linear
F : V → U, F 0 : V → U and G : U → W, G0 : U → W
Then for any scalars k ∈ K
(1) G(F + F 0 ) = GF + GF 0
(2) (G + G0 )F = GF + G0 F
(3) k(GF ) = (kG)F + G(kF )
9.
(1) Let U : F 2 → F 2 s.t U (a, b) = (a + b, 0), ∀a, b ∈ F and
T : F 2 → F 2 s.t T (a, b) = (a, −a), ∀a, b ∈ F
then U T (a, b) = U (a, −a) = (0, 0) , i.e U T = T0 and T U (a, b) = T (a + b, 0) =
(a + b, −a − b) 6= (0, 0), i.e T U 6= T0
µ
¶
µ
¶
1 1
1 0
(2) A =
,B=
0 0
−1 0
then AB = 0 but BA 6= 0
10.
(⊇)Since Aij = δij Aij , if i 6= j, then Aij = 0
thus A is a diagonal matrix
(⊆) Since A is a diagonal matrix,
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2.3.
if i 6= j, then Aij = 0, thus Aij = δij Aij
and if i = j, then δij = 1, therefore Aij = δij Aij
11.
(⊆) ∀v ∈ V , T (V ) ∈ R(T ), since T (T (V )) = {0}, T (V ) ∈ N (T )
∴ R(T ) ⊆ N (T )
(⊇) ∀v ∈ V , T (V ) ⊆ R(T ),
∴ R(T ) ⊆ N (T )
∴ T 2 (v) ⊆ T (T (v) ⊆ T (N (T )) = 0
∴ T 2 = T0
12.
(a) Assume U T : V → Z is one-to-one and let T (V1 ) = T (v2 ) for v1 , v2 ∈ V
Then U (T (v1 )) = U (T (v2 )) i.e (U T )(v1 ) = (U T )(v2 )
Since U T is one-to-one, v1 = v2
∴ T is one-to-one
(Example)
V = R2 , W = Z = R3
T : V → W T (a, b) = (a, b, 0)
U : W → Z U (a, b, c) = (a, b, 0)
(b) Assume that U T : V → Z is onto and let z ∈ Z
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2.3.
∃v ∈ V s.t (U T )(v) = z
Let w = T (v)
Then w ∈ W and U (W ) = U (T (v)) = (U T )(v) = z
∴ U : W → Z is onto
(Example)
V = W = R3 , W = R2
T : V → W T (a, b, c) = (a, b, 0)
U : W → Z U (a, b, c) = (a, b)
U T is one-to-one but U is not
(c) (1) Let v1 , v2 ∈ V and assume (U T )(v1 ) = (U T )(v2 )
If U (T (v1 )) = U (T (v2 )), then T (v1 ) = T (v)2 (∵ U is one-to-one)
and v1 = v2 (∵) T is one-to-one
(Example)
V = W = R2 , W = R3
T : V → W T (a, b) = (a, b, 0)
U : W → Z U (a, b, c) = (a, b)
U T = iR2 is one-to-one an onto
(2) Let z ∈ Z
Since U is onto, ∃w ∈ W s.t z = U (w)
and T is onto, ∃v ∈ V s.t w = T (v))
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2.3.
Thus z = U (w) = U (T ()v) = (U T )(w)
∴ UT is onto
Therefore if U and T are one-to-one and onto, then so is UT.
13.
n
P
(a) tr(AB) =
(AB)ii =
i=1
n
PP
(A)ij (B)ji =
i=1 j=1
n
n
P P
P
( (B)ji (A)ij ) =
(BA)jj =
j=1 i=1
j=1
tr(BA)
n
P
(b) tr(A) =
(A)ii =
n
P
(At )ii = tr(At )
i=1
i=1
14.
(a) z = (a1 , a2 , · · · , ap )t =
So Bz = B(
p
P
aj ej ) =
j=1
p
P
p
P
aj ej
j=1
aj (Bej ) =
p
P
aj vj
j=1
j=1
n
P
(b) (AB)j = (AB)ej = A(Bej ) = Avj =
(c) wA = (
m
P
i=1
ai ei )A =
m
P
i=1
ai (ei A) =
m
P
bkj (uk ), vj = (b1j , · · · , vnj )t
k=1
ai ui
i=1
(d) (AB)(i) = ei (AB) = (ei A)B = ui B =
n
P
bik (vk ), ui = (bi1 , · · · , vin )
k=1
15.
M = (γij )m×n , A = (ajk )n×p
P
P
P
P
If A(j) =
bk A(k) ⇒ M A(j) = M ( bk A(k) ) =
bk M A(k) =
bk (M A)(k)
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16.
(a) If rank(T ) = rank(T 2 )
i.e. dimR(T ) = dimR(T 2 ) ∴ dimN (T ) = dimN (T 2 )
(i) But dimR(T ) ≤ dimV < ∞ ⇒ R(T ) = R(T 2 ) (; T = T 2 )
(ii) and dimN (T 2 ) ≤ dimV < ∞ ⇒ N (T ) = N (T 2 )
(∵) If v ∈ R(T 2 ) ⇒ v = T 2 (z) for some z ∈ V
⇒ v = T (T (z)) ∈ R(T )
∴ R(T 2 ) ⊆ R(T )
(a) If w ∈ R(T ) ∩ N (T )
⇒ w = T (V ) for some v ∈ V
Then T 2 (v) = T (w) = 0
∴ v ∈ N (T 2 ) = N (T )
∴ w = T (v) = 0
(b)
Since R(T ) ⊇ R(T 2 ) ⊇ · · ·
we have dimV ≥ rankT ≥ rankT 2 ≥ · · · ≥ 0
So ∃k(≥ 1), rank(T k ) = rank(T k−1 ) ⇒∗ R(T k ) = R(T k+1 ) = · · · = R(T 2k )
d(∗ )
If w ∈ R(T k ) ⇒ w = T k (v) = T k+ (v1 )
and T k+1 (v1 ) = T (T k (v1 )) = T (T k+1 (v2 )) = T k+2 (v2 )
c
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2.3.
i.e. R(T k ) = R(T 2k )
Let u := T k
then u2 := T 2k
∴ R(u) = R(u2 )
Using the similar way to this, then V = R(U ) ⊕ N (U )
∴ V = R(T k ) ⊕ N (T k )
17.
For every x ∈ V , x = T (x) + (x − T (x)) and
we are going to show that V = {y | T (y) = y} ⊕ N (T )
Since T 2 (x) = T (T (x)) = T (x), T (x) ∈ {y | T (y) = y}
and T (T (x − T (x)) = T (x) − T 2 (x) = T (x) − T (x) = 0
∴ x − T (x) ∈ N (T )
∴ V = {y | T (y) = y} + N (T )
and if ∃x ∈ {y | T (y) = y} ∩ N (T ), then x = 0
(∵) Since x ∈ {y | T (y) = y}, ∴ T (x) = x
and x ∈ N (T ), T (x) = 0
∴x=0
∴ V = {y | T (y) = y} ⊕ N (T )
18.
Let A be an m × n matrix, B be an n × p matrix and C be an p × q matrix
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2.3.
We are going to show that (AB)C=A(BC), for 1 ≤ i ≤ m, 1 ≤ j ≤ q
p P
p
p
n
n
P
P
P
P
P
( (A)il Blk )Ckj =
Ail ( Blk Ckj ) =
((AB)C)ij = ( (AB)ik Ckj ) =
Ail (BC)lj =
k=1
k=1 l=1
l=1
k=1
(A(BC))ij
19.
(B 3 )kk > 0 ⇔ Bki Bij Bjk = 1 for some i 6= j differ from k
⇔ k belongs to a clique
20.

0 2 0

2 0 1
(a) Since B 3 = 
0 1 0
3 0 2
∴ @ clique

2 0 3

0 0 0
(b) Since B 3 = 
3 0 2
3 0 3
∴ 1,3,4 belong to clique

3
0
 , f orallBii3 = 0
2
0

3
0

3
2
21.
For convenience, let A + A2 = (bij )n×n and for all i(1 ≤ i ≤ n),
let D(i) = {j|aij = 1}
Choose a k such that D(k) is maximal in the set {D(i)|1 ≤ i ≤ n}
n
P
We will show that bkj = akj +
aki aij > 0 for all j 6= k
i=1
For a fixed j(6= k), if akj = 1 then bkj > 0
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2.3.
Now suppose akj = 0, then ajk = 1
So k ∈ D(j)
If (for the case akj = 0) aij = 0 for all i ∈ D(k), then
aji = 1 and hence D(k) ⊆ D(j)
But k ∈ D(j), k ∈
/ D(k)
This is a contradiction to the choice of k
Thus aij = 1 for some i ∈ D(k), and this proves the property bkj = akj +
n
P
aki aij > 0
i=1
22.
1,2 and 3 dominate all the others in at most two stages, while 1,2, and 3 are
dominated by all the others in at most two stages
23.
n(n − 1)/2
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2.4.
2.4. Invertibility and Isomorphisms
1.
(a) F, ([T ]βα )−1 = [T −1 ]αβ
(b) T
(c) F
A ∈ M atm×n (F ), LA : F n → F m
⇒ [LA ]βα = A in case of α, β are the standard bases
(d) F, dim(M2×3 (F )) 6= F 5
(e) Pn (F ) ' Pm (F ) iff n=m
(⇐) clear
(⇒)∃T : Pn (F ) → Pm (F ) is isomorphic
Since T is one-to-one and onto,
dim(Pn (F )) = rank(T ) + nullity(T ) = dim(Pm (F ))
∴n+1=m+1
∴n=m
(f) F (In case A and B are n × n matrices, it’s true)
(g) T
(h) T (Exercise 8)
(i) T
2. T is invertible iff [T ]γβ
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2.4.
(a) T : R2 → R3 , T (a1 , a2 ) = (a1 − 2a2 , a2 , 3a1 + 4a2 )
For β = {(1, 0), (0, 1)} and γ = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} bases for R2 , R3 , respectively
T (1, 0) = (1, 0, 3) = 1 · (1, 0, 0) + 0 · (0, 1, 0) + 3 · (0, 0, 1)
T (0, 1) 
= (−2, 1,
4) = −2 · (1, 0, 0) + 1 · (0, 1, 0) + 4 · (0, 0, 1)
1 −2
[T ]γβ = 0 1  is not a aquare matrix
3 4
γ
So [T ]β is not invertible
∴ T is not invertible
(b) T : R2 → R3 , [T ]γβ ∈ M3×2 (F ) is not a square matrix
∴ T is not invertible
(c) T : R3 → R3 ,
T (1, 0, 0) 
= (3, 0, 3),
T (0, 1, 0) = (0, 1, 4), T (0, 0, 1) = (−2, 0, 0)
3 0 2
∴ [T ]γβ = 0 1 0 is invertible
3 4 0
∴ T is invertible
(cf ) If T (a1 , a2 , a3 ) = (3a1 − 2a2 , a2 , 3a1 + 4a2 ) = (0, 0, 0, )
then (a1 , a2 , a3 ) = (0, 0, 0)
∴ kerT = (0)
(d) T : P3 (R) → P2 (R), T (p(x)) = p0 (x)
[T ]γβ ∈ M3×4 (R) is not a square matrix
∴ T is not invertible
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2.4.
¶
a b
(e) T : M2×2 (R) → P2 (R), T
= a + 2bx + (c + d)x2
c d
[T ]γβ ∈ M3×4 (R) is not a square matrix
µ
∴ T is not invertible
µ
a b
c d
¶
(f) T : M2×2 (R) → M2×2 (R), T


1 1 0 0
1 0 0 0 

[T ]γβ = 
0 0 1 0 is invertible
0 0 1 1
∴ T is invertible
µ
¶
a+b
a
=T
c
c+d
µ
¶ µ
¶
a b
0 0
(cf ) If T
=
c d
0 0
then a = b = c = d = 0
∴ kerT = (0)
∴ T is invertible
3.
(a) F 3 P3 (F ) (∵)dimF 3 6= dimP3 (F )
(b) F 4 ∼
= P3 (F )
(c) M2×2 (R) ∼
= P3 (R)
(d) V = {A ∈ M2×2 (R) | tr(A) = 0} R4
µ
¶ µ
¶ µ
¶
1 0
0 1
0 0
(∵)dimV =| {
,
,
} |= 3
0 −1
0 0
1 0
dimR4 =| {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)} |= 4
∴ dimV 6= dimR4
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4.
Let A and B n × n invertible matrix
Since ∃A−1 and B −1 , (AB)(B −1 A−1 ) = In = (B −1 A−1 )(AB)
∴ B −1 A−1 is an inverse of AB
∴ AB is invertible and (AB)−1 = B −1 A−1
5.
Let A is invertible
Since ∃A−1 s.t AA−1 = A−1 A = In
(A−1 )t At = (AA−1 )t = (In )t = In
At (A−1 )t = (A−1 A)t = (In )t = In
∴ (A−1 )t is an inverse of At
∴ (At )−1 = (A−1 )t
6.
If A is invertible and AB=0, then ∃A−1 s.t A−1 A = AA−1 = In
∴ B = IB = (A−1 A)B = A−1 0 = 0
∴B=0
7.
(a) Suppose A2 = 0
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Assume A is invertible, then A−1 AA = A−1 0 ⇒ A = 0
It’s contradict to A is invertible
∴ A is not invertible
(b) Suppose AB=0 for some 0 6= B ∈ Mn×n (F )
Assume A is invertible, then A−1 AB = A−1 0 ⇒ B = 0
It’s contradict to B 6= 0
∴ A can’t be invertible
8. (a) [T ]ββ = [T ]β
(b) A = [LA ]β
(⇐) A = [LA ]β ∈ Mm×n (F ), where β the standard basis of F n
Since LA is invertible, so A is invertible
(⇒) Since A = [LA ]β ∈ Mm×n (F ) is invertible
∴ LA is invertible
9.
Let A, B ∈ Mn×n (F ) s.t AB is invertible
(a) Let LA , LB , LAB : F n → F n be the left multiplication
Then clearly LA LB = LAB and LAB is invertible since AB is invertible
∴ LA is onto, LB is one-to-one
By theorem 2.5, LA and LB are both invertible
∴ A and B are invertible
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2.4.

µ
¶
1 3
1 0 0
1 3


(b) Let A =
, B = −2 1 then AB =
is invertible
2 1 −1
−1 7
1 0
But A and B are not invertible
µ
(Example) (b)
µ
1 0
Let A =
0 1

1
but BA = 0
0

¶
0
0
0
1
0

µ
¶
1 0
1
0
= I,
, B = 0 1 then AB =
0 1
0 0

0
0 6= I
0
¶

10.
(a) AB = In ⇒ A and B are invertible by exercise 9
(b) ∃C ∈ M atn×n (F ) s.tCA = In , so
C = CIn = C(AB) = (CA)B = In B = B
that is, BA = CA = In = AB
∴ B = A−1 or A + B −1
(c) Let V be of finite dimensional vector space and let T : V → V s.t T R = IV
(1) T is invertible
T, R : V → V linear s.t. T R = 1V
Since T R = 1V , T is onto
So T is invertible
Similarly R is ono-to-one
and hence R is invertible
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(2) R = T −1
Since T is invertible, ∃ T −1 : V → V
⇒ T −1 = T −1 1V = T −1 (T R) = (T −1 T )R = IR = R
∴ R = T −1
µ
¶
f (1) f (2)
11. T : P (R) → M2×2 (R) is linear by T (f ) =
f (3) f (4)
We are going to show that T (f ) = 0 ⇒ f = 0(the zero polynomial)
3
In this case, f (1) = 0(i.e.f (c0 ) = b0 ), f (2) = 0, f (3) = 0, f (4) = 0
∴ ∀f (ci ) = 0, i = 0, 1, 2, 3
3
3
P
P
f (ci )fi (x) = 0
bi fi (x) =
∴ f (x) =
i=0
i=0
∴ f is the zero polynomial
∴ T is one-to-one
12. φβ (vi ) = [vi ]β = ei = (0, · · · , 1, · · · , 0)t
[φβ ]γβ = In , when γ is the standard basis for F n
∴ φβ is an isomorphism
or
φβ : V → F n is onto
⇒ φβ is an isomorphism because dimV = dimF n
13. ∼ is an equivalence relation on the class of vector space over F
(i) ∼ is reflexive
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2.4.
∀V ∈ C, V ∼ V
(∵) IV : V → V s.t IV = v, ∀v ∈ V is an isomorphism
(ii) ∼ is symmetric
If V ∼ W , then W ∼ V
(∵) If T : V → W is isomorphic then ∃T −1 : W → V is isomorphic
∴W ∼V
(iii) ∼ is transitive
If V ∼ W and W∼ Z, then V ∼ Z
(∵) Let T : V → W and U : W → Z are isomorphic, then U T is isomorphic
14.
¶
a a+b
| a, b, c ∈ F }
Let V = {
0
cµ
¶
a a+b
3
T : V → F s.t T
= (a, b, c)
0
cµ
¶
µ
¶
µ
¶
1 1
0 1
0 0
For the basis for V, {v1 =
, v2 =
, v3 =
}
0 0
0 0
0 1
∃!T (vi ) = wi is linear, wi ∈ F 3 i = 1, 2, 3
µ
and since dimV = dimF 3 , V is isomorphic to F 3
∴ T is an isomorphism from V to F 3
15.
T is isomorphic iff T (β) is a basis for W
(⇒) Section 2.1 exercise 14 (c) (p75)
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2.4.
(⇐) T (β) = {T (v1 ), T (v2 ), · · · , T (vn )}
∀w ∈ W
P
P
P
w=
ai T (vi ) = T ( ai vi ) = T (v), where v =
ai vi ∈ V
∴ T is onto
∴ T is invertible since dimV = dimW = n
16. Φ : Mn×n (F ) → Mn×n (F )
c ∈ Mm×n(F ) , ∃A = B −1 CB ∈ Mm×n (F ) s.t. Φ(A) = C
∴ Φ is onto
∴ Φ is an isomorphism
17. V is finite dimensional and T : V → V is isomorphic
let v0 ∈ V
(a) T (V0 ) ≤ W as a subspace
(i) T (v1 ) + T (v2 ) = T (v1 + v2 ) ∈ T (V0 )
(ii) T (av) = aT (v) ∈ T (V0 )
(b) Since T is an isomorphism, rank(T ) = dimW and nullity(T ) = 0
therefore nullity(T |V0 ) = 0
dim(V0 ) = rank(T |V0 ) + nullity(T |V0 ) = dim(T (V0 ))
,where T |V0 : V0 → T (V0 ) (⊆ W )
the restriction
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2.4.
18.


