Homework #4
May 30, 2011
1. For which sides (find a condition on b1 , b2 , b3 ) are these systems solvable?
1
(a) 2
1
4
8
4
1
(b) 2
1
b1
4 x
9 1 b2 b3
4 x2
2
x1
b1
4 x2 b2 2 x3
b3
Solution:
(a)
1
2
1
4
8
4
2
4
2
b1
1
b2 Ñ 0
b3
0
Therefore, the linear system is solvable when b2
b2 2b1 and b3 b1 .
(b)
1
2
1
4
0
0
b1
b2 2b1 b3 b1
2b1 0 and b3
4 b1
1 4
9 b2 Ñ 0 1
4 b3
0 0
Therefore, the linear system is solvable when b3
2
0
0
b1 .
b1
0, in other words,
b1
b2 2b1 b3 b1
2. Suppose Ax b and Ay c are both solvable. Then Az b c is solvable. What is z? This translates
into: If b and c are in the column space colpAq, then b c is in colpAq.
Solution: If we add two linear systems, we get
Ap x
therefore, z
x
yq b
c
y.
3. True or false (with a counterexample if false):
(a) The vectors b that are not in the column space colpAq form a subspace.
(b) If colpAq contains only the zero vector, then A is the zero matrix.
(c) The column space of 2A equals the column space of A.
1
(d) The column space of A I equals the column space of A.
Solution:
(a) False, since any subspace should have zero vector. (Any matrix can be a counterexample.)
(b) True, since if A contains any nonzero column vector, that vector should be in colpAq.
(c) True. The proof is as follows.
• colp2Aq € colpAq.
Let x P colp2Aq, i.e., x p2Aqy with some y. Then x Ap2y q therefore x P colpAq.
• colpAq € colp2Aq.
Let x P colpAq, i.e., x Ay with some y. Then x p2Aqpy {2q therefore x P colp2Aq.
(d) False. For A I, colpAq Rn , but colpA I q colpOq t0u.
4. Suppose column 1 + column 3 + column 5 = 0 in a 4 by 5 matrix with four pivots. Which column is
sure to have no pivot (and which variable is free)? What is the special solution? What is the nullspace?
Solution:
• Since the matrix has five columns and four pivots, only one column has no pivot.
Suppose the column 5 has a pivot. Let the k-th entry of the column 5 has the leading (nonzero)
entry. Obviously, the k-th entry of the column 1 & 3 are all zeros. (Otherwise it cannot be a
leading entry, by definition.) Therefore, the column 5 cannot have a pivot.
• Let the matrix A. Then
1
0
A
1
0
column 1 + column 2 + column 3 0
1
therefore p1, 0, 1, 0, 1q is a special solution.
(The special solution here means a nonzero solution of the homogeneous linear system. Since
this is not explained, this part won’t be graded.)
1
• nullpAq 0
span 1
0
1
5. Why does no 3 by 3 matrix have a nullspace that equals its column space?
Page 2
Solution: Let A be a 3 3 matrix. By rank theorem,
rankpAq
nullitypAq 3.
Since rankpAq and nullitypAq are all positive integers, rankpAq cannot be the same as nullitypAq
therefore colpAq nullpAq.
6. If the nullspace of A consists of all multiples of x p2, 1, 0, 1q, how many pivots appear in U (row echelon
form)? What is R (reduced row echelon form)?
Solution: Let the last entry of x be the free variable. Then,
x1
2x4 ,
x2
x4 ,
0,
x3
and x4
t.
In other words, the homogeneous linear system
Ax 0
can be reduced to the reduced row echelon form (R)
1
0
0
0
..
.
0
1
0
0
..
.
0
0
1
0
..
.
0
0
0
2
1
.
0
0
..
.
0
Therefore, there are three pivots in U .
7. Find a basis for each of the three subspace associated with
0
A 0
0
Solution: Let
1
1
0
1
L : 1
0
2
2
0
3
4
1
0
1
1
1
4
6 1
0
2
0
0
0 0
0
1
0
1
1
0
0
0 and U : 0
1
0
1
0
0
2
0
0
1
0
0
2
0
0
3
1
0
3
4 , 0
3
1
0
4
2
0
4
2 .