0 1 0 0
A = [T ]γβ = 0 0 2 0
0 0 0 3  
1

LA πβ (p(x)) = πγ T (p(x)) = 4
9
 
 

 1
1
0 1 0 0  
1  



LA πβ (p(x)) = 0 0 2 0   = 4
2
9
0 0 0 3
3
Since T (p(x)) = p0 (x) = 1 + 4x + 9x2
So LA πβ (p(x)) = πγ T (p(x))
19.


1 0 0 0
0 0 1 0 

(a) A = [T ]γβ = 
0 1 0 0 
0 0 0 1
   
1
1
2 3
  
(b) LA πβ (M ) = A 
3 = 2
4µ
¶4
1
3
Since T (M ) = M t =
2 4
We have πγ T (M ) = (1 3 2 4)t
20.
Let A = (aij )m×n be an incidence matrix associated with a dominance relation
Then A + A2 has a row[column] in which every entry, except for the orthogonal,
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2.4.
is positive
(∵) For convenience, let A + A2 = (bij )n×n
and for ∀i (1 ≤ i ≤ n), let D(i) = {j | aij = 1}
Choose a k such that D(k) is maximal in the set {D(i) | 1 ≤ i ≤ n}
n
P
aki aij > 0 for all j 6= k
We will show that bkj = akj +
i=1
For a fixed j(6= k), if akj = 1 then bkj > 0
Now suppose akj = 0, then aij = 1 ⇒ k ∈ D(j)
If (for the case akj = 0), aij = 0 for all i ∈ D(k), then
aij = 1 and hence D(k) ⊆ D(j)
But k ∈ D(j), k is not in D(k)
This is a contradiction to the choice of k
Thus aij = 1 for some i ∈ D(k) and
this proves the property bkj = akj +
n
P
aki aij > 0
i=1
21. {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} is a basis for dimL(V, W )
(a) Since dim L(V, W ) = mn,
we need to show that the given set is linearly independent
m P
n
P
aij Tij = 0, aij ∈ F
Let
i=1 j=1
∀k(1 ≤ k ≤ n),
m P
n
m
P
P
0=(
aij Tij )(vk ) =
aik wi
i=1 j=1
i=1
and ξ = {w1 , · · · , wn } is a basis for W
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2.4.
∴ aik = 0, ∀i, k
∴ ∀aij = 0
∴ It’s linearly independent
(b)
β = {v1 , v2 , · · · , vn }, γ = {w1 , w2 , · · · , wn }
{Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} a basis for L(V, W ) by (a)
∀k, Tij (vk ) wi = 0w1 + · · · + 1wi + · · · + 0wn if j = k
= 0w1 + · · · + 0wn otherwise
∴ [Tij ]γβ = M ij
(c) Now Φ : L(V, W ) → Mm×n (F ) is defined by Φ(Tij ) = M ij for all i, j
Since {M ij } = Φ(T ij ) is a basis of Mm×n (F )
Φ is a n isomorphism by the exercise 15(p.108)
22.
(i) T is well-defined and linear(check!)
T (f + g) = T (f ) + T (g)
T (αf ) = αT (f )
(ii) T is one-to-one
Suppose that f ∈ ker(T ) and let f (x) =
where fi (x) =
(x−c0 )···(x−cn )
,
(ci −c0 )···(ci −cn )
n
P
bi fi (x),
i=0
bi = f (ci ), ∀i
Then 0 = T (f ) = (f (c0 ), f (c1 ), · · · , f (cn )) = (b0 , b1 , · · · , bn )
Since ∀ bi = 0, f ≡ 0
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(iii) T is onto
∀a = (f (c0 ), · · · , f (cn )) ∈ F n+1 , ∃f ∈ Pn (F ) s.t T (f ) = a
23. T (σ) = T ({an }) = a0 + a1 x + · · · + an xn , where am 6= 0, ∀m > n
(i) T is well-defined
(ii) If T (σ) = 0, since x0i s are linearly independent
∴ ∀ai = 0
∴ σ = {0}
∴ T is one-to-one
(iii) T is onto
n
P
ai xi ∈ Pn (F ),
∀f (x) =
i=0
∃σ = {an } = {a1 , · · · , an , 0, · · · , 0} ∈ V s.t T (σ) = f (x)
24. (a) T is well-defined
v + N (T ) = v 0 + N (T )
⇒ v − v 0 ∈ N (T )
⇒ T (v − v 0 ) = 0
⇒ T (v) − T (v 0 ) = 0
∴ T (v) = T (v 0 )
⇒ T (v + N (T )) = T (v) = T (v 0 ) = T (v 0 + N (T ))
(b) T is linear
T (α(v+N (T ))+(v 0 +N (T ))) = T (αv+v 0 +N (T )) = T (αv+v 0 ) = αT (v)+T (v 0 ) =
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αT (v + N (T )) + T (v 0 + N (T ))
(c) T is an isomorphism
(i) If T (v + N (T )) = 0, then v + N (T ) = N (T ) i.e. v ∈ N (T )
(∵) Since 0 = T (v + N (T )) = T (v) ∴ v ∈ N (T )
T (v + N (T )) = 0 ⇒ 0 = T (v + N (T )) = T (V ) ⇒ v ∈ N (T )
∴ T is one-to-one
(ii) Clearly T (V + N (T )) ⊆ T (V )
If v ∈ T (V ), then v = T (u) for some u ∈ V
and so v = T (u) = T (u + N (T )))
∴ v ∈ T (V + N (T ))
∴ T is onto
(d) T = T η
T (v) = T (v + N (T )) = T (η(v)) = T η(v)
∀v ∈ V, T η(v) = T (v + N (T )) = T (v)
∴ T = Tη
25.
(i) Ψ : C(S, F ) → V is well-defined
(ii) Ψ is onto
P
∀v ∈ V. v =
ai si , si ∈ S
Define f (si ) = ai , ∀i
(then ai = 0 for all but a finite number of i)
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2.4.
then ∀v, ∃f ∈ C(S, F ) s.t. Ψ(f ) =
P
f (si )si =
P
ai si , si = V
(iii) Ψ is one-to-one
P
Ψ(f ) = 0 ⇒
f (si )si = 0, ∀si ∈ S ⇒ f ≡ 0
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2.5.
2.5. The change of Coordinate Matrix
1.
(a) F ([x0j ]β )
(b) T (Since Q = [IV ]ββ 0 , Q is an isomorphism)
(c) T
(d) F (B = Q−1 AQ)
(e) T ([T ]γ = Q−1 [T ]β Q)
2.Q = [IV ]ββ 0
(a) IV (a1 , a2 ) = (a1 , a2 ) = a1 e1 + a2 e2
IV (b1 , b2 ) = (b1 , b2 ) = b1 e1 + b2 e2
µ
¶
a1 b1
∴Q=
a2 b2¶
µ
4 1
(b) Q =
µ2 3 ¶
3 −1
(c) Q =
µ5 −2 ¶
2 −1
(d) Q =
5 4
3. Q = [IV ]ββ 0
(a) IV (a2 x2 + a1 x + a0 ) = a2 x2 + a1 x + a0
IV (b2 x2 + b1 x + b0 ) = b2 x2 + b1 x + b0
IV (c2 x2 + c1 x + c0 ) = c2 x2 + c1 x + c0
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2.5.


a2 b2 c2
∴ Q = a1 b1 c1 
a0 b0 c0 

a0 b0 c0

(b) Q = a1 b1 c1 
 a2 b2 c2 
0 −1 0

1
0 0
(c) Q =
−3 1 1 
2
1 1

3 −2 1
(d) Q =
 −1 3 1
5 −6 3

(e) Q = 0 4 −1
 3 −1 2
−2 1 2
(f) Q =  3 4 1
−1 5 2
4. [T ]β 0 = Q−1 [T ]β Q
¶
¶
µ
µ
2 1
1 1
, [T ]β =
Q=
1 −3
1 2
µ
[T ]β 0 =
2 −1
−1 1
¶µ
¶µ
¶ µ
¶
2 1
1 1
8 13
=
1 −3
1 2
−5 −9
5. [T ]β 0 = Q−1 [T ]β Q
µ
¶
µ
¶
1 1
0 1
Q=
, [T ]β =
1 −1
0 0
µ
1 1
[T ]β 0 = 1/2
1 −1
¶µ
0 1
0 0
¶µ
¶
µ
¶
1 1
1 −1
= 1/2
1 −1
1 −1
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2.5.
6. [LA ]β = Q−1 AQ
µ
¶
µ
¶
1 1
2 −1
−1
(a) Q =
,Q =
1 2
−1¶ µ1 ¶ µ
µ
¶ µ
¶
2 −1
1 3
1 1
6 11
−1
[LA ]β = Q AQ =
=
−2 4
µ
¶ −1 1 µ 1 1 ¶ 1 2
1 1
1 1
(b) Q =
, Q−1 = 1/2
1 −1
µ
¶1µ −1 ¶ µ
¶ µ
¶
1 1
1 2
1 1
3 0
−1
[LA ]β = Q AQ = 1/2
=
1
−1
2
1
1
−1
0 −1




1 1 1
1
1 −1
−1



1 −1 0 
(c) Q = 1 0 1 , Q =
1 1 2 
−1  
0
1

 
1
1 −1
1 1 −1
1 1 1
2
2
−1







1 −1 0
2 0 1
1 1 1 = −2 −3
[LA ]β = Q AQ =
−1
0
1
1
1
0
1
1



 1 1 2
1
1 1
1 1 −2
−1



1 −1 1 , Q = 6 3 −3 0 
(d) Q =
−2 0 
1
2 
 2 2


1 1 −2
13 1 4
1
1 1
6
−1







1 13 4
1 −1 1 = 24 0
[LA ]β = Q AQ = 6 3 −3 0
2 2
2
4 4 10
−2 0 1
0
7. In R2 , let L be the line y = mx, m 6= 0
(a) T is the reflection of R2 about L
¶
µ ¶ µ
1
−m
0
} : an ordered basis for R2
β ={
,
1
m
Since T (1, m)t = (1, m)t = 1(1, m)t + 0(−m, 1)t
T (−m, 1)t = (m, −1)t = 0(1, m)t + −1(−m, 1)t
µ
¶
1 0
∴ [T ]β 0 =
0 −1
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
2
−4
2

0 5
18 0 
−6 31
2.5.
Let β be the standard ordered basis for R2 and Q be the matrix that changes
β 0 -coordinates into β-coordinates
µ
¶
1 −m
β
and Q−1 [T ]β Q = [T ]β 0
Then Q = [IV ]β 0 =
mµ 1
¶µ
¶µ
¶
1 −m
1 0
1 m
−1
1/(1 + m2 )
∴ [T ]β = Q[T ]β 0 Q =
m
1
0
−1
−m
1
µ
¶
1 − m2
2m
2
= 1/(1 + m )
2m µ m2 − 1
µ ¶
¶µ ¶
µ
¶
x
1 − m2
2m
x
(1 − m2 )x + 2my
2
2
∴T
= 1/(1+m )
=1/(1+m )
2
y
2m
m
−
1
y
2mx + (m2 − 1)y
µ
¶
cos 2θ sin 2θ
(cf ) tan θ = m ⇒ T =
sin 2θ − cos 2θ
(b) T is the projection on L along the line perpendicular to L
µ ¶ µ
¶
1
−m
0
β ={
,
}
m
1
Since T (1, m)t = (1, m)t = 1(1, m)t ) + 0(−m, 1)t
T (−m, 1)t = (0, 0)t = 0(1, m)t ) + 0(−m, 1)t
¶
µ
1 0
∴ [T ]β 0 =
0 0
¶
µ
¶
¶µ
¶µ
µ
1 m
1 m
1 0
1 −m
2
2
−1
1/(1+m ) = 1/(1+m )
[T ]β = Q[T ]β 0 Q =
m m2
−m 1
0¶ 0
m
µ 1
µ ¶
x + my
x
=1/(1 + m2 )
∴T
2
y
µmx +2m y
¶
cos θ
sin θ cos θ
(cf ) tan θ = m ⇒ T =
sin θ cos θ
sin2 θ
8. Let V and W be finite-dimensional vector spaces,
T : V → W be linear
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β, β 0 be ordered bases for V and γ, γ 0 be ordered bases for W
0
Then [T ]γβ 0 = P −1 [T ]γβ Q
0
0
(∵) [T ]γβ 0 = [IW · T · IV ]γβ 0 = [IW ]γγ [T ]γβ [IV ]ββ 0 = P −1 [T ]γβ Q,
0
where P = [IW ]γγ 0 , Q = [IV ]ββ 0
9. (i) ∀A ∈ Mn×n (F ), A is similar to A
(∵)∃I s.t A = I −1 AI
(ii) If A is similar to B, then ∃Qs.t A = Q−1 BQ
∴ B = QAQ−1 = (Q−1 )−1 A(Q−1 )
∴ B is similar to A
(iii) If A is similar to B and B is similar to C
then ∃Q, P s.t A = Q−1 BQ, B = P −1 CP ∴ A = Q−1 BQ = Q−1 )(P −1 CP )Q =
(P C)−1 C(P Q)
∴ A is similar to C
∴ ”is similar to” is an equivalence relation on Mn×n (F )
10. Since A is similar to B, A = Q−1 BQ,
tr(A) = tr(QQ−1 B) = tr(B)
11.
(a) Let Q = [IV ]βα , R = [IV ]γβ
Then RQ = [IV ]γβ [IV ]βα = [IV ]γα
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(b) Q−1 = ([IV ]βα )−1 = [IV−1 ]αβ = [IV ]αβ
∴ Q−1 changes β 0 −coordinates into β−coordinates
12.
β : the standard ordered basis for F n
[LA ]γ = [IF n ]γβ [LA ]ββ [IF n ]βγ
Let [IF n ]βγ = Q, then [LA ]γ = Q−1 AQ
13. x0j =
n
P
Qij xi , j = 1, · · · , n
i=1
(1)
By the theorem 2.6(p.72), there is a unique linear operator T : V → V s.t.
T (xj ) = x0j for all j = 1, 2, · · · , n
Clearly [T ]ββ 0 = Q
Since Q is invertible, T is an isomorphism by the theorem 2.8
So β 0 = T (β) is a basis for V
(2)
Since x0j =
n
P
i=1
Qij xi (1 ≤ j ≤ n), [x0j ]β is the j-th column of Q
∴ Q = [IV ]ββ 0 changes β 0 -coordinates into β-coordinates
14.
If A, B ∈ Mm×n (F ) and P ∈ Mm×m (F ), Q ∈ Mn×n (F ) are invertible and
B = P −1 AQ,
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then ∃ an n-dimensional vector space V and an m-dimensional vector space W
(both over F), ordered bases β and β 0 for V and γ and γ 0 for W, and a linear
0
transformation T : V → W s.t A = [T ]γβ , B = [T ]γβ 0
(∵)
Let V = F n , W = F m , T = LA
β = {x1 , x2 , · · · , xn }, γ = {y1 , y2 , · · · , ym } : the standard ordered bases for F n
and F m , respectively
P
Define x0j =
Qij ej for 1 ≤ j ≤ n
then the set β 0 = {x01 , · · · , x0n } is a basis for V
and Q = [IF n ]ββ 0
P
Define wj0 =
Pij ei for 1 ≤ j ≤ m
0
then the set γ 0 = {w10 , · · · , wm
} is a basis for W
and P = [IF m ]γγ 0
Now [T ]γβ = [LA ]γβ = A and
0
[T ]γβ 0 = [IF m ]γγ [T ]γβ [IF n ]ββ 0 = P −1 AQ = B
0
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2.6. Dual Spaces
1.
(a) F
(linear transformation from V into its field of scalars F is called a linear functional)
(b) T
(f : F → F, [f ] ∈ M at1×1 (F ))
(c) T
dimV ∗ =dim(L(V, F ))=dimV dimF =dimV
∴V 'V∗
(d) T
For a vector space V , we can define the dual space of V i.e. (L(V, F )) = V ∗
Then V is the dual space of V ∗ ((V ∗ )∗ = V )
∴ Every vector space is the dual of some vector space
(e) (example) V = R2 , F = R
µ ¶
µ ¶
1
0
β = {e1 =
, e2 =
}:
0
1
So V ∗ = L(V, R) : 1 × 2 matrices
and e∗1 = (1, 0), e∗2 = (0, 1) i.e. β ∗ = {e∗1 = (1, 0), e∗2 = (0, 1)} Now if we define
T : V → V ∗ by T (e1 ) = (1, 1), T (e2 ) = (1, −1)
Since {T (e1 ), T (e2 )} = T (β) a basis of V ∗ , then clearly T is an isomorphism
But T (β) = {T (e1 ) = (1, 1) 6= e∗1 , T (e2 ) = (1, −1) 6= e∗2 } 6= β ∗
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(example 2)
V = R, F = R, f : V → F i.e.V ∗ = F and β ∗ = {1∗ } (∵ β = {1})
But T : R → R is an isomorphism
a 7→ 2a T (β) = {T (1)} = {2id} 6= β ∗ = {id}
(f) T
T : V → W, T t : W ∗ → V ∗ by T t (g) = gT
(T t )t : (V ∗ )∗ → (W ∗ )∗
(g) T
V ' W ⇔ T : V → W : an isomorphism ⇔ ∃[T ]γβ : invertible ⇔
∗
([T ]γβ )t = ([T t ]γβ ∗ ) : invertible ⇔ T t : W ∗ → V ∗ : an isomorphism ⇔ V ∗ ' W ∗
(h) F
f : Dn (R) → R by f (g(x)) = g 0 (x), ∀g(x) = Dn (R)
but in case g(x) = x2 , f (g(x)) = g 0 (x) = 2x is not in R
∴ It’s not a linear functional
2.
(a) p(x), g(x) ∈ P (R), α ∈ R
f (αp(x) + g(x)) = 2(αp0 (0) + g 0 (0) + αp00 (1) + g 00 (1)) = 2αp0 (0) + αp00 (0) + 2g 0 (0) +
g 00 (1) = αf (p(x)) + f (g(x))
∴ f is a linear functional
(b) (x1 , y1 ), (x2 , y2 ) ∈ R2 , α ∈ R
f (α(x1 , y1 ) + (x2 , y2 )) = f (αx1 + x2 , αy1 + y2 ) = 2(αx1 + x2 ) + 4(αy1 + y2 ) =
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α(2x1 + 4y1 ) + (2x2 + 4y2 ) = αf (x1 , y1 ) + f (x2 , y2 )
(c) A, B ∈ M2×2 (F ), α ∈ F
f (αA + B) = tr(αA + B) = tr(αA) + tr(B) = αtr(A) + tr(B) = αf (A) + f (B)
(d)
f ((x1 , y1 , z1 ) + (x2 , y2 , z2 )) = f (x1 + x2 , y1 + y2 + z1 + z2 ) = (x1 + x2 )2 + (y1 +
y2 )2 + (z1 + z2 )2 6= (x21 + y12 + z12 )2 + (x22 + y22 + z22 )2 = f (x1 , y1 , z1 ) + f (x2 , y2 , z2 )
∴ f is not a linear functional
(e)
f (αA + B) = αA11 + B11 = αf (A) + f (B)
∴ f is a linear functional
3.
(a) β = {v1 = (1, 0, 1), v2 = (1, 2, 1), v3 = (0, 0, 1)}
since fi (vj ) = δij
1 = f1 (v1 ) = f1 (e1 + e2 ) = f1 (e1 ) + f1 (e3 )
0 = f1 (v2 ) = f1 (e1 + 2e2 + e3 ) = f1 (e1 ) + 2f1 (e2 ) + f1 (e3 )
0 = f1 (v3 ) = f1 (e3 )
∴ f1 (e3 ) = 0, f1 (e1 ) = 1, f1 (e2 ) = −1/2
∴ f1 (x, y, z) = xf1 (e1 ) + yf1 (e2 ) + zf1 (e3 ) = x − 1/2y
0 = f2 (v1 ) = f2 (e1 ) + f2 (e3 )
1 = f2 (v2 ) = f2 (e1 + 2f2 (e2 ) + f2 (e3 )
0 = f2 (v3 ) = f2 (e3 )
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∴ f2 (e2 ) = f2 (e3 ) = 0, f2 (e1 ) = 0
∴ f2 (x, y, z) = 1/2y
0 = f3 (v1 ) = f3 (e1 ) + f3 (e3 )
0 = f3 (v2 ) = f3 (e1 + 2f3 (e2 ) + f3 (e3 )
1 = f3 (v3 ) = f3 (e3 )
∴ f3 (e1 ) = −1, f3 (e2 ) = 0, f3 (e3 ) = 1
∴ f3 (x, y, z) = −x + z
(b) β = {1, x, x2 }
f1 (a + bx + cx2 ) = af1 (e1 ) + bf1 (e2 ) + cf1 (e3 ) = a
f2 (a + bx + cx2 ) = af2 (e1 ) + bf2 (e2 ) + cf2 (e3 ) = b
f3 (a + bx + cx2 ) = af3 (e1 ) + bf3 (e2 ) + cf3 (e3 ) = c
4. {f1 , f2 , f3 } is linearly independent
(af1 + bf2 + cf3 )(x, y, z) = 0(x, y, z)
af1 (x, y, z) + bf2 (x, y, z) + cf3 (x, y, z)
= a(x − 2y) + b(x + y + z) + c(y − 3z)
= (a + b)x + (−2a + b + c)y + (b − 3c)z = 0, ∀(x, y, z) ∈ V
∴ a=b=c=0
∴ {f1 , f2 , f3 } is linearly independent in V ∗
5.
∗
(i) If Vcol → Vrow
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∗
Vrow → Vcol
µ ¶
¡
¢ c
a b
= 1 × 1(scalar)
d
µ ¶
a
V = P1 (R) = {a + bx or
}
µ ¶ b µ ¶
µ ¶
1
0
a
( ∵ In P1 (R), 1 =
,1 =
∴ a + bx ⇒
)
0
1
b
µ ¶
a
1
f1 (a + bx) = (1, 2 )
= a + 21 b
b
µ ¶
a
f2 (a + bx) = (2, 2)
= 2a + 2b
b
∴ af1 + bf2 = a(1, 21 ) + b(2, 2) = (a + 2b, 21 a + 2b) = 0 = (0, 0)
∴ a=b=0
7.
(a) T t (f ) = g, where g(a + bx) = −3a − 4b
¶
µ
∗
−1
1
β
(b) [T t ]γ ∗ =
µ −2 1¶
−1 −2
(c) [T ]γβ =
1
1
8.
Now let π a plane in R3 through the origin
Then ∃ 0 6= (a, b, c) ∈ R3 ⇒ π = {(x, y, z) | ax + by + cz = 0}
Define f : R3 → R by (x, y, z) 7→ ax + by + cz : linear functional
Then f ∈ V ∗ and kerf = π
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9.
β = {x1 , x2 , · · · , xn }
γ = {y1 , y2 , · · · , yn } : standard bases of F n and F m
10. {p0 = 1, p1 = (x − c1 ), · · · , pn = (x − c1 )(x − c2 ) · · · (x − cn )}
(a) Define fi (p(x)) = p(ci )
n
P
Suppose that
αi fi = 0
then
n
P
i=0
αi fi (pj (x)) =
i=0
n
P
αi (pj (ci )) = αi = 0
i=0
∴ αi = 0, ∀i = 0, 1, · · · , n
∴ {f0 , f1 , · · · , fn } : linearly independent
Since dimV = dimV ∗ = n + 1,
{f0 , f1 , · · · , fn } : a basis for V ∗
(b) (i) By the corollary of theorem 2.26 and (a),
xbi (fj ) = fj (xi ) = δij
then, fj (pi (x)) = pi (cj ) = δij
(ii) Consider ∃qi s.t fj (qi (x)) = qi (cj ) = δij , ∀i = 0, 1, · · · , n,
Let Ri (x) = pi (x) − qi (x), then Ri (cj ) = pi (cj ) − qi (cj ) = δij − δij = 0, ∀j =
0, 1, · · · , n
∴ Ri (x) = 0
(∵) dimRi ≤ n and Ri (x) has n + 1 roots
∴ qi = p i
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2.6.
(c) Assume that ∃h(x) ∈ Pn (x) s.t. h(ci ) = ai , ∀i
Since {P0 (x), P1 (x), · · · , Pn (x)} a basis for Pn (x)
n
P
∴ h(x) =
bi Pi (x), (∀bi ∈ F )
i=0
aj = h(cj ) =
∴ h(x) =
n
P
n
P
bi Pi (cj ) = bj , ∀j
i=0
bj Pj (x) =
n
P
ai Pj (x) = q(x)
j=0
j=0
(d) Let c0 , · · · , cn be distinct scalars in F
The polynomials p0 (x), · · · , pn (x) defined by
pi (x) =
Y x − ck
∈ Pn (F )
ci − ck
k=0
(1)
k6=i
Since pi (cj ) = δij and {p0 , · · · , pn } is linearly independent
{p0 , · · · , pn } : a basis for Pn (F )
n
P
aipi (x), (ai ∈ F )
p(x) =
i=0
p(cj ) = aj , ∀j
n
n
P
P
p(ci )pi (x)
aipi (x) = p(x) =
∴ p(x) =
(e)
Rb
a
i=0
p(t)dt =
n
P
p(ci )di ,
i=0
i=0
Rb
di = a
pi (t)dt
(∵)p(t) = p(c0 )p0 (t) + · · · + p(cn )pn (t)
Rb
Rb
p(t)dt = a (p(c0 )p0 (t) + · · · + p(cn )pn (t))dt
a
Rb
Rb
= p(c0 ) a p0 (t)dt + · · · + p(cn ) a pn (t)dt
= p(c0 )d0 + · · · + p(cn )dn
n
P
=
p(ci )di
i=0
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Rb
(a)
Trapezoidal rule - a p(t)dt ≈ (b − a) f (b)+f
2
Rb
Simpson’s rule - a f (t)dt ≈ (b−a)
(f (a) + 4f ( a+b
) + f (b))
6
2
11.
[
For ∀x ∈ V, x 7→ ψ2 T (x) = T
(x)
x 7→ T tt ψ1 (x) = T tt (b
x) = x
bT t
[
To show commuting, T
(x) = x
bT t in W ∗∗ : W ∗ → F
∀g ∈ W ∗ , i.e. g : W → F a linear functional
[
Show (b
xT t )(g) = T
(x)(g)
[
T
(x)(g) = g(T (x)) = (gT )(x) = widehatx(gT ) = x
b(T t g) = x
bT t (g)
∴ ψ2 T = T tt ψ1
12. ψ(β) = β ∗∗
β = {x1 , x2 , · · · , xn } a basis for V
⇒ β ∗ = {x∗1 , x∗2 , · · · , x∗n } a basis for V ∗
, where x∗i : V → F a linear functional s.t. fi (xj ) = δij
Since V ∗∗ is the dual space of V ∗
∗∗
∗∗
∗∗
∃ β ∗∗ = {x∗∗
1 , x2 , · · · , xn } a basis for V
∗
∗
∗∗ ∗
s.t. x∗∗
i : V → F linear functional s.t. xi (xj ) = δij
bi , ∀i, ψ(β) = {x1 , x2 , · · · , xn }
Show x∗∗
i = x
∗
x
b(xi∗ ) = δij = x∗∗
i (xj ), ∀i, j
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∗∗
i.e.xbi = x∗∗
i on a basis β
⇒ xbi = x∗∗
i , ∀i
∴ ψ(β) = β ∗∗
13. S 0 = {f ∈ V ∗ |f (x) = 0, ∀x ∈ S}, S ⊆ V
(a) S 0 is a subspace of V ∗
(i) 0(x) = 0, ∀x ∈ S ∴ 0 ∈ S0
(ii) ∀f, g ∈ S 0 , α ∈ F ⇒ αf + g ∈ S 0
(αf + g)(x) = αf (x) + g(x) = α0 + 0 = 0, ∀x ∈ S
∴ αf + g ∈ S 0
(b)
Let {x1 , x2 , · · · , xm } be a basis of W
If x is not in W , then {x1 , x2 , · · · , xm , x} = {x1 , x2 , · · · , xm } ∪ {x} is linearly
independent
∃ β = {x1 , x2 , · · · , xm , x = xm+1 , · · · , xn } a basis of V
⇒ β ∗ = {f1 , f2 , · · · , fm , fm+1 , · · · , fn } a basis of V ∗
fm+1 (W ) = 0 and fm+1 (x) = fm+1 (xm+1 ) = 1
∴ fm+1 ∈ W 0 , fm+1 (x) 6= 0
(c) (S 0 )0 =spanψ(S)
(⇐) If vb ∈spanψ(S), then ∃x1 , x2 , · · · , xn ∈ S, s.t. vb = a1 xb1 + a2 xb2 + · · · + an x
cn
for some a1 , a2 , · · · , an ∈ F
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Now ∀f ∈ S 0
n
n
n
P
P
P
vb(f ) = ( ai xbi )(f ) =
ai xbi (f ) = ai f (xi ) =
ai 0 = 0
i=1
So vb ∈ (S 0 )0
i=1
i=1
∴ spanψ(S) ⊆ (S 0 )0
(⇒)
(Step 1)
First note that for ∀ subset S of V
S 0 = (spanS)0
d For convenience, let spanS = W ,
then clearly S ⊆ W, so W 0 ⊆ S 0
Now let f ∈ S 0 and x ∈ W ; then
x = a1 x1 + a2 x2 + · · · + an xn for some xi ∈ S, ai ∈ F
⇒ f (x) = a1 f (x1 ) + a2 f (x2 ) + · · · + an f (xn ) = 0
∴ f ∈ W0
∴ S0 ⊆ W 0 c
(Step 2)
Show if W ≤ V as a subspace, then (W 0 )0 = ψ(W )
d (⊆) Forv ∈ V, vb ∈ W 0 )0 ⇒ vb(f ) = 0 for all f ∈ W 0
⇒ f (V ) = 0
⇒ v ∈ W by the exercise 13(b)
⇒ vb ∈ ψ(W )
(⊇) If v ∈ W, vb ∈ ψ(W )
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vb(f ) = f (v) = 0 for all f ∈ W 0
⇒ vb ∈ (W 0 )0
This implies that ψ(W ) ⊆ (W 0 )0
(Step 3)
Clearly ψ(W ) = spanψ(S), where W = span(S)
(∵) If v ∈ W
v = a1 x1 + a2 x2 + · · · + an xn (∀ai ∈ F, xi ∈ S)
n
P
then ψ(v) = vb = a1 xb1 + a2 xb2 + · · · + an x
cn =
Ai ψi (xi ) ∈ span(ψ(S))
i=1
∴ vb ∈ ψ(S)
vb ∈ span(ψ(S))
Now ψ(S) ⊆ ψ(W ) and ψ(W ) is a subspace of W ∗∗
span(ψ(S)) ⊆ span(ψ(W )) = ψ(W )
∴ ψ(W ) = span(ψ(S))