0
• Row space
rowpAq rowpU q span
0
1
2
• Column space
0
Since the basis of colpU q is composed of its 2nd and 4th columns,
1
3
colpAq span 1 , 4.
0
1
Page 3
0
1
2
.
• Null space
If we further reduce U ,
0
0
0
1
0
0
2
0
0
3
1
0
4
0
2 Ñ 0
0
0
1
0
0
2
0
0
2
0
1
0
2
0
Therefore, the solution of the homogeneous system Ax 0 is
#
2x3 2x5
2x5 0
x2
x4
Therefore,
0
2 x5 0 .
2
1
0
2
x3 1 0 0
1
2x3 2x5 0
Ñ x x3 x1 0
2x5
0
x5
0
0
x1
1
0
0
0 2 2 nullpAq span 0 , 1 , 0 .
0 0 2
0
0
1
8. What are the dimensions of the three subspaces for A, B, and C if I is the 3 by 3 identity matrix and
O is the 3 by 2 zero matrix?
A I
O
and B
I
OT
I
OT
and C
0
0
O .
Solution:
• rowpAq span
1
0
0
0
0 , 0
1
0
0
0 , 0
0
1
Therefore,
dimprowpAqq dimpcolpAqq rankpAq 3 and dimpnullpAqq 5 rankpAq 2.
• rowpB q span
1
0
Therefore,
• rowpC q t 0
Therefore,
0
1
0
0 , 0
1
0
0
1
0 , 0
0
1
0
0
1
dimprowpB qq dimpcolpB qq 3 and nullpB q 6 3 3.
0
u
dimprowpC qq dimpcolpC qq 0 and nullitypC q 2 0 2.
9. A is an m by n matrix of rank r. Suppose there are right sides b for which Ax b has no solution.
(a) What are all inequalities ( or
(b) How do you know that A
T
¤) that must be true between m, n, and r?
y 0 has solutions other than y 0?
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Solution:
(a) Obviously, r cannot be larger than m and n. If r m, rankpAq r dimpcolpAqq therefore
colpAq Rr Rm . In other words, for any b P Rm there exists a linear combination of the
columns of A such that Ax b. Therefore, r should be strictly smaller than m: r m.
(b) By the rank theorem,
rankpAT q
Since r
nullitypAT q r
nullitypAT q m.
m by (a), nullitypAT q ¡ 0 therefore there is a non-trivial solution for AT y 0.
10. Without multiplying matrices, find bases for the row and column spaces of A:
1
A 4
2
2 3
5
1
7
0
1
3
.
2
How do you know from these shapes that A is not invertible?
Solution: Let
2
3
5 and C :
1
7
1
B : 4
2
0
1
3
.
2
• rowpAq
We first show that rowpAq rowpC q.
– rowpAq € rowpC q
For any y P rowpAq, there exists x P R3 such that xA y. Therefore,
y
xA xpBC q pxB qC
P rowpC q.
– rowpC q € rowpAq
For any y P rowpC q, there exists x P R2 such that xC y. On the other hand, zB x
always has a solution since rowpB q R2 . Therefore,
y xC pzB qC z pBC q zA
hence y P rowpAq.
Now, to find the basis of rowpC q,
hence y
Therefore,
3
1
0
1
3
2
Ñ
1
1
0
1
rowpAq rowpC q span
• colpAq
We first show that colpAq colpB q.
Page 5
1
2
1
Ñ
0
1
0
0
1
1 , 0
1
.
1
1
1
.
– colpAq € colpB q
For any y P colpAq, there exists x P R3 such that Ax y. Therefore,
Ax BCx B pCxq
y
P colpB q.
– colpB q € colpAq
For any y P colpB q, there exists x P R2 such that Bx y. On the other hand, Cz x
always has a solution since colpC q R2 . Therefore,
y Bx B pCz q pBC qz Az
hence y P colpAq.
Now, to find the basis of colpB q,
hence y
1
4
2
1
2
5 Ñ 0
0
7
1
2
3 Ñ 0
0
3
Therefore,
2
1
0
2
1
colpAq colpB q span 4 , 5.