Finally (S 0 )0 = (W 0 )0 = ψ(W ) = span(ψ(S))
(d) W1 = W2 ⇔ W10 = W20
(⇒) clear
(⇐) If W1 6= W2 , then x ∈ W2 , x is not in W1
by (b), ∃ f ∈ W10 , f (x) 6= 0
i.e. f is not in W20
∴ W10 6= W20
(e)
(⇒)
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(W1 + W2 )0 ⊆ W10 , (W1 + W2 )0 ⊆ W20
∴ (W1 + W2 )0 ⊆ W10 ∩ W20
(⇐) clear
14.
Let dimW = k and {x1 , · · · , xk } : a basis for W
Extend it to {x1 , · · · , xk , xk+1 , · · · , xn } : a basis for V
Let {f1 , · · · , fn } be the basis for V ∗
We are going to show that {fk+1 , · · · , fn } is a basis for W 0
If f ∈ W 0 we have f (xi ) = 0, i ≤ k
n
P
∴ f=
f (xi )fi
i=k+1
∴ {fk+1 , · · · , fn } spans W 0
∴ Since dimW = k and dimV = n, then dimW 0 = n − k
15.
(⇒) Suppose that φ ∈ N (T t )
i.e. T t (φ) = φT = 0
If u ∈ R(T ),then u = T (v) for some v ∈ V
hence φ(u) = φ(T (v)) = (φT )(v) = 0(v) = 0, ∀u ∈ R(T )
∴ φ ∈ (R(T ))0
(⇐)
If σ ∈ (R(T ))0 , σ(R(T )) = 0
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then, ∀v ∈ V, (T t (σ))(v) = (σT )(v) = σ(T (v)) = 0(v) = 0
∴ T t (σ) = 0
∴ σ ∈ N (T t )
16.
rank(LAt ) = rank(LA )
Let dimV = n, dimW = m and rank(T ) = r
by the exercise 14, dim(R(T )) + dim(R(T ))◦ = dimW
∴ dim(R(T ))◦ = m − r
by the exercise 15 and the dimension theorem,
N (T t ) = (R(T ))◦ , dim(W ∗ ) = nullity(T t ) + rank(T t )
∴ rank(T t ) = dim(W ∗ ) − nullity(T t ) = m − (m − r) = rank(T ) − − − −(∗)
Since rank(T ) = rank(LA ) and rank(T t ) = rank(LAt ) − − − −(∗)
rank(LAt ) = rank(LA )
Q. rank(T t ) = rank(LAt )
(∵) When T = LA : F n → F m left multiplication
∗
rank(LA ) = rank(T ) = rank(T t ) = rank([T t ]βγ ∗ )
= rank([T ]γβ )t = rankAt = rankLAt
17.
(⇒) ∀ ∈ W ◦
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Since T t f = f T and T (W ) ⊆ W,
f T (W ) ⊆ f (W ) = 0
∴ T tf ∈ W ◦
∴ T t (W ◦ ) ⊆ W ◦
∴ W ◦ is T t -invariant
(⇐) If T (W ) * W, ∃ w ∈ W s.t. T (w) ∈ W
by the exercise 13, ∃f ∈ W ◦ s.t. f (T (w)) 6= 0
T t f (w) = f T (w) 6= 0
i.e. ∃f ∈ W ◦ s.t. T t f ∈
/ W◦
∴ W is T -invariant
18.
Φ : V ∗ → L(S, F )
(Actually L(S, F ) ≡ L(V, F ))
(i) Clearly L(S, F ) is a vector space over F and
(ii) Φ is a linear map
d f, g ∈ L(S, F ),
Φ(f + g) = (f + g) |s = f |s +g |s = Φ(f ) + Φ(g)
Φ(αf ) = αf |s = αfs = αΦ(f ) c
(iii) ∀f ∈ kerΦ, Φ(f ) = fs = 0
∴ f ∈ S ◦ = (spanS)◦ = V ◦ = {0}
∴ kerΦ = {0}
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∴ Φ is one-to-one
(iv) By the exercise 34 in section 2.1
(∀fs : S → F, ∃! f : V → F a linear map s.t. f (x) = fs (x), ∀x ∈ S)
∀ fs ∈ L(S, F ), ∃! f ∈ V ∗ s.t. Φ(f ) = fs
∴ Φ is onto
19.
(i) Choose y ∈ V, y ∈
/ W and let γ : a basis of W
then γ ∪ {y}: linearly independent
by (section 1.7 or) Maximal principle,
∃ β : a basis of V s.t. γ ∪ {y} ⊆ β
Define a function : g : β → F s.t. g(x) = 0 ∀x ∈ β, x 6= y
g(y) = 1
then by the exercise 18, ∃ f ∈ L(V, F ) = V ∗ s.t. f |β = g(∵ γ ⊆ β)
i.e. f is the function we desired.
20.
(a) T : V → W : linear map ⇒ T t : W ∗ → V ∗ given by T t (g) = gT, ∀g ∈ W ∗
(⇒) Suppose that T is onto
Let g ∈ KerT t ⇒ T t (g) = 0 ⇒ gT = 0 in V ∗
i.e. gT (v) = (gT )(v) = 0, ∀v ∈ V
Since T is onto, g(w) = gT (v) = 0
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i.e. g(w) = 0(∵ g ∈ KerT t ), ∀w ∈ W
∴ g = 0 in V ∗
∴ T t is one-to-one
(⇐) Suppose that T t is one-to-one
Let W1 = R(T ) : the range of T
If W1 6= W ⇒ by the exercise 19, ∃0 6= g ∈ W ∗ , g(W1 ) = 0 (i.e. g ∈ W ◦ 1)
⇒ (gT )(V ) = g(T (V )) = g(W1 ) = 0
⇒ (T t )g(V ) = 0 ⇒ T t g = 0 in W ∗
∴ g ∈ KerT t
Since T t is one-to-one, g = 0
∴ W1 = W
∴ T is onto
(b)
(⇐) Suppose that T is one-to-one
Let f ∈ V ∗
Suppose that T is one-to-one
W = W1 ⊕ W2 , where W1 , W2 ≤ W and W1 = R(T ) ∼
=V
So the map U : W1 ⊕ W2 → V, U (w1 ⊕ w2 ) = v, where w1 = T (v) is a well-defined
linear map
T (v) + w2 7→ v
Let g = f U , then ∃ g ∈ W ∗ and T t (g) = gT = f
d(∵) (gT )(v) = g(T (v)) = (f U )(T (v)) = f (U (T (v))) = f (v), ∀v ∈ V
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∴ gT = f
i.e. T t g = f c
∴ T t is onto
(⇒) Suppose that T t : W ∗ → V ∗ onto
show T is one-to-one
Assume on the contrary T is not one-to-one
⇒ ∃0 6= v ∈ V s.t. T (v) = 0
∃ f ∈ V ∗ s.t. f (v) = 1 (by the exercise 18)
Now since T t is onto
∃g ∈ W ∗ s.t. f = T t (g) = gT
⇒ 1 = f (v) = (gT )(v) = g(T (v)) = g(0) = 0
∴ T must be one-to-one
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2.7.
2.7. Homogeneous Linear Differential Equations with Constant Coefficients
1.
(a) T (p.137 corollary to Theorem 2.32)
(b) T (p.132 Theorem 2.28)
(c) F
(d) F (Any solution is a linear combination of eat and tk eat )
(e) T
(∵) If x and y are solutions of p(D) = 0,
then p(D)(αx + βy) = αp(D)x + βp(D)y = 0 + 0 = 0, α, β ∈ F
∴ αx + βy is a solution of p(D) = 0
(f) F
(∵) It’s different with the multiplicity of ci (p.137 and 139, Theorem 2.33 and
2.34) (g) T (p.131)
2.
(a) F
a
∞
Let S = { 1+t
2 | a ∈ R} ⇒ S: 1-dimensional subspace of C
But there is no homogeneous linear differential equation with constant coefficients
(b) F
Let {t, t2 } is the solution of y 00 + ay 0 + by = 0
0 + a + bt = 0 ⇒ a = b = 0
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then y 00 + ay 0 + by = 0 becomes y 00 = 0
(t2 )00 = 2 = 0
(cf) y 000 = 0 ⇒ D3 = 0 ⇒ t = 0
e0t , te0t , t2 e0t i.e.1, t, t2
∃ y 000 = 0
(c) T
Let x is a solution to the homogeneous linear differential equation with constant
coefficients P (D)y = 0
Since P (D)x = 0, P (D)x0 = P (D)(Dx) = P (D)Dx = DP (D)x = D(0) = 0
∴ x0 is also a solution to the equation
(d)T
Let p(D)x = 0 and q(D)y = 0
p(D)q(D)(x + y) = p(D)q(D)x + p(D)q(D)y = q(D)(p(D)x) + p(D)(q(D)y) =
q(D)(0) + p(D)(0) = 0 + 0 = 0
(e) F
Let p(t) = t2 + 2t + 1 = 0 ∴ p(D) = D2 + 2D + 1
q(t) = t3 − 1 ∴ q(D) = D3 − 1
{e−t } : a basis for the solution space of p(D)
{et } : a basis for the solution space of q(D)
p(D)q(D) = D5 + 2D4 + D3 − D2 − 2D − 1
p(D)q(D)y = y (5) + 2y (4) + y (3) − y (2) − 2y 0 − y
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p(D)(e−t ) = 0, q(D)(et ) = 0
But p(D)q(D)(e−t et ) 6= 0
3.
(a) Given the differential equation is y 00 + 2y 0 + y = 0 and its auxiliary polynomial
is p(t) = t2 + 2t + 1 = (t + 1)2
Hence, e−t and te−t are solutions to the differential equation because c = −1 is a
zero of p(t)
∴ {e−t , −te−t } is a basis for the solution space
So any solution y is to the given differential equation is of the form
y(t) = b1 e−t + b2 te−t for unique b1 and b2
(b)
Since y 000 = y 0 , the auxiliary polynomial is t3 − t = 0
∴ t = 0, −1, 1
∴ {1, e−t , et }
(c) y (4) − 2y (2) + y = 0,
the auxiliary polynomial is t4 − 2t2 + 1 = (t2 − 1)2 = (t + 1)2 (t − 1)2
∴ t = −1, 1
∴ {e−t , te−t , et , tet }
(d)=(a)
(e)
Since y (3) − y (2) + 3y (1) + 5y = 0,
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the auxiliary polynomial is t3 − t2 + 3t + 5 = (t1 )(t2 − 2t + 5)0
∴ t = −1, 1 + 2i, 1 − 2i
∴ {e−t , et e2it , et e−2it }
∴ {e−t , et cos2t, et sin2t}
4.
(a)
p(t) = t2 − t − 1 = 0
t=
√
√
1+ 5 1− 5
, 2
2
∴ {e
√
(1+ 5)t
2
,e
√
(1− 5)t
2
}
(b)
p(t) = t3 − 3t2 + 3t − 1 = (t − 1)3 = 0
∴ {et , tet , t2 et }
(c)
p(t) = t3 + 6t2 + 8t = t(t2 + 6t + 8) = 0
∴ t = 0, −2, −4
∴ {1, e−2t , e−4t }
5.
∀f, g ∈ C ∞ , α ∈ F,
(αf + g)(n) = αf (n) + g (n) ∈ C ∞ , ∀n
∴ C ∞ is a subspace of F(R, C)
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6.
(a)
∀f, g ∈ C ∞ , α ∈ F,
D(αf + g) = (αf + g)0 = αf 0 + g 0 = αD(f ) + D(g)
∴ D : C ∞ → C ∞ is a linear operator
(b)
∀f, g ∈ C ∞ , α ∈ F,
Define L = p(D) = an Dn + an−1 Dn−1 + · · · + a1 D1 + a0 I
L(αf + g) = an (αf + g)n + an−1 (αf + g)n−1 + · · · + a0 I
= an (αf (n) + g (n) ) + an−1 (αf (n−1) + g (n−1) ) + · · · + a0 I
= α(an f (n) + an−1 f (n−1) + · · · + a1 f 0 + a0 ) + (an g (n) + an−1 g (n−1) + · · · + a1 g 0 + a0 )
= αL(f ) + L(g)
∴ Any differential operator is a linear operator on C ∞
(cf) If D : a linear operator,
then DD : a linear operator
..
.
Dn : a linear operator
∴ p(D = Dn + an−1 Dn−1 + cdots + a1 D + a0 I) : a linear operator
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2.7.
7.
We only need to show that { x+y
, x−y
} is linearly independent
2
2i
If α (x+y)
+ β (x−y)
= 0, (α, β ∈ F )
2
2i
then ( α2 +
⇒
α
2
+
β
2i
β
)x
2i
= 0,
β
)y
2i
+ ( α2 −
α
2
−
β
2i
=0
=0
∴α=β=0
∴ { x+y
, x−y
} is a basis
2
2i
8.
Let t1 = a + ib(a, b ∈ R, b 6= 0) is a zero of p(t)
then, e(a+ib)t is a solution of p(D)
Using exercise 7 in this chapter,
If {e(a+ib)t , e(a−ib)t } is a basis, then so is { 12 (e(a+ib)t + e(a−ib)t ), 2i1 (e(a+ib)t − e(a−ib)t )}
Since 21 (e(a+ib)t + e(a−ib)t ) = eat cosbt, 2i1 (e(a+ib)t − e(a−ib)t ) = eat sinbt
∴ {eat cosbt, eat sinbt} is a basis
(cf) Let p(t) = α(t2 − 2at + a2 + b2 ), t1 = a + ib, t2 = a − ib
then p(D)y = y 00 − 2ay 0 + (a2 + b2 )y = 0, y1 = eat cosbt, y2 = eat sinbt
check p(D)y1 = 0, and p(D)y2 = 0
9.
u ∈ N (Ui ) ⇒ Ui (u) = 0
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So U1 · · · Un (u) = U1 · · · Ui−1 Ui+1 · · · Un Ui (u) = U1 · · · Ui−1 Ui+1 · · · Un (0) = 0
∴ u ∈ N (U1 · · · Un )
10.
Suppose that b1 ec1 t + · · · + bn ecn t = 0, ci ’s are distinct
Apply the mathematical induction on n,
If n = 1, then b1 ec1 t = 0 ∴ b1 = 0
Assume that this assertion is true for n − 1
We are going to prove that this is also true for n
(D − cn I)(b1 ec1 t + · · · + bn ecn t ) = 0
⇒ b1 c1 ec1 t +· · ·+bn−1 cn−1 ecn−1 t +bn cn ecn t −(b1 cn ec1 t +· · ·+bn−1 cn ecn−1 t +bn cn ecn t ) =
0
⇒ b1 (c1 − cn )ec1 t + b2 (c2 − cn )ec2 t + · · · + bn−1 (cn−1 − cn )ecn−1 t = 0
By the induction hypothesis and ci − cn 6= 0,
∀i = 1, · · · , n − 1, bi = 0
∴ {ec1 t , · · · , ecn t } is linearly independent
Since the solution space is n-dimensional, the given set is a basis for the solution
space of the differential equation
11.
Suppose that
k nP
i −1
P
cij tj eci t = 0
i=1 j=0
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Let Pi (t) =
nP
i −1
cij tj
j=0
Then we have P1 (t)ec1 t + P2 (t)ec2 t + · · · + Pk (t)eck t = 0
Assume that not all cij are zero, then ∃Pi 6= 0
Say, Pk
Divide the equation by ec1 t
P1 (t) + P2 (t)e(c2 −c1 )t + · · · + Pk (t)e(ck −c1 )t = 0 · · · (1)
Upon differentiating (1) sufficiently many times we can reduces P1 (t) to 0
Q2 (t)e(c2 −c1 )t + · · · + Qk (t)e(ck −c1 )t = 0, and degQi =degPi
and Qk does not vanish identically
Continuing this process, Rk (t)e(ck −c1 )t = 0, and degRk =degPk
and Rk does not vanish identically
But Rk (t)e(ck −c1 )t = 0 implies Rk = 0
It’s a contradiction to Pk 6= 0
∴ Pk (t) = 0, ∀x ∈ I
∴ All cij ’s are zero
12.
(i) g(D)(V ) ⊆ N (h(D))
(ii) dim N (h(D)) = dim g(D)(V )
Suppose degg(t) = k, deg(h(t)) = m (n = k + m)
Consider the linear map g(DV ) : V → V
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By the dimension theorem,
dim V = dim R(g(DV )) + dim N (g(DV ))
= dim R(g(DV )) + dim N (g(D)) (∵ N (g(D)) ⊆ V )
= dim(g(D)(V )) + dim N (g(D)) (∵ R(g(DV )) = g(D)(V ))
= dim g(D)(V ) + k
∴ dim g(D)(V ) = n − k = m = dim N (h(D))
13.
(a) Ontoness of P (D) : C ∞ → C ∞
Since C is algebraically closed
P (D) = α(D − c1 )(D − c2 ) · · · (D − cn ), where α 6= 0, c1 , · · · , c1 ∈ C
Let v ∈ C ∞ : by lemma 1, ∃ u1 ∈ C ∞ s.t.(D − c1 )u1 = v
and ∃ u2 ∈ C ∞ (D − c2 )u2 = u1 · · · continuing this process,
we get u1 , u2 , · · · , un ∈ C ∞ s.t. (D − ci )ui = ui−1 (2 ≤ i ≤ n)
Put u = α1 un : P (D)u = α(D − c1 )(D − c2 ) · · · (D − cn−1 )(D − cn )(u)
= (D − c1 )(D − c2 ) · · · (D − cn−1 )(D − cn )un = v
14.
By induction on n, p(t): a polynomial of degree n(≥ 1)
A solution x(t) of p(D)y = 0 − − − (∗)
We may assume w.l.o.g that p(t) monic
For n = 1, (∗) becomes y 0 − ay = 0 ⇒ x(t) = ceat (c ∈ C)
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if x(t0 ) = 0 ⇒ ceat0 = 0 ⇒ c = 0 ∴ x(t) = 0
Assume it is true for n − 1(n > 1) and degp(t) = n
This case p(t) = q(t)(t − c), q(t) of degree n − 1, c ∈ C
let z = q(D)x
then by (∗), we have (D − cz = (D − c)q((D)))x = p(D)x = 0
∴ zis a solution to (D − c)y = 0
By hypothesis, x(t0 ) = x0 (t0 ) = · · · = xn−2 (t0 ) = xn−1 (t0 ) = 0 for fixed t0 ∈ R
⇒ ∀t ∈ R, z(t) = x(n−1) (t) + an−1 x(n−2) (t) + · · · + a1 x0 (t) + a0 x(t) ⇒ z(t0 ) = 0
⇒ z(t0 ) = 0 ⇒ z 0 (t0 ) = 0 (∵ z 0 (t0 ) − cz(t0 ) = 0, z 0 (t0 ) = 0)
⇒ z(t) = 0, ∀t
⇒ q(D)x = z = 0
15.
Φ : V → C n , Φ(x) = (x(t0 ), x0 (t0 ), · · · , xn−1 (t0 ))T , ∀x ∈ V
(a)
(i) Φ is linear
x, y ∈ V, α ∈
F
 