7
2
A is not invertible since, while A is a 3 3 matrix, rankpAq rankpB q 3.
11. (Fill (a),(b), and (c) below.)
If AB C, the rows of C are combinations of the rows of (a). So the rank of C is not greater than the
rank of (b). Since B T AT C T , the rank of C is also not greater than the rank of (c).
Solution:
(a) B
(b) B
(c) A
12. Which of these transformations satisfy T pv
(a) T pv q v {kvk
(b) T pv q v1
v2
w q T pv q
v3
(c) T pv q pv1 , 2v2 , 3v3 q
(d) T pv q largest component of v
Solution:
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T pwq and which satisfy T pcv q cT pv q?
(a) Satisfies neither.
p3, 4q and w p1, 0q,
?
?
T pv wq T pp4, 4qq p4, 4q{4 2 p1, 1q{ 2 T pv q T pwq p3, 4q{5 p1, 0q p8{5, 4{5q
• For v
• For v
(b) T pv q 1
1
(c) T pv q 0
0
p1, 1q and c 2,
?
?
? ?
T pcv q p2, 2q{2 2 p1, 1q{ 2 cT pv q 2p1, 1q{ 2 2p1, 1q
1
1 v: Satisfies both.
0
2
0
0
0 v: Satisfies both.
3
(d) Satisfies neither.
• For v
• For v
p3, 2, 0q and w p0, 2, 3q,
T pv wq 4 T pv q
p1, 2q and c 1,
T pwq 3
36
cT pv q 2 T pcv q 1
13. Suppose T is reflection across the x axis and S is reflection across the y axis. The domain V is the xy
plane. If v px, y q what is S pT pv qq? Find a simpler description of the product ST .
Solution:
therefore
rT s 1
0
S pT pv qq rST sv
0
1
1
and rS s 0
0
1
rS srT sv 01 01
v
xy
.
Therefore, ST is a reflection with respect to the origin.
14. Suppose T is reflection across the 450 line, and S is reflection across the y axis. If v
T pv q p1, 2q. Find S pT pv qq and T pS pv qq. This shows that generall ST T S.
Solution:
therefore
and
rT s 0
1
S pT pv qq rST sv
T pS pv qq rT S sv
1
0
and rS s rS srT sv rT srS sv Page 7
0
1
0
0
1
0
1
1
1
v
0
1
0
y
x
y
v
x
.
p2, 1q then
15. Let l1 and l2 are lines through the origin. And let T1 and T2 are reflections across the line l1 and l2 ,
respectively. Show that the product T1 T2 is a rotation. (Hint: See problem 26 on p.222 of our textbook.)
Solution: Let θ be the angle with respect to the x-axis and Rθ be the rotation with respect to the
origin by θ. Then, the reflection across the line with its angle θ, Tθ , is defined by
rTθ s rRθ s
1
0
cos θ
sin θ
0
1
rRθ s
sin θ
cos θ
cos θ
sin θ
sin θ
cos θ
cos2 θ sin2 θ
2 sin θ cos θ
cos 2θ
sin 2θ
1
0
0
1
cos θ
sin θ
cospθq
sinpθq
sin θ
cos θ
2 sin θ cos θ
sin2 θ cos2 θ
sinpθq
cospθq
sin 2θ
cos 2θ .
Now let θ1 and θ2 be the angles of l1 and l2 , respectively. Then, by the trigonometric identities
(http://en.wikipedia.org/wiki/List_of_trigonometric_identities),
rT1 T2 s rTθ Tθ s rTθ srTθ s
2θ1
sin 2θ1
cos 2θ2
sin 2θ2
cos
sin 2θ1 cos 2θ1 sin 2θ2 cos 2θ2
2θ1 cos 2θ2 sin 2θ1 sin 2θ2 cos 2θ1 sin 2θ2 sin 2θ1 cos 2θ2
cos
sin 2θ1 cos 2θ2 cos 2θ1 sin 2θ2 sin 2θ1 sin 2θ2 cos 2θ1 cos 2θ2
cospθ1 θ2 q sin 2pθ1 θ2 q
sin
2pθ1 θ2 q
cospθ1 θ2 q
rRθ θ s.
1
1
2
1
2
2
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