(x + y)(t0 )
x(t0 ) + y(t0 )
 (x + y)0 (t0 )   x0 (t0 ) + y 0 (t0 ) 
=
 = Φ(x) + Φ(y)
Φ(x + y) = 

 

···
···
n−1
n−1
n−1
(x + y) (t0 )
x (t0 ) + y (t0 )




(αx)(t0 )
x(t0 )
 (αx)0 (t0 ) 
 0

 = α  x (t0 )  = αΦ(x)
Φ(αx) = 


 ··· 
···
n−1
(αx) (t0 )
xn−1 (t0 )
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(ii) Φ(x) = 0 ⇒ x = {0}
If x(t0 ) = x0 (t0 ) = · · · = xn−1 (t0 ) = 0, then x = 0
Since dimC V = dimC Cn = n, Φ is an isomorphism
n
(iii) Since dimVC = dimCC = n, Φ is onto
(b)
Since Φ is an isomorphism,
Let c = (c0 , c1 , · · · , cn−1 ) ∈ C n s.t x(t0 ) = c0 and xk (t0 ) = ck , k = 1, · · · , n − 1
By (a),∃!x ∈ V
16.
00
(a) θ + gl θ = 0
t2 +
g
l
= 0, t1 =
pg
pg
i,
t
=
−
i
2
l
l
∴ θ = c1 cos gl t + c2 sin gl t
(b) θ(0) = θ0 > 0, θ0 (0) = 0
θ(0) = c1 = θ0
p
θ0 (0) = c2 gl = 0
∴ c2 = 0
p
∴ θ = θ0 cos gl t
2π
(c) The period of the system is √
g = 2π
l
pg
l
17.
00
y +
k
y
m
=0
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q
q
k
t
m
y(t) = c1 cos
+ ic2 sin
k
t
m
18.
00
(a) my + ry 0 + ky = 0, r > 0
Since the auxiliary polynomial is p(t) = mt2 + rt + k = 0,
∴ t1 =
√
−r+ r2 −4km
,
2m
t2 =
√
−r− r2 −4km
2m
∴ y(t) = c1 et1 t + c2 et2 t
(b)
y(0) = c1 + c2 = 0 ∴ c1 = −c2
y 0 (0) = c1 t1 + c2 t2 = v0
c2 =
v0
, c1
t2 −t1
=
−v0
t2 −t1
0
0
∴ y(t) = ( t2−v
)et1 t + ( t2v−t
)et2 t , (t2 − t1 =
−t1
1
(c)
√
r2 −4km
)
m
√
t1 t
y(t) = c1 e
t→∞⇒e
t2
+ c2 e t = e
−r
t
2m
−r
t
2m
(c1 e
r 2 −4km
t
2m
√
−
+ c2 e
r 2 −4km
t
2m
)
→0
∴ limt→∞ y(t) = 0
19.
∵ C is algebraically closed
20.
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2.7.
(a) Theorem 2.27
If n = 1, then x0 + a0 x = 0 ⇒ x0 = −a0 x
00
Since x has a derivative x0 , x0 must has a derivative x = −a0 x0
Assume that this assertion is true for an n − 1th-order homogeneous linear differential equation with constant coefficients
Then x(n) = x(n−1) x = −a0 x(n−2) x = −a0 x(n−1)
So x(k) exists for every positive integer k
(b)
y1 = ec+t , y2 = ec et (for ∀c ∈ R)
(i) Let x = (ec+t − ec et ) ⇒ x0 = ec+t − ec et = x
So x is a solution to the equation y 0 − y = 0 with x(0) = 0
So by the exercise 14, x(t) ≡ 0 ∀t
∴ ec+t = ec et putting t = d ∈ R
ec+d = ec ed
(ii) ec e−c = e0 = 1 = ec e1c
ec (e−c −
1
)
ec
∴ e−c =
1
ec
= 0 ⇒ ec 6= 0
Since C is algebraically closed
(c) Theorem 2.28
Any homogeneous linear differential equation with constant coefficients can be
rewritten as P (D)y = 0, where p(t) is the auxiliary polynomial associated with
the equation.
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Therefore the set of all solutions to a homogeneous linear differential equation
with constant coefficients coincides with the null space of P (D), where p(t) is the
auxiliary polynomial associated with the equation.
(d) Theorem 2.29
f (t) = ect = ea+ibt = eat (cosbt + isinbt) = eat cosbt + ieat sinbt
f 0 (t) = aeat cosbt − beat sinbt + iaeat sinbt + ibeat cosbt
= (a + ib)eat cosbt + i(a + ib)eat sinbt
= (a + ib){eat (cosbt + isinbt)}
= cect
(e)
(xy)0 = (u1 u2 + i(u1 v2 + u2 v1 ) − v1 v2 )0
= {(u1 u2 − v1 v2 ) + i(u1 v2 + u2 v1 )}0
= (u01 u2 + u1 u02 − v10 v2 − v1 v20 ) + i(u01 v2 + u1 v20 + u02 v1 + u2 v10 )
= {u01 u2 + i(u01 v2 + u2 v10 ) − v10 v2 } + {u1 u02 + i(u1 v20 + u02 v1 ) − v1 v20 }
= (u01 + iv10 )(u2 + iv2 ) + (u1 + iv1 )(u02 + iv20 )
= x0 y + xy 0
(f)
Let x = u + iv
x0 = u0 + iv 0 = 0
∴ u0 = 0, v 0 = 0
∴ u, v is a constant function
∴ x = u + iv is a constant function
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3.1.
§3.
Elementary Matrix Operations and Systems
of Linear Equations
3.1. Elementary Matrix Operations and Elementary Matrices
1. (a) T
(b) F


1 0 −2
(∵) I3 Ã 0 1 0  , by −2 × C1 + C3 ⇒ C3
0 0 1
(c) T
(∵) In à In , by 1 × C1 ⇒ C1
(d) F
(e) T
(∵) Theorem 3.2
(f) F




1 0 0
1 0 0
(∵) Let E1 = 0 −1 0 , E2 = 0 1 0 
0 0 −1
0 0 1
then, E1 + E2 is not an elementary matrix
(g) T
(h) F
(i) T
Let A Ã B = EA,
then E is invertible and its inverse is also an elementary matrix
154
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3.1.
∴ B Ã A = E −1 B
2.
(i) A Ã B, by −2 × C1 + C2 → C2
(ii) B Ã C, by −1 × R1 + R2 → R2
(iii) C Ã I3 , by
1
2
× R2 → R2
R2 ↔ R3
R3 + R2 → R2
1
4
× R2 → R2
− 1 × R2 + R3 → R3
− 3 × R3 + R1 → R1
3.


0 0 1
(a) E = 0 1 0
1 0 0
I3 ⇒ (R1 ↔
E ⇒ (R1 ↔ R3 ) ⇒ I3
 R3 ) ⇒ 
0 0 1
∴ E −1 = 0 1 0
0
 1 0 
1 0 0

(b) E = 0 3 0
0 0 1
I3 ⇒ (3 × R2 → R2 ) ⇒ E ⇒ ( 31 × R2 → R2 ) ⇒ I3
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3.1.


1 0 0
∴ E −1 = 0 13 0
 0 0 1
1 0 0

0 1 0
(c) E =
−2 0 1
I3 ⇒ (−2 ×
3 → R3 ) ⇒ E ⇒ (2 × R1 + R3 → R3 ) ⇒ I3
 R1 + R
1 0 0
∴ E −1 = 0 1 0
2 0 1
4.
(i) E is of type 1
E = (e1 , · · · , ej , · · · , ei , · · · , en )
F = (e1 , · · · , ej , · · · , ei , · · · , en ) = E
(ii) E is of type 2
E = (e1 , · · · , aej , · · · , en )
F = (e1 , · · · , aej , · · · , en ) = E
(iii) E is of type 3
E = (e1 , · · · , ei , · · · , ej + aei , · · · , en )
F = (e1 , · · · , ei + aej , · · · , ej , · · · , en ) = E t
5.
(⇒)
(i) E is of type 1
E = (e1 , · · · , ej , · · · , ei , · · · , en )
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3.1.
E t = (e1 , · · · , ej , · · · , ei , · · · , en )
(ii) E is of type 2
E = (e1 , · · · , aej , · · · , en )
E t = (e1 , · · · , aej , · · · , en )
(iii) E is of type 3
E = (e1 , · · · , ei , · · · , ej + aei , · · · , en )
E t = (e1 , · · · , ei + aej , · · · , ej , · · · , en )
(⇐) Using the fact (E t )t = E, then it is clear
6.
(i) if B = EA, then B t = (EA)t = At E t
(ii) if B = AE, then B t = (AE)t = E t At
7.
(1) Elementary column operation
(i) E is of type 1
B = (A(1) , · · · , A(j) , · · · , A(i) , · · · , A(n) )
E = (e1 , · · · , ej , · · · , ei , · · · , en )
⇒ AE = (Ae1 , · · · , Aej , · · · , Aei , · · · , Aen ) = B
∴ B = AE
(ii) E is of type 2
B = (A(1) , · · · , aA(j) , · · · , A(n) )
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3.1.
E = (e1 , · · · , aej , · · · , en )
⇒ AE = (Ae1 , · · · , aAej , · · · , Aen ) = B
∴ B = AE
(iii) E is of type 3
B = (A(1) , · · · , A(i) , · · · , A(j) + aA(i) , · · · , A(n) )
E = (e1 , · · · , ei , · · · , ej + aei , · · · , en )
⇒ AE = (Ae1 , · · · , Aei , · · · , Aej + aAei , · · · , Aen ) = B
∴ B = AE
(2) Elementary row operation
(i) E is of type 1
B = (A(1) , · · · , A(j) , · · · , A(i) , · · · , A(m) )
E = (e1 , · · · , ej , · · · , ei , · · · , em )
⇒ EA = (e1 A, · · · , ej A, · · · , ei A, · · · , em A) = B
∴ B = EA
(ii) E is of type 2
B = (A(1) , · · · , aA(j) , · · · , A(m) )
E = (e1 , · · · , aej , · · · , em )
⇒ EA = (e1 A, · · · , aej A, · · · , em A) = B
∴ B = EA
(iii) E is of type 3
B = (A(1) , · · · , A(i) , · · · , A(j) + aA(i) , · · · , A(m) )
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3.1.
E = (e1 , · · · , ei , · · · , ej + aei , · · · , em )
⇒ EA = (e1 A, · · · , ei A, · · · , ej A + aei A, · · · , em A) = B
∴ B = EA
8.
(i) E is of type 1
E = (e1 , · · · , ej , · · · , ei , · · · , en )
E −1 = (e1 , · · · , ej , · · · , ei , · · · , en )
(ii) E is of type 2
E = (e1 , · · · , aej , · · · , en )
E −1 = (e1 , · · · , a1 ej , · · · , en )
(iii) E is of type 3
E = (e1 , · · · , ei , · · · , ej + aei , · · · , en )
E −1 = (e1 , · · · , ei , · · · , ej − aei , · · · , en )
If P Ã Q = EP , then P = E −1 Q
Let E 0 = E −1
Since E 0 is invertible and is an elementary matrix, Q Ã P = E 0 Q
9. In ⇒ E30 , E300 , E3000 , E2 ⇒ E
,where E30 : −1 × Ri + Rj ⇒ Rj i.e E = (e1 , · · · , ei , · · · , ej − ei , · · · , en )
E300 : 1 × Rj + Ri ⇒ Ri i.e E = (e1 , · · · , ei + (ej − ei ), · · · , ej − ei , · · · , en )
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3.1.
E3000 : −1 × Ri + Rj ⇒ Rj i.e E = (e1 , · · · , ej , · · · , −ei , · · · , en )
E2 : −1 × Rj ⇒ Rj i.e E = (e1 , · · · , ej , · · · , ei , · · · , en )
∴ E is of type 1
10. a is a nonzero scalar
In ⇒ E2 ⇒ E
,where E2 :
1
a
× Ri → Ri i.e E = (e1 , · · · , a1 ei , · · · , en )
∴ E is of type 2
11. In ⇒ E3 ⇒ E
,where E3 : −a × Ri + Rj → Rj i.e E = (e1 , · · · , ei , · · · , ej − aei , · · · , en )
∴ E is of type 3
12.
By induction on n ≥ 1
If n = 1 o.k.
Assume that n > 1
If the first column of A is zero, then A = (O | B), where B = ()m×(n−1)
So by induction hypothesis, B can be transformed by row operation of type 1
and 3
If the first column of A is not zero, we may assume that a11 6= 0
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3.1.


a11 a12 · · · a1n
0



So A Ã  ..
 , where C = ()(m−1)×(n−1)
 .

C
0
By induction
 hypothesis, C Ã U.T.M
a11 a12 · · · a1n

0


∴ A Ã  ..
 : U.T.M

 .
D
0
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3.2.
3.2. The Rank of a Matrix and Matrix Inverse
1. (a) F (Theorem 3.5)
(b) F
(∵) If A ∈ Mm×n (F ), B ∈ Mn×n (F ) and B is invertible,
then the rank(AB)=rank(A)
(Example)
µ
¶
0 1
A=B=
and rank(A) = rank(B) = 1
µ 0¶ 0
0 0
AB =
and rank(AB) = 0
0 0
(c) T
(d) T (p.153, Corollary to the Theorem 3.4)
(e) F (p.153, Corollary to the Theorem 3.4)
(f) T (p.153, Theorem 3.4 and Theorem 3.5)
(g) T (p.161)
(h) T
(∵) ∀A ∈ Mm×n (F ), rank(A) = dimR(LA ) ≤ n
(i) T
2.


1 1 0
(a) A = 0 1 1 , rank(A)=2
0 0 0
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3.2.

1

(b) A = 0
µ0
1
(c) A =
µ0
1
(d) A =
0

1
0
(e) A = 
0
0

1
0
(g) A = 
0
0

1 0
−1 1 , rank(A)= 3
0 ¶1
0 2
, rank(A)= 2
1 2¶
2 1
, rank(A)= 1
0 0

2 0 1 1
0 1 1 −2
 , rank(A)= 3
0 0 0 1
0 0 0 0

1 0 1
0 0 0
 , rank(A)= 1
0 0 0
0 0 0
3.
∀ A ∈ Mm×n (F ), rank(A) = 0 iff A is the zero matrix
(⇐) clear
(⇒) let A = (e1 , 0, · · · , 0), e1 6= 0
Since rank(A) = 0, e1 is dependent
∴ ∃a ∈ F s.t e1 = a0
It’s contradict to e1 6= 0
4. 





1 1 1 2
1 0 − 21 1
1 0 0 0
(a) 2 0 −1 2 ⇒ 0 1 32 1 ⇒ 0 1 0 0 , 2
1 1 1 2
0 0 0 0
0 0 0 0
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3.2.






2 1
2 0
12 0
(b)−1 2 ⇒ 0 1 ⇒  0 1 , 2
2 1
0 0
0 0
5. (A | In ) Ã (In | B), B = A−1
µ
¶
¶
µ
1 0 -1 2
1 2 1 0
⇒
(a)
1 µ
1 0 1 ¶
0 1 1 -1
−1 2
∴ A−1 =
, rank(A) = 2
1
−1
µ
¶
µ
¶
1 2 1 0
1 2 1 0
(b)
⇒
2 4 0 1
0 0 -2 1
−1
rank(A)



 = 1, so @A
1 0 -5 3 -2 0
1 2 1 1 0 0
(c)  1 3 4 0 1 0  ⇒  0 1 3 -1 1 0 
2 3 -1 0 0 1
0 0 0 -3 1 1
−1
rank(A)
 = 2, so @A



0 -2 4 1 0 0
1 0 0 1/6 15/9 -1/3
(d)  1 1 -1 0 1 0  ⇒  0 1 0 -5/18 -4/9 2/9 
2 4 -5 0 0 1
0 0 1 1/9 -2/9 1/9
1
15
1
−3
6
9
5
− 94 29 
∴ rank(A) = 3, A−1 = − 18
1
2
1
9

 −9  9

0 -2 4 1 0 0
1 0 0 1/6 -1/3 1/2
0
-1/2 
(e)  1 1 -1 0 1 0  ⇒  0 1 0 1/2
2 4 5 0 0 1
0 0 1 1/6 1/3 1/2
1
1
1
−3 2
6
0 − 12 
∴ rank(A) = 3, A−1 =  12
− 16 13  12



0
1
0
0 2 1 1 0 0
1 0 1
(f)  1 0 1 0 1 0  ⇒  0 1 0 -1/2 -1/2 0 
1 1 1 0 0 1
0 0 0 -1/2 -1/2 1
−1
∴ rank(A) = 2, , so @A
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3.2.

1 2 1 0
 2 5 5 1
(g) 
 -2 -3 0 3
3 4 -2 -3
∴ rank(A) = 4, A−1

1 0 1
 1 1 -1
(h) 
 2 0 1
0 -1 1
1
2
0
-3
∴ rank(A) = 4, A−1
6.
1 0 0
0 1 0
0 0 1
0 0 0

−51
 31
=
−10
−3
1 0 0
0 1 0
0 0 1
0 0 0

−3/2
 1/2
=
 −3
1/2


0
1


0 
0
⇒ 
 0
0 
0
1

15 7
2
−9 −4 −7

3
1
2
1
1
1


0
1
 0
0 
 ⇒ 
 0
0 
1
0
−1/10 13/10
1/2
−7/5
−1/5 16/10
1/10 −3/10
0 0 0
1 0 0
0 1 0
0 0 1
-51 15
31 -9
-10 3
-3 1
7
-4
1
1

2
-7 

2 
1
0 0 0 -3/2 -1/10 13/10
1 0 0 1/2
1/2
-7/5
-1/5 16/10
0 1 0 -3
0 0 1 1/2 1/10 -3/10

1/10
1/2 

1/5 
−1/10


−1 2
2
(a) [T ]β =  0 −1 4  , rank[T ]β = 3
0
0 −1
∴T is invertible


−1 −2 10
([T ]β )−1 =  0 −1 4 
0
0 −1
∴ T −1 (ax2
+ bx + c)= −ax2 − (4a + b)x − (10a + 2b + c)
0 1 0
(b) [T ]β = 0 1 2 , rank[T ]β = 2
0 0 0
∴T is not invertible


1 2 1
(c) [T ]β = −1 1 2 , rank[T ]β = 3
1 0 1
∴T is invertible
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
1/10
1/2 

1/5 
-1/10
3.2.


1/6 −1/3 1/2
0
−1/2
([T ]β )−1 =  1/2
−1/6 1/3
1/
1
1
1
−1
∴ T (a, b, c) = ( 6 a − 3 b + 2 c, 12 a − 12 c, − 16 a + 13 b + 12 c)
7.


1 2 1
A = 1 0 1 Ã I3 = E6 · · · E1 A
1 1 2
−1
∴A
=
· ·
E6−1
 E1 ·





1 0 0
1 0 0
1 −1 0
1 0 0
1 0 0
1 0 1
= 1 1 0 0 1 0 1 1 0 0 −2 0 0 1 0 0 1 0
0 0 1
1 0 1
0 0 1
0 0 1
0 −1 1
0 0 1
8.
cA = (cIm )A
Since cIm is invertible
∴ rank(cA) = rank((cIm )A)
9.
If B is obtained from a matrix A by an elementary column operation, then there
exists an elementary matrix E such that B = AE
Since E is invertible and hence rank(B) = rank(A)
10. If A is the zero matrix, then r = 0
rank(A) = rank(D) = 0
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3.2.
Suppose that A is a nonzero matrix
By means of at most one type 1 row operation, at most one type 2 row operation,
and at most (m − 1) type 3 row operations
this matrix can be transformed into (1, 0, · · · , 0)T
∴ rank(A) = rank(D) = 1
11. (By theorem 3.6)
µ
¶
I
O
k
B 0 Ã D0 =
, rankB 0 = k and
O O


µ
¶
1 O
Ik+1 O


=
B Ã D=
O Ik O
O O
O O
∴ rankB = k + 1 = r
∴ rankB 0 = r − 1
12. By induction on n ≥ 1
If n = 1, it is clear
Assume n > 1
¡
¢
If the first column of A is zero, then A = O B , where B ∈ Mm×(n−1)
So by induction hypothesis, B can be transformed by row operation of type 1
and 3
If the first column of A is not zero,
we may assume that a11 6= 0
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3.2.

a11 a12 · · · a1n
 0

So A Ã  ..
 .
C
0
By by induction hypothesis, C



, where C ∈ M(m−1)×(n−1)

à U.T.M
∴ A Ã U.T.M
13. (b) and (c)
Since rank(A) = rank(At ), the maximal number of linearly independent columns
of A equals to the maximal number of linearly independent columns of At ,
i.e. the maximal number of linearly independent rows of A
∴ colrank(A) = rowrank(A)
(cf ) Let S the solution space for AX = 0, then by the dimension theorem and
rank(LA ) = colrank(A),
dimS = n − colrank(A)
If r = dim(row space of A), then the solution space S has a (n − r) vectors,
i.e. dimS = n − rowrank(A)
∴ colrank(A) = rowrank(A)
14. T, U : V → W
(i) w ∈ R(T + U ) ⇒ ∃ v ∈ V s.t. w = (T + U )(v)
⇒ w = T (v) + U (v) ∈ R(T ) + R(U )
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3.2.
(ii) rank(T + U ) ≤ dim(R(T ) + R(U ))
≤ dim(R(T )) + dim(R(U ))
= rank(T ) + rank(U )
(c) By (b), rank(A + B) = rank(LA+B )
= rank(LA + LB )
≤ rank(LA ) + rank(LB )
= rank(A) + rank(B)
15.
Let A = (a1 , a2 , · · · , ap )n×p and B = (b1 , b2 , · · · , bq )n×q
(A | B) = (a1 , a2 , · · · , ap , b1 , b2 , · · · , bq )n×(p+q)
M (A | B) = (M a1 , M a2 , · · · , M ap , M b1 , M b2 , · · · , M bq )n×(p+q)
(M A | M B)
16. (Theorem 3.4 (b))
Let V = R(LA ) = LA (F n )
Then V ≤ F m and dimV = dimLP (V )
because LP is an isomorphism (by the exercise 17 in section 2.4)
rank(A) = dimLA (V ) = dimV
= dimLP (V )
= dimLP LA (F n )
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3.2.
= dimLP A (F n )
rank(P A)
17.


1 0 0
Let P AQ = 0 0 0, then
 0 0 0
1 ¡
¢
A = P −1 0 1 0 0 Q−1
0
18.
Let A = (a1 , a2 , · · · , an )m×n where ak the k−th column of A and let Ak =
(0, · · · , ak , · · · , 0)m×n
Thus ∀k, rank(Ak B) ≤ 1 and AB = A1 B + · · · + An B
19.
By the theorem 3.7 rank(AB) ≤ rank(A) and
by the Sylvester inequality, rank(AB) ≥ rank(A)
∴ rank(AB) = rank(A) = m
(cf )the Sylvester inequality
rank(A) + rank(B) ≤ rank(AB) + n,
where n is the number of columns of A and also the number of rows of B
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3.2.
20.
(a)
A Ã D = EA = () : the reduced row echelon form of A
⇒ {v1 = (), v2 = ()} is a basis of N ull(LA )
Thus AM = (0)5×5 , where M = (v1 , v2 , 0, 0, 0)
(b)
Suppose B = (b1 , b2 , · · · , b5 )5×5 s.t. AB = (0)4×5
⇒ b1 , b2 , · · · , b5 ∈ N (LA ) with null LA = 2
Thus rankB = dimspan(b1 , b2 , · · · , b5 ) ≤ dimN ull(LA ) = 2
21.
A = (aij )m×n , rank(A) = m
A Ã D = AQ = (Im | O)m×n , Q ∈ Mn×n : invertible matrix
µ ¶
I
Let B = QM , where M = m
O n×m
µ ¶
I
then AB = (QM ) = (AQ)M = (Im | O) m = Im
O
22.
B = (bij )n×m , rank(B) = m
µ ¶
I
, Q ∈ Mn×n : invertible matrix
B Ã D = QB = m
O n×m
Let A = M Q, where M = (Im | O)m × n
then AB = M (QB) = M D = Im
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3.3.
3.3. Systems of Linear equations - Theoretical aspects
1.
(a) F
p.170, Example 1 (c)
(b) F
(c) T
Any homogeneous system has at least one solution, namely, the zero vector
(d) F
p.174, Theorem 3.10
(e) F
(f) F
p.172, Theorem 3.9
(g) T
If A is invertible, then AX = 0 has no nonzero solutions
(h) T
2. Let K be the solution set of the given system and A is the coefficient matrix of the system
µ
¶
µ
¶
1 3
1 3
(a) A =
Ã
2 6
0 0
rank(A) = 1, dim(K) = 2 − 1 = 1
½µ ¶¾
−3
: a basis for K
1
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3.3.
¶
µ
¶
1 1 −1
1 0 0
(b) A =
Ã
0 1 0
4 1 −2
rank(A) = 2, dim(K) = 3 − 2 = 1
µ
t
Since
(1,
2, 3) is a solution to AX = 0,
 
 1 
2 : a basis for K


3 µ
¶
µ
¶
1 2 −1
1 0 0
(c) A =
Ã
2 1 1
0 1 0
rank(A) = 2, dim(K) = 3 − 2 = 1
t
Since

 (−1,
1, 1) is a solution to AX
 −1 
 1  : a basis for K


1 


1 0
2 1 −1



0 1
Ã
(d) A = 1 −1 1
0 0
1 2 −2
rank(A) = 2, dim(K) = 3 − 2 = 1
= 0,

0
0
0
t
Since
(0,
1, 1) is a solution to AX = 0,
 
 0 
1 : a basis for K


1
(e) A = (1, 2, −3, 1) Ã (1, 0, 0, 0)
rank(A) = 1, dim(K) = 4 − 1 = 3
x1 = −2x2 + 3x3 − 4x4
Note
} is linearly
independent vectors in K
 that
 {v1 , v2 , v3

−2
3
−1



      

1
0
0
  ,   ,   : a basis for K
 0  1  0 





0
0
1
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3.3.
¶
µ
¶
1 2
1 0
(f) A =
Ã
0 1
1 −1
rank(A) = 2, dim(K) = 2 − 2 = 0
½µ ¶¾
0
: a basis for K
0
(So the given system is inconsistent)
µ
¶
µ
¶
1 2 1 1
1 0 0 0
(g) A =
Ã
0 1 −1 1
0 1 0 0
rank(A)
 = 2,
 = 4 − 2 = 2
 dim(K)
−3
1




    
 1  , −1 : a basis for K

  1   0 



0
1
µ
3.
¶
¾
¶
µ
−3
5
|t ∈ R
+t
(a) A =

 0 
1

1
−1


(b) A =  1  + t  1  |t ∈ R



 1 
 1 
−1
2


(c) A =  1  + t  1  |t ∈ R



 1 
 1 
0
2


(d) A =  0  + t  1  |t ∈ R


1
 −1



1
−2
3




 −1 
 0
0




(e) A = 
 0  + r 0  + s 1



0
0
½µ 0 ¶¾
1
(f) A =
0
½µ
174


−1




 0 
 + t
 |r, s, t ∈ R

 0 



0


PNU-MATH
3.3.



0




0 
+r
(g) A = 
 0 




1


−3
1


1 
−1
+ s
 0
1 
0
1






 |r, s ∈ R




4.
(a)
µ
¶
−5 3
(1) A =
µ2 ¶ −1µ
¶
−11
−1 4
(2) x = A
=
3
5
(b)

 1
0 13
3
(1) A−1 =  19 13 − 92 
− 49 23 
− 91 

5
3
−1  

1 =
0
(2) x = A
4
−2
−1
5.
AX = B, A = (a1 , · · · , an ) ∈ Mn×n , ai is the nth column of A
If ai is expressed by other column vectors of A, i.e. ai ’s are linearly independent,
then the given system has infinitely many solutions
(Example) A = (a1 , · · · , ai , · · · , kai , · · · , an ), aj = kai , k ∈ F
6.
 11 

 
1
 2

T −1 ({(1, 11)}) = − 92  + t −1 |t ∈ R


0
2
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3.3.
7. The systems in parts (b), (c), and (d) have solutions
(a) rank(A)
=3 


 = 2, rank(A|b)
1 0 0 0 0
1 0 0 0
A Ã 0 0 1 0 , (A|b) Ã 0 0 1 0 0
0 0 0 0
0 0 0 0 1
(b) rank(A) = rank(A|b) = 2
(c) rank(A) = rank(A|b) = 3
(d) rank(A) = rank(A|b) = 4
(e) rank(A) = 2, rank(A|b) = 3
8.
(a) v ∈ R(T ) T (−2, 3, 0) = (1, 3, 2)
(b) v ∈ R(T ) T (1, 1, 0) = (2, 1, 1)
9.
LA : F n → F m , A = (a1 , · · · , an )
Let x = (x1 , · · · , xn )t is a solution to Ax = b
⇔ b = Ax = a1 x1 + · · · + an xn , xi ∈ F
⇔ b ∈ span(a1 , · · · , an ) = R(LA )
11.


1
Ap = p à p = 0.75
1
The farmer, tailor, and carpenter must have incomes in the proportions 4:3:4
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3.3.
12.
0.60p1 + 0.30p2 = p1
0.40p1 − 0.30p2 = 0
µ
¶
0.75
∴ p=
1
13.
There must be 7.8 units of the first commodity and 9.5 units of the second
µ ¶
7.8
−1
x = (I − A )d =
9.5
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3.4.
3.4. Systems of Linear equations - Computational aspects
1.
(a) F (a finite row operations)
(b) T (P.182 Corollary)
(c) T (P.158 Corollary 1 to Theorem 3.6)
(d) T (p.187 Theorem 3.14)
(e) F
The system has a solution if and if only the echelon form of the augmented matrix
M does not have a row of the form (0, · · · , 0, b) with b 6= 0
(f) T
rank(A) = rank(A | b) ⇔ (A | b) is consistent
If the system (A | b) is consistent and rank(A) = r, then the dimension of the
solution set is n − r.
(g) T
Since A is row equivalent to A0 i.e.A0 = EA rank(A) = rank(EA0 ) = rank(A0 )
2. 

(a) 



(b) 


4


−3

1 



9
−5




4
−3
+r
|r∈R

0
1
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3.4.




(c) 








(d) 








(e) 








(f) 











(g) 








2




3 
−2 


1




9
13



 −15 
22 



+
r
|
r
∈
R
1 
 0 
− 26



18
1
13





4
4
1



 1 
 0 
0 
+r
 + s
 | r, s ∈ R
 0 
 2 
1 



0
0
1 



1
−3






−2
3 


|r∈R
+r
0 
1 



1
0






−23
1
−23









0 
1
0










7  + r  0  + s  6  | r, s ∈ R


 0 
 9 
9 



0
0
1
3.
(a)
(⇐) If (A0 | b0 ) has such a row, say k − th row, then
the corresponding k − th equation in the system A0 x = b0 is
0x1 + 0x2 + · · · + 0xn = ck , ck 6= 0 ;
which has no solutions i.e. A0 x = b0 is inconsistent
∴ rank(A0 ) 6= rank(A0 | b0 )
(⇒) Assume that (A0 | b0 ) has no such a row, then b0 ∈ R(LA0 )
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3.4.
∴ rank(A0 ) = rank(A0 | b0 )
(b)
If two matrices are row equivalent, they have the same solution set
Since (A0 | b0 ) is equivalent to (A | b), so
Ax = b is consistent
⇔ A0 x = b0 is consistent
⇔ rank(A0 ) = rank(A0 | b0 )
By (a), (A0 | b0 ) has no such a row
(cf) (a) (⇒) b0 ∈ R(LA0 ) (∵ T heorem3.16(b))
4.
(a)

1
 0
0
Since
0
1
0
(A0

0 -1/2 4/3
0 1/2 1/3 
1 -1/2 0
| b0 ) contains no row in which the only nonzero entry lies in the last
column,
therefore


  Ax = bis consistent
4/3
1/2






  −1/2 
1/3

+
 r | r ∈ R : the solution set
 0   1/ 






0
1


1/2






−1/2

 : a basis for the solution set
 1/2 





1
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3.4.
(b)


1 1 0 -1/2 1
 0 0 1 -1/2 1 
0 0 0
0
0
0
0
Since
|b0 ), Ax
= b is consistent


 rank(A
 ) = rank(A

1
−1
1/2






  1 
 0 
0

+
r + 
 s | r, s ∈ R : the solution set
 1   0 
 1/2 






0
0
1

 

1/2 
−1







0
1

,
 : a basis for the solution set

 0   1/2 



0
1
(c)


1 1 0 -1/2 7/4
 0 0 1 -1/2 1/4 
0 0 0
0
-3/4
0
Since rank(A ) 6= rank(A0 | b0 ), Ax = b is inconsistent
5.


1 0 2 0 −2
B = 0 1 −5 0 −3
0 0 0 1 6
Let A = (a1 , a2 , · · · , a5 ) and B = (b1 , b2 , · · · , b5 )
Since b3 = 2e1 − 5e2 , it follows that a3 = 2a1 − 5a2
Moreover
, the same result shows that a5 = −2a1 − 3a2 + 6a4
 b5 = −2e1 − 3e2 + 6e3
1
0 2 1
4
∴ A = −1 −1 3 −2 −7
3
1 1 0 −9
6.
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PNU-MATH
3.4.


1 −3 0 4 0 5
0 0 1 3 0 2 

B=
0 0 0 0 1 −1
0 0 0 0 0 0
Let A = (a1 , a2 , · · · , a6 ) and B = (b1 , b2 , · · · , b6 )
Since b2 = −3e1 ,
b4 = 4e1 + 3e2 ,
b6 = 5e1 + 2e2 − e3
it follows that a2 = −3a1 ,
a4 = 4a1 + 3a3 ,
a6 = 5a1+ 2a3 − a5

1 −3 −1 1
0
3
−2 6
1 −5 1 −9

∴ A=
−1 3
2
2 −3 2 
3 −9 −4 0
2
5
7.




1 0 −4 −4 0
2
1 −8 1 −3
−3 4 12 37 −5 ⇒ 0 1 0
7 0
0 0 0
0 1
1 −2 −4 −17 8
∴ {u1 , u2 , u5 } is a basis for W
9.



0 −1 −1 1
1
1


2
2
3

0
2

1
1
9 ⇒ 

0
 1 −2 −2 4 
0
−1 2
2 −1
0
0
1
0
0
0
0
1
0
0
0

0
0

1

0
0
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3.4.
½µ
0 −1
−1 1
(cf)

0
−1

−1
1
(a) (i)
¶ µ
¶ µ
¶¾
1 2
2 1
,
,
: a basis for W
2 3
1 9

· · · −1
··· 2 

··· 2 
· · · −1
S⊆V

1
0

(ii) S : linearly independent A Ã 
0
0
0
{u1 , u2 , · · · , u5 }
0
1
0
0
0

0 1 0
0 −1 0

1 −1 0

0 0 0
0 0 1
10.
(a) It’s a singleton set, so it’s linearly independent
(b)
Since x1 = 2x2 − 3x3 + x4 − 2x5 , assign parametric values to x2 , x3 , x4 and x5
Let x2 = t1 , x3 = t2 , x4 = t3 and x5 = t4 , then the vectors in V have the form
(x1 , x2 · · · , x5 ) = t1 (2, 1, 0, 0, 0)+t2 (−3, 0, 1, 0, 0)+t3 (1, 0, 0, 1, 0)+t4 (−2, 0, 0, 0, 1)
Hence
β = {(2, 1, 0, 0, 0), (−3, 0, 1, 0, 0), (1, 0, 0, 1, 0), (−2, 0, 0, 0, 1)} is a basis for V
The matrix whose columns consist of the vectors in S followed by those β is
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3.4.

0
1

1

1
0

2 −3 1 −2
1 0 0 0

0 1 0 0

0 0 1 0
0 0 0 1

1 0 0 1 0
0 1 0 −1 0



0
0
1
−1
0
and its reduced row echelon form is 


0 0 0 0 0 
0 0 0 0 1
Thus {(0, 1, 1, 1, 0), (2, 1, 0, 0, 0), (−3, 0, 1, 0, 0), (−2, 0, 0, 0, 1)} is a basis for V con
taining S
11.
(a) It’s a singleton set, so it’s clear
(b)
The

1
2

1

0
0
matrix
2 −3
1 0
0 1
0 0
0 0
whose columns
consist of the vectors in S followed by those β is

1 −2
0 0

0 0

1 0
0 1


1 0 1 0 0
0 1 −2 0 0



0
0
0
1
0
and its reduced row echelon form is 


0 0 0 0 1 
0 0 0 0 0
Thus {(1, 2, 1, 0, 0), (2, 1, 0, 0, 0), (1, 0, 0, 1, 0), (−2, 0, 0, 0, 1)} is a basis for V containing S
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3.4.
12.
(a) If a(0, −1, 0, 1, 1, 0) + b(1, 0, 1, 1, 1, 0) = (0, · · · , 0). a, b ∈ F
then a = b = 0
∴ S is linearly independent
(b)
Since x1 = x3 − x4 + x5 − 3x6
x2 = x3 + x4 − 2x5 − 2x6
Let x3 = t1 , x4 = t2 , x5 = t3 and x6 = t4 , then the vectors in V have the
form
(x1 , x2 · · · , x6 ) = t1 (1, 1, 1, 0, 0, 0)+t2 (−1, 1, 0, 1, 0, 0)+t3 (1, −2, 0, 0, 1, 0)+t4 (−3, −2, 0, 0, 0, 1)
Hence
β = {(2, 1, 0, 0, 0), (−3, 0, 1, 0, 0), (1, 0, 0, 1, 0), (−2, 0, 0, 0, 1)} is a basis for V
The
 matrix whose columnsconsist of the vectors in S followed by those β is
0 1 1 −1 1 −3
−1 0 1 1 −2 −2


0 1 1 0
0
0


1 1 0 1

0
0


1 1 0 0
1
0
0 0 0 0
0
1
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3.4.


1 0 −1 0 −1 0
0 1 1 0 0 0 


0 0 0 1 −1 0


and its reduced row echelon form is 

0
0
0
0
0
1


0 0 0 0 0 0 
0 0 0 0 0 0
Thus {(0, −1, 0, 1, 1, 0), (1, 0, 1, 1, 1, 0), (−1, 1, 0, 1, 0, 0), (−3, −2, 0, 0, 0, 1)} is a basis for V containing S
13.
(a) If a(1, 0, 1, 1, 1, 0) + b(0, 2, 1, 1, 0, 0) = (0, · · · , 0). a, b ∈ F
then a = b = 0
∴ S is linearly independent
(b)
The

1
0

1

1

1
0
matrix whose columns
 consist of the vectors in S followed by those β is
0 1 −1 1 −3
2 1 1 −2 −2

1 1 0
0
0

1 0 1
0
0

0 0 0
1
0
0 0 0
0
1


1 0 0 0
1 0
0 1 0 0
0 0


0 0 1 −1 0 0


and its reduced row echelon form is 

0
0
0
1
−1
0


0 0 0 0
0 1
0 0 0 0
0 0
Thus {(1, 0, 1, 1, 1, 0), (0, 2, 1, 1, 0, 0), (1, 1, 1, 0, 0, 0), (−3, −2, 0, 0, 0, 1)} is a basis
for V containing S
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3.4.
15.
Let B is a reduced row echelon form to A, then A is row equivalent to B,
i.e. A ∼ B
If C is an another reduced row echelon form to A, then C ∼ A ∼ B
Since row equivalent matrices have the same row space and they are finite dimensional, so must be identical
(cf) If C is an another reduced row echelon form to A, then C ∼ A ∼ B
Since ∃ E s.t. C = EB and E is invertible, we have E −1 C = B,
Comparing with their rank,
the row space C and B are the subspace of each other
So C = B
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4.1.
§4.
Determinants
4.1. Determinants of Order 2
1. (a) F det(A + B) 6= det(A) + det(B)
(b) T (p.200. Theorem 4.1)
(c) F (p.201. Theorem 4.2)
(d) F
µ ¶
µ ¶
u
u
The area of the paralleogram determined by u and v equals O
det
v
v
(e) F (p.203)
2. (a) 30
(b) -17
(c) -8
3. (a) -10+15i
(b) 36+41i
(c) -24
4. (a) 4
(b) 10
(c) 14
(d) 26
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4.1.
5.
µ
¶
µ
¶
a11 a12
a21 a22
Let A =
, then B =
a21 a22
a11 a12
So det(B) = a12 a21 − a11 a22 = −(a11 a22 − a12 a21 ) = − det(A)
6.
µ
¶
µ
¶
a a
b b
Let A =
, and B =
b b
a a
By the exercise 15, det(B) = − det(A)
Since det(A) = det(B)
∴ det(A) = 0
7.
µ
¶
µ
¶
a11 a12
a11 a21
t
Let A =
, then A =
a21 a22
a12 a22
t
det(A ) = a11 a22 − a21 a12 = det(A)
∴ det(At ) = det(A)
8.
µ
If A =
9.
Let A =
¶
a11 a12
, then det(A) = a11 a22 : the product of the diagonal entries of A
0 a22
µ
a11 a12
a21 a22
¶
¶
µ
b11 b12
and B =
b21 b22
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4.1.
µ
then AB =
a11 b11 + a12 b21 a11 b12 + a12 b22
a21 b11 + a22 b21 a21 b12 + a22 b22
¶
det(AB)
= (a11 b11 + a12 b21 )(a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 )(a21 b11 + a22 b21 )
= (a11 a22 − a12 a21 )(b11 b22 − b12 b21 )
= det(A) det(B)
10.
C=
µ
a22 −a12
−a21 a11
¶
µ
a11 a22 − a12 a21
0
(a) CA = AC = (det A)I =
0
a11 a22 − a12 a21
(b) det(C) = a22 a11 − a12 a21 = det(A)
µ
¶
a22 −a21
t
(c) The classical adjoint of A =
= Ct
−a12 a11
(d) If A is invertible, then det(A) 6= 0
¶
1
1
)C} = {( det(A)
)C}A = I
by (a), A{( det(A)
1
∴ A−1 = ( det(A)
)C
11.
Let A ∈ M2×2 (F )
(1) If A has rank less than 2, then by the assumption (ii) δ(A) = 0
In this case det(A) equals to zero
∴ δ(A) = det(A)
(2) If A has rank 2, then A is invertible
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4.1.
hence A = Ek Ek−1 · · · E1 for some k
Since δ(I) = 1,δ(E) = det(E), ∀E: an elementary matrix
Hence we have δ(A) = δ(Ek Ek−1 · · · E1 )
= δ(Ek )δ(Ek−1 ) · · · δ(E1 )
= det(Ek ) det(Ek−1 ) · · · det(E1 )
= det(Ek Ek−1 · · · E1 )
= det(A)
12.
(⇐) Let u = (a1 , a2 ), then v = (a1 cos θ − a2 sin θ, a1 sin θ + a2 cos θ), 0 < θ < π
µ ¶
µ ¶
u
u
2
2
= (a1 + a2 ) sin θ and det
Since det
>0
v
µ ¶ v
µ ¶
u
u
∴ O
= 1 (⇒) Since O
= 1, sin θ > 0
v
v
∴ 0<θ<π
∴ {u, v} forms a right-handed coordinate system
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4.2.
4.2. Determinants of Order n
1.
(a) F
(b) T (Theorem 4.4)
(c) T (Corollary to Theorem 4.4)
(d) T (Theorem 4.5)
(e) F (det(B) = k det(A))
(f) F (det(B) = det(A))
(g) F (If A ∈ Mn×n (F ) and rankA = n ⇒ A : invertible⇒ det A 6= 0)
(h) T
2. k = 33
3. k = 42
4. k = 2
5. det(A) = −12
6. det(A) = −13
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4.2.
7. det(A) = −12
8. det(A) = −13
9. det(A) = 22
10. det(A) = 4 + 2i
11. det(A) = −3
12. det(A) = 154
13. det(A) = −8
14. det(A) = −168
15. det(A) = 0
16. det(A) = 36
17. det(A) = −49
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4.2.
18. det(A) = 10
19. det(A) = −28 − i
20. det(A) = 17 − 3i
21. det(A) = 95
22. det(A) = 100
23.

a11 a12 · · ·

a22 · · ·

Let A = 
..

.

a1n
a2n 

.. 
. 
ann
By expanding along
 the first column,
 we have
a22 a23 · · · a2n

a33 · · · a3n 


det A = a11 det 
.. 
..

.
. 
ann


a33 a34 · · · a3n

a44 · · · a4n 


= a11 a22 det 
.. 
.
.

.
. 
ann
..
.
= a11 a22 · · · ann
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4.2.
24.
If A has a row consisting entirely of zeros, then det(A) = 0
Let i − th row of A is the zero row
n
P
(−1)i+j Aij det(A˜ij ) and ∀ Aij = 0, (j = 1, 2, · · · , n)
Since det(A) =
j=1
∴ det(A) = 0
25.
Let A = (a1 , a2 , · · · , an ), ai : rows of A, i = 1, 2, · · · , n
If A0 is obtained by multiplying a row of A by a nonzero scalar k,
then det(A0 ) = k det(A)
det(kA) = det(ka1 , ka2 , · · · , kan )
= k det(a1 , ka2 , · · · , kan )
= k 2 det(a1 , a2 , ka3 , · · · , an )
..
.
= k n det(a1 , a2 , · · · , an )
= k n det(A)
26.
Since det(−A) = det(−In A) = det(−In ) det(A) = (−1)n det(A)
∴ det(−A) = det(A), n = 2k, k ∈ N
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4.2.
27.
Let A = (a1 , · · · , ai , · · · , aj , · · · , an ) and B = (a1 , · · · , aj , · · · , ai , · · · , an )
(ai : columns of A and ai = aj )
By the row-interchanging property, we have det(B) = − det(A)
Since ai = aj , det(B) = det(A)
∴ det(A) = 0
28.
(i) E1 is of type 1
det(E1 ) = − det(In ) = −1
(ii) E2 is of type 2
det(E2 ) = k det(In ) = k
(iii) E3 is of type 1
det(E3 ) = det(In ) = 1
29.
(i) E1 is of type 1
det(E1 ) = det(E1t ) = −1
(ii) E2 is of type 2
det(E2 ) = det(E2t ) = k
(iii) E3 is of type 1
det(E3 ) = det(E3t ) = 1
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4.2.
30.
det(B) = det(an , an−1 , · · · , a2 , a1 )
= (−1) det(a1 , an−1 , · · · , a2 , an )
= (−1)2 det(a1 , a2 , an−2 , · · · , a3 , an−1 , an )
..
.
= (−1)[n/2] det(a1 , a2 , · · · , an )
= (−1)[n/2] det(A)
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4.3.
4.3. Properties of Determinants
1.
(a) F (p.223)
(b) T (Theorem 4.7 p. 223)
(c) F
(d) T
(e) F (Theorem 4.8 p. 224)
(f) T
(g) F
(h) F
2.
µ
¶
µ ¶
a11 a12
b
(a) A =
and b = 1
a21 a22
b2
Since det(A) = a11 a22 − a12 a21 6= 0, Cramer’s rule applies
Using the notation of theorem 4.9, we have
x1 = det(M1 )/ det(A) = (b1 a22 − a12 b2 )/(a11 a22 − a12 a21 )
x2 = det(M2 )/ det(A) = (b2 a11 − a21 b1 )/(a11 a22 − a12 a21 )
∴ x = (x1 , x2 )
= ((b1 a22 − a12 b2 )/(a11 a22 − a12 a21 ), (b2 a11 − a21 b1 )/(a11 a22 − a12 a21 ))
3.
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4.3.


 
2 1 −3
5



A = 1 −2 1 , b = 10 , det(A) = −25
3 4 −2
0



5 1 −3
2 5 −3
det(M1 ) = det 10 −2 1  , det(M2 ) = det 1 10 1  , det(M3 ) =
3 0 −2

 0 4 −2
2 1 5

det 1 −2 10
3 4 0
x1 = det(M1 )/ det(A) = −100/ − 25 = 4
x2 = det(M2 )/ det(A) = 75/ − 25 = −3
x3 = det(M3 )/ det(A) = 0/ − 25 = 0
∴ x = (4, −3, 0)
5. (4, −3, 0)
7. (0, −12, 16)
8.
Since det(At ) = det(A),

t

t

t
a1
a1
a1
 .. 
 .. 
 .. 
 . 
 . 
 . 






 ar−1 
ar−1 
ar−1 






from the theorem 4.3, det u + kv  = det  u  + k  v 






 ar+1 
ar+1 
ar+1 
 . 
 . 
 . 
 .. 
 .. 
 .. 
an
an
an
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4.3.
that is,
det(a1 , · · · , ar−1 , u + kv, ar+1 , · · · , an )
= det(a1 , · · · , ar−1 , u, ar+1 , · · · , an ) + k det(a1 , · · · , ar−1 , v, ar+1 , · · · , an )
, u, v, a0i s : column vectors in F n
9.
(⇒) SInce the determinant of an upper triangular matrix is the product of its
diagonal entries
So A is invertible, det(A) 6= 0
thus its diagonal entries are nonzero
(⇐) By hypothesis, det(A) 6= 0
By the corollary to the theorem 4.7, A is invertible
10.
det(M k ) = det(0) = 0
det(M k ) = det(M · · · M ) = det(M )k = 0
∴ det(M ) = 0
11.
(i) n is odd
det(M t ) = (−1)n det(M ) = − det(M )
Since det(M t ) = det(M ), det(M ) = − det(M )
200
PNU-MATH
4.3.
∴ det(M ) = 0
(ii) n is even
det(M t ) = (−1)n det(M ) = det(M )
12.
det(QQt ) = det(Q) det(Qt ) = 1
Since det(Qt ) = det(Q), det(Q) = ±1
13.
(a) det(M ) = det(M )
Let M = A + iB, then
det(M ) = det(A − iB) = det(A) − i det(B) = det(M )
(b)
Since Q is unitary, det(QQ∗ ) = det(I) = 1
1 = det(Q) det(Q∗ )
= det(Q) det(Q∗)
= det(Q) det(Qt )
= det(Q)det(Qt )
= det(Q)det(Q)
=| det(Q) |
14.
201
PNU-MATH
4.3.
(⇐) Since det(B) 6= 0, rank(B) = n
Let B 0 be the reduced row echelon form of B,
then rank(B 0 ) = n
Moreover B 0 = In
By the theorem 3.16, bi ’s consist of a basis for F n , i = 1, · · · , n
(⇒) Since β is a basis, B = In
∴ det B = 1 6= 0
15.
If A, B are similar, then
∃ Q ∈ Mn×n (F ) : invertible s.t. A = Q−1 BQ
∴ det(A) = det(Q−1 BQ)
= det(Q−1 ) det(B) det(Q)
= det(Q)−1 det(B) det(Q)
= det(B)
16.
Suppose that det(A) is not invertible
then det(A) = 0
Since 1 = det(AB) = det(A) det(B) = 0
It’s a contradiction
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4.3.
17.
Since n is odd, det(−B) = (−1)n det(B) = − det(B)
Let AB = −BA
then det(A) det(B) = det(−B) det(A) = − det(B) det(A)
∴ 2 det(B) = 0
Since char(F ) 6= 2, det(B) = 0
∴ B is not invertible
18.
(i) If A is of type 2, then det(A) = k
Since AB is a matrix obtained by multiple of some row of B by the nonzero scalar
k,
det(AB) = k det(B) = det(A) det(B)
(ii) If A is of type 3, then det(A) = 1
Since AB is a matrix obtained by adding a multiple of some row of B to another
row,
det(AB) = det(B) = det(A) det(B)
19.
Let A = (aij ) be an (2 × 2) lower triangular matrix, then det(A) = a11 a22
Proceeding inductively, suppose that this assertion is true for any (k × k) lower
triangular matrix, then
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4.3.
¯
¯
¯
¯ a11 0
¯
¯ a22 0
0
0
¯
¯
¯
¯ a21 a22 0
¯
¯
0 ¯
¯
¯ a32 a33 0
0
0
det(A) = ¯ ..
=
a
A
,
where
A
=
¯
¯ ..
11
.
..
..
¯ .
¯
¯ .
.
¯
¯
¯
¯an1 an2 · · · ann ¯
¯an1 an2 · · ·
Since det(A0 ) = a22 a33 · · · ann , so det(A) = a11 a22 · · · ann
¯
¯
¯
0 ¯¯
¯
¯
¯
ann ¯
20.
Reduce C to upper triangular form with elementary column operations
21.
Reduce C to upper triangular form with elementary row operations
then gain reduce A to upper triangular form with elementary column operations
22.

1 c0
1 c1

(a) [T ]γβ =  .. ..
. .
1 cn
(b) By the exercise

 
c20
a0
 a1 
c21 

 
..  , f =  .. 
.
.
2
2
cn · · · cn
an
22 in section 2.4, T is an isomorphism
c20 · · ·
c21 · · ·
..
.
so M = [T ]γβ is invertible
Thus det(M ) 6= 0
(c)
Proceed the following column operations ;
−C1 × c0 + C2 ⇒ C2
− C2 × c0 + C3 ⇒ C3
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4.3.
..
.
− Cn × c0 + Cn+1 ⇒ Cn+1
¯then we have
¯1
0
0
¯
¯1 (c1 − c0 ) c1 (c1 − c0 )
¯
¯1 (c2 − c0 ) c2 (c2 − c0 )
¯
¯ ..
..
..
¯.
.
.
¯
¯1 (cn − c0 ) cn (cn − c0 )
¯
¯
0
¯
n−1
c1 (c1 − c0 ) ¯¯
cn−1
(c2 − c0 ) ¯¯
2
¯
..
¯
.
¯
n−1
· · · cn (cn − c0 )¯
¯
¯
¯1 c1 c2 · · · cn−1 ¯
1
1
¯
¯
¯1 c2 c2 · · · cn−1 ¯
2
2
¯
¯
= (c1 − c0 )(c2 − c0 ) · · · (cn − c0 ) ¯ .. .. ..
.. ¯
¯. . .
. ¯¯
¯
¯1 cn c2n · · · cnn−1 ¯
¯
¯
¯1
¯
0
0
·
·
·
0
¯
¯
¯1 (c2 − c1 ) c2 (c2 − c1 ) · · · cn−2 (c2 − c1 ) ¯
2
¯
¯
= (c1 − c0 )(c2 − c0 ) · · · (cn − c0 ) ¯ ..
¯
..
..
..
¯.
¯
.
.
.
¯
¯
n−2
¯1 (cn − c1 ) cn (cn − c1 ) · · · cn (cn − c1 )¯
¯
¯
¯1 c2 c2 · · · cn−2 ¯
2
2
¯
¯
¯ .. .. ..
¯
.
.
= (c1 − c0 ) · · · (cn − c0 )(c2 − c1 ) · · · (cn − c1 ) ¯ . . .
¯
.
¯
¯
2
n−2
¯1 cn cn · · · cn ¯
= ···
Q
= 0≤i<j≤n (cj − ci )
···
···
···
23.
a)
So k is the largest number of linearly independent columns of A
thus rankA(A) = k
(b)
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4.3.
Since rank(A) = k, there exists {a1 , a2 , · · · , ak } a linearly independent set of
columns of A
So A can be written as

a11 · · · a1k 0 · · ·
 ..
.. ..
 .
. .

0
.. 
.
an1 · · · ank 0 · · · 0
Since dim(row space A) = dim(column space of A)
Therefore

a11 · · ·
 ..
 .


A = ak1 · · ·



O
a1k
..
.
akk


O 





O 
and det(A) 6= 0
24.


t
0 0 ··· 0
a0
−1 t 0 · · · 0
a1 


 0 −1 t · · · 0
a2 


A + tI =  ..

.. ..
..
..
 .

. .
.
.


0
0 0 ··· t
an+2 
0
0 0 · · · −1 t + an−1
Let A be a (2 × 2) matrix, i.e. n = 2
by cofactor expansion
along
¯
¯ the first column,
¯ t
a0 ¯¯
det(A + tI) = ¯¯
= t2 + a1 t + a0
−1 t + a1 ¯
Assume that this assertion holds for (n − 1),
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4.3.
let A ∈ Mn×n (F )
¯
¯ t
¯
¯−1
¯
¯
det(A+tI) = t ¯ ...
¯
¯0
¯
¯0
¯
¯0 1 0
¯
¯
¯
0 ··· 0
a1 ¯
¯0 t 1
¯
t ··· 0
a2 ¯¯
¯ .. .. ..
¯
¯. . .
..
..
..
+
¯
¯
.
.
.
¯0 0 0
¯
¯
0 ··· t
an−2 ¯¯
¯0 0 0
¯
¯
0 · · · −1 t + an−1
¯a0 a2 a3
n−1
n−2
= t(t
+ an−1 t
+ · · · + a2 t + a1 ) + a0
···
···
0
0
..
.
···
···
···
1
t
an−2
¯
¯
¯
¯
¯
¯
¯
¯
0 ¯¯
1 ¯¯
t + an−1 ¯
0
0
..
.
= tn + an−1 tn−1 + · · · + a2 t2 + a1 t + a0
25.
(a)
| B |=
n
P
cjk Ajk
j=0
c1k A1k + c2k A2k + · · · + cjk + · · · + cnk Ank
= c1k 0 + c2k 0 + · · · + cjk 1 + · · · + cnk 0
= cjk
(b)
Apply Cramer0 srule to Ax = ej
we have 
det(M
 i ) = aji andxi =
 det(Mi )/ det(A) i = 1, · · · , n
x1
c1
 x2 
 c2 
 
 
Since A  ..  = ej , so A  ..  = det(A)ej
.
.
xn
cn
(c)
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4.3.
By (b),
AC = A(c1 ), c2 , · · · , cn )
= (Ac1 , Ac2 , · · · , Acn )
(det(A)e1 , det(A)e2 , · · · , det(A)en )
det(A)(e1 , e2 , · · · , en )
det(A)I
(d)
Since det(B) = det(r1 , · · · , ei , · · · , rn ) = rki
so (r1i , r2i , · · · , rni )A = det(A)ei
Let CA = (r1 , r2 , · · · , rn )A
= (r1 A, r2 A, · · · , rn A)
= (det(A)e1 , det(A)e2 , · · · , det(A)en)
det(A)I
If det(A) 6= 0,
by (b), AC = det(A)I
by the above proof, CA = det(A)I
∴ A−1 = det(A)−1 C
(d)
A = ()n×n , det(adjA) = (det A)n−1
(proof ) A = 0 or rankA = n o.k.
A 6= 0, rankA 6= n
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4.3.
() AX=0
= n − rankA eqn − 1
adjA AX = 0
⇒ adjA
⇒ det(adjA) = 0 = (det A)n−1
In fact,
(1)rankA = n ⇒ rank(adjA) = n
(2)rankA = n − 1 ⇒ rank(adjA) = 1
(3)rankA ≤ n − 2 ⇒ rank(adjA) = 0 i.e. adjA = (0) : zero matrix (?)
So rankA − rank(adjA) ≤ n − 2
26.
(a)
µ
¶
µ
¶
A11 A12
A22 −A12
⇒
A21 A22
−A21 A11
(b)




16 0 0
4 0 0
0 4 0 ⇒  0 16 0 
0 0 16
0 0 4
(c)




−4 0 0
10 0
0
 0 2 0 ⇒  0 −20 0 
0 0 5
0
0 −8
(d)
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4.3.




3 6 7
20 −30 20
0 4 8 ⇒  0 15 −24
0 0 5
0
0
12
(e)




−3i
0
0
1−i
0
0
 4
4
−1 + i
0 
3i
0 ⇒ 
10 + 16i −5 − 3i 3 + 3i
2i 1 + 4i −1
(f)


6 22
12
12 −2 24 
21 −39 −27
(g)


18
28 −6
−20 −21 37 
48
14 −16
(h)


−i
−8 + i −1 + 2i
 1 − 5i 9 − 6i
−3i 
−1 + i
−6
−3 + i
27.
A 6= (0), det(A) = 0
A = (aij )m×n ) = (A1 , · · · , An )
Since det(A) = 0, rank(A) < n
P
∃ k(1 ≤ k ≤ n), A(k) =
bj A(j) , A(j) is the jth column of A
j6=k
Without loss of generality, we may assume k = n
ei(n−1) = b1 B1 + · · · + bn−2 Bn−2 + bn−1 Bn−1 )
(A
n−2
ei(n−1) = P bj Bj + bn−1 A
ein , where Bj = ()(n−1)×(n−1) is the matrix obtained
A
j=1
ein by replacing the last column with the jth column (1 ≤ k ≤ n − 2)
from A
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4.3.

c11 · · ·
c11 · · ·
ei(n−1) ) = 
Actually Ci(n−1) = (det A
 ..
..
 .
.
c11 · · ·


b1 c1n b2 c1n · · · bn−1 c1n c1n
 b1 c2n b2 c2n · · · bn−1 c2n c2n 


=  ..
..
..
.. 
 .
.
.
. 
b1 cnn b2 cnn · · · bn−1 cnn cnn
= (b1 cn , · · · , bn−1 cn , cn )

c1(n−1) c1n
c1(n−1) c1n 

.. 
. 
c1(n−1) c1n
∴ det(adj(A)) = 0
28.



(a) T (y + z) = det 




= det 

(y + z)(t)
(y + z)0 (t)
..
.
(n)
(y)
(t)


T (ky)(t) = det 




= k det 

···
···

yn (t)
yn0 (t) 

.. 
. 
···
(n)
· · · yn (t)
(y + z) (t)
y1 (t) · · · yn (t)
(z)(t)
y1 (t)
 (z)0 (t)
y10 (t) · · · yn0 (t) 
y10 (t)


..
..  + det 
..
..

.
···
. 
.
.
(n)
(n)
(n)
(n)
y1 (t) · · · yn (t)
(z) (t) y1 (t)

k(y)(t)
y1 (t) · · · yn (t)
0
k(y) (t)
y10 (t) · · · yn0 (t) 

..
..
.. 
.
.
···
. 
(n)
(y)(t)
(y)0 (t)
..
.
y1 (t)
y10 (t)
..
.
k(y)(n) (t)
(y)(t)
y1 (t)
(y)0 (t)
y10 (t)
..
..
.
.
(n)
(n)
y1 (t)
···
···
···
···

yn (t)
yn0 (t) 

.. 
. 
(n)
yn (t)
(n)
(n)
y1 (t) · · · yn (t)
· · · yn (t)
· · · yn0 (t) 

.. 
···
. 
(y)(n) (t) y1 (t) · · ·
(n)
yn (t)
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4.3.



(b) Let M (y) = 

(y)(t)
(y)0 (t)
..
.
y1 (t)
y10 (t)
..
.
(n)
(n)
y1 (t)
(y)
(t)
⇔ T (y) = 0
⇔ rankM
(y) = m 

y1

 y10 



⇔  ..  ∈ span(

 . 
···
···
···
···

yn (t)
yn0 (t) 

.. 
. 
(n)
yn (t)

yn
 yn0 




 , · · · ,  .. )
 . 

(n)
(n)
(n)
yn
y2
y1
⇔ y(t) ∈ span(y1 , y2 , · · · , yn )
y2
y20
..
.


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4.4.
4.4. Summary-Important Facts about Determinants
1. (a) T
(b) T
(c) T
(d) F (det(B) = − det(A))
(e) F (det(B) = k det(A))
(f) T
(g) T
(h) F (det(At ) = det(A))
(i) T
(j) T
(k) T
2. (a) 22 (b) −29 (c) 2 − 4i (d) −24 + 6i
3. (a) −12 (b) −13 (c) −12 (d) −13 (e) 22 (f) 4 + 2i (g) −2 (h) 154
4. (a) 36 (b) −100 (c) −49 (d) −10 (e) −28 − i (f) 17 − 3i (g) 95
5 6. 20 and 21 in 4.3
213
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4.4.
214
PNU-MATH
4.5.
4.5. A characterization of the Determinants
1. (a) F
(b) T
(c) T
(d) F (δ(B) = −δ(A))
(e) F (δ(I) = 1)
(f) T (p.238, 239)
2.
Determine all the 1 − linear functions δ : M1×1 (F ) → F
Identity function
3. No

4. No


a1
Let A =  a2  , v = (b1 , b2 , b3 )
a3 + tv
 
 
a1
a1



δ(A) = k 6= k + tv = δ a2 + tδ a2 
a3
v

a1
Let A = a2 + kv  , v = (b1 , b2 , b3 )
a3
 
 
a1
a1



δ(A) = a2 6= a2 + ka2 = δ a2 + kδ v 
a3
a3
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4.5.
5. Yes


 
a1 + kv
a1



a2
(i) δ
= (A11 + kb1 )A23 A32 = A11 A23 A32 + k(b1 A23 A32 ) = δ a2  +
a3
  a3
v

k a2 
a3
 

a1
a1



(ii) δ a2 + kv = A11 (A23 + kb3 )A32 = A11 A23 A32 + k(A11 b3 A32 ) = δ a2  +
a3
  a3
a1
kv
a3


 
a1
a1
(iii) δ  a2  = A11 A23 (A32 + kb2 ) = A11 A23 A32 + k(A11 A23 b2 ) = δ a2  +
a3
  a3 + kv
a1
k a2 
v
6.No

a1
δ  a2  = A11 + A23 + (A32 + kb2 ) 6= (A11 + A23 + A32 ) + k(A11 + A23 + b2 ) =
+ kv  
a3 
a1
a1
δ a2  + k a2 
a3
v
7. Yes


 
a1 + kv
a1



a2
(i) δ
= (A11 + kb1 )A21 A32 = A11 A21 A32 + k(b1 A21 A32 ) = δ a2  +
a3
a3
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4.5.
 
v

k a2 
a3

 
a1
a1



(ii) δ a2 + kv = A11 (A21 + kb1 )A32 = A11 A21 A32 + k(A11 b1 A32 ) = δ a2  +
a3
  a3
a1

k v
a3
 


a1
a1



a2
(iii) δ
= A11 A21 (A32 + kb2 ) = A11 A21 A32 + k(A11 A21 b2 ) = δ a2  +
a3
  a3 + kv
a1
k a2 
v
8.No

 
a1
a1



a2
δ
= A11 (A31 + kb1 )(A32 + kb2 ) 6= A11 A31 A32 + k(A11 b1 b2 ) = δ a2  +
+ kv
a3
a3 
a1
k a2 
v
9.No

 
a1
a1
2
2
2
2
2
2
2
2 2



a2
δ
= A11 A22 (A33 + kb3 ) 6= A11 A22 A33 + k(A11 A22 b3 ) = δ a2  +
+ kv
a3
a3 
a1
k a2 
v
10. Yes
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4.5.


a1 + kv
(i) δ  a2  = (A11 + kb1 )A22 A33 − (A11 + kb1 )A21 A32
a3
= A
A21 A32 + k(b1 A22 A33 − b1 A21 A32 )
11 A22
A33 −A11
a1
v
= δ a2  + k a2 
a3
a3

a1
(ii) δ a2 + kv  = A11 (A22 + kb2 )A33 − A11 (A21 + kb1 )A32
a3
= A
A
A
A21 A32 + k(A11 b2 A33 − A11 b1 A32 )
11 22
 33 −A11
a1
a1



a
=δ
+k v
2
a3
a3


a1

a2  = A11 A22 (A33 + kb3 ) − A11 A21 (A32 + kb2 )
(iii) δ
a3 + kv
= A
A21 A32 + k(A11 A22 b3 − A11 A21 b2 )
11 A22
A33 −A11
a1
a1
= δ a2  + k a2 
a3
v
11.
(i) Corollary 2 to the theorem 4.10(p.241)
Let M = (a1 , a2 , · · · , an )t , a0i s : rows of M
Since rank(M ) < n,
some row of M , say r, is a linear combination of the other rows
That is, ∃ c1 , · · · , cr−1 , cr+1 , · · · , cn ∈ F s.t.
ar = c1 a1 + · · · + cr−1 ar−1 + cr+1 ar+1 + · · · + cn an
If M 0 is obtained from M by adding −ci times row i to row r for each i 6= r,
218
PNU-MATH
4.5.
then row r of M 0 consists entirely of zeros, so δ(M 0 ) = 0
But by the corollary 1 to the theorem 4.10, δ(M 0 ) = δ(M )
∴ δ(M ) = 0
(i) Corollary 3 to the theorem 4.10
By the theorem 4.10 (a), δ(E1 ) = −δ(I)
By the n-linearity, δ(E2 ) = kδ(I)
By the corollary 1 to the theorem 4.10, δ(E3 ) = δ(I)
12. Theorem 4.11
(i) A = E1 ⇒ δ(E1 ) = −1
δ(AB) = −δ(B) = δ(A)δ(B)
(ii) A = E2 ⇒ δ(E2 ) = k
δ(AB) = kδ(B) = δ(A)δ(B)
(ii) A = E3 ⇒ δ(E3 ) = 1
δ(AB) = δ(B) = δ(A)δ(B)
13.
µ ¶
b
∀A ∈ M2×2 (F ), A = (a1 , a2 ), v = 1 , ai : columns of A
b2
(i) det(a1 , a2 + kv) = det(a1 , a2 ) + k det(a1 , v)
(ii) det(a1 + kv, a2 ) = det(a1 , a2 ) + k det(v, a2 )
14.
219
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4.5.
¶
A11 A12
Let A =
, a : rows of A, v = (b1 , b2 )
A21¶ A22 µ i
µ
¶
a1 + kv
A11 + kb1 A12 + kb2
(i) δ
=δ
= (A11 +kb1 )A22 a+(A11 +kb1 )A21 b+
a2
A21
A22
(A12 + kb2 )A22 c + (A12 + kb2 )A21 d
µ ¶
µ ¶
a1
v
δ
+δ
= A11 A22 a + A11 A21 b + A12 A22 c + A12 A21 d + k(A22 b1 a + A21 b1 b +
a2
a2
A22 b2 c + A21 b1 d)
µ
¶
µ
¶
a1
A11
A12
(ii) δ
= δ
= A11 (A22 + kb2 )a + A11 (A21 +
a2 + kv
A21 + kb1 A22 + kb2
kb1 )b + A12 (A22 + kb2 )c + A12 (A21 + kb1 )d
µ ¶
µ ¶
a1
a
δ
+ δ 1 = A11 A22 a + A11 A21 b + A12 A22 c + A12 A21 d + k(A11 b2 a + A11 b1 b +
a2
v
A12 b2 c + A12 b1 d)
µ
15.
(⇐)Exercise 14
µ
¶
A11 A12
(⇒) Let A =
∈ M2×2 (F )
A
21 A22
µ
¶
µ
¶
A11 A12
A11 A12
δ(A) = δ
+δ
A¶
0µ
0¶ A22
21
µ
µ
¶
µ
¶
A11 0
0 A12
A11 0
0 A12
=δ
+δ
+δ
+δ
A21 µ
0
A
0
0
A
21
22
¶
µ
¶
µ 0 ¶A22
µ
¶
1 0
0 1
1 0
0 1
= A11 A21 δ
+ A12 A21 δ
+ A11 A22 δ
+ A12 A22 δ
1 0
1 0
0 1
0 1
Since δ is 2 − linear function,
µ
¶
µ
¶
µ
¶
µ
¶
1 0
0 1
1 0
0 1
δ
=δ
= 0, δ
= 1, δ
= −1
1 0
0 1
0 1
1 0
Let a = 1, b = 0, c = 0, d = −1
then δ(A) = A11 A21 · 0 + A12 A21 · (−1) + A11 A22 · 1 + A12 A22 · 0
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4.5.
16.
If δ(I) = t, then
δ(E1 ) = −t = t det(E1 )
δ(E2 ) = kt = t det(E2 )
δ(E3 ) = t = t det(E3 )
∴ δ(E) = −t det(E)
17.

a1
..
.


a1
..
.


a1
..
.



















(aδ1 + bδ2 ) ar + kv  = (aδ1 ) ar + kv  + (bδ2 ) ar + kv 
 . 
 . 
 . 
 .. 
 .. 
 .. 
an
a
a
 
 
n
  n
a1
a1
a1
a1
 .. 
 .. 
 .. 
 .. 
.
.
.
.
 
 
 
 
a
v
a
= a(δ1  r  + kδ1  ) + b(δ2  r  + kδ2  v )
.
.
.
.
 .. 
 .. 
 .. 
 .. 
a
an
a
a
 n
 
 n
 n
a1
a1
a1
a1
 .. 
 .. 
 .. 
 .. 
.
.
.
.
 
 
 
 
= aδ1  ar  + bδ2  ar  + k(aδ1  v  + bδ2  v )
.
.
.
.
 .. 
 .. 
 .. 
 .. 
an
an
an
an
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4.5.
 
 
a1
a1
 .. 
 .. 
.
.
 
 
= (aδ1 + bδ2 )  ar  + k(aδ1 + bδ2 )  v 
.
.
 .. 
 .. 
an
an
18. V : the set of all m × n matrices with entries from a field F is a vector
space, and W ⊆ V
(i) By the exercise 17,
∀ δ1 , δ2 , δ1 + δ2 ∈ W and cδ1 ∈ W
(ii) By the example in p.238, O ∈ W
19.
Let M = (a1 , a2 , · · · , an )t , a0i s : rows of M
Say ai and aj are identical rows in M
If M 0 is obtained from M by interchanging ai and aj ,
then δ(M 0 ) = −δ(M )
Since ai = aj , δ(M 0 ) = δ(M )
therefore 2δ(M ) = 0
Since char(F ) 6= 2, δ(M ) = 0
20.
In Z2 , 2δ(M ) = 0 ; δ(M ) = 0
222
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