MATH 314 Fall 2011
Section 001
Practice Final - Solutions
1. Check your notes or book for answer to problem 1.
2. Give examples of:
(a) Four different vector spaces each having dimension 10.
Solution: R10 , P9 , M2×5 , M5×2 , M1×10 , M10×1 , the set of symmetric
4 × 4 matrices, etc.
(b) A set of linearly independent polynomials in P2 which is not a basis.
Solution: {1, x} is a set of two linearly independent polynomials in P2
which is not a basis of P2 because any basis of P2 has 3 elements whereas
the set {1, x} only has 2 elements. Your example may be different.
(c) A basis of P2 obtained by extending the set in part (b).
Solution: {1, x, x2 } is the standard basis that extends the independent
set {1, x}. Your example may be different.
M2×2 but do not form a basis.
1
0 0
0 0
1 1
,
,
,
} is a set of ma0
1 0
0 1 0 0 1 0
0 1
0 0
0 0
trices that span M2×2 because in fact the subset {
,
,
,
}
0 0
0 0
1 0
0 1
is enough to span M2×2 . A is not a basis of M2×2 beacause any basis
of M2×2 has 4 elements, whereas A has 5 elements.
(d) A set of matrices that span
1 0
0
Solution: A = {
,
0 0
0
Your example may be different.
(e) A subspace W of M2×2 of such that the dimension of W is 2.
a b
1 0
0 1
Solution: W =
, a, b ∈ R = Span
,
or
b a
0 1
1 0
a 5a + b
1 5
0 1
W =
, a, b ∈ R = Span
,
. Because
7b
a
0 1
7 0
the given sets are bases (you need to check this!) these are 2-dimensional
subspaces of M2×2 . Your example may be different.
1


1 −1 0
−3 1 −1


.
5
2
4
3. Let A = 


 0 −2 −1
5
3
5
(a) Find bases for Row(A) and N ull(A).
Solution:




1 −1 0
1 0 0
−3 1 −1
0 1 0

 RREF


 −→ R = 0 0 1
5
2
4
A=




 0 −2 −1
0 0 0
5
3
5
0 0 0
so a basis for Row(A) is 1 0 0 , 0 1 0 , 0 0 1 (we can see
from here that Row(A) = R3 ). N ull(A) = (Row(A))⊥ = (R3 )⊥ = {~0}
because the only vector in R3 that is orthogonal to the whole R3 is ~0.
(b) Find bases for Col(A) and N ull(AT ). Solution:




1 −3 5 0 5 | 0
1 0 0 3 4 | 0
RREF
[AT |0] = −1 1 2 −2 3 | 0 −→ 0 1 0 1 3 | 0
0 −1 4 −1 5 | 0
0 0 1 0 2 | 0
   



−3
−3s
−
4t
−4 











−1 −3





 −s − 3t 






T






hence N ull(A ) =  −2t  , s, t ∈ R = Span  0  , −2







 1   0 


s












0
t
1
T
and this is in fact a basis. A basis
for
 Col(A)
   isabasis for Row(A )
0
0 
 1











0 1 0

    
so a possible choice of basis is 
0 , 0 , 1. (Another choice


3 1 0






4
3
2
would be the three original columns of A.)
(c) Find the rank of A and the nullity of A. Check the rank theorem.
Solution: rank(A) = 3 and nullity(A) = dim(N ull(A)) = 0. The rank
theorem for A asserts: rank(A) + nullity(A) = 3 and indeed 3 + 0 = 3.
2
(d) Find the rank and the nullity of AT . Check the rank theorem for AT .
Solution: rank(AT ) = rank(A) = 3 and nullity(AT ) = dim(N ull(AT )) =
2. The rank theorem for AT asserts: rank(AT ) + nullity(AT ) = 5 and
indeed 3 + 2 = 5.

 
 
2
−1 




4. Consider W = Span ~x1 = 1 , ~x2 = 1  as a subspace of R3 .


1
−2
(a) Find dim(W ) and say what W is as a geometric object.
   
2
−1



Solution: the two vectors 1 , 1  are linearly independent (not a
1
−2
   
2
−1



multiple of each other) and span W , therefore the set 1 , 1  is a
1
−2
basis of W and so dim(W ) = 2. Any 2-dimensional subspace of R3 is
a plane.
(b) Find and orthonormal basis for W .
Solution: Apply Gramm-Schmidt to find an orthogonal basis:
  
 

 
2
−1
0
2
·~
x2
1 =  3/2 .
~v =  1  − −3
~v1 = ~x1 = 1; ~v2 = ~x2 − ~~vv11 ·~
v1 1
6
1
−2
−3/2
1
Then normalize:
 
 
2
0
~v2
~v1
1  
1  
~q1 =
1
= √ 1 ; ~q2 =
=√
||~v1 ||
||~v2 ||
6 1
2 −1
 
1

(c) Find the projection of the vector ~v = 2 onto W .
0
Solution:
√
 
  

8/
6 √
2
0
√
4
2
projW (~v) = (~q1 ·~v)~q1 +(~q2 ·~v)~q2 = √ 1+ √  1  = 4/√6 + 2/√2
6 1
2 −1
4/ 6 − 2/ 2
3
(d) Find the distance from the point (1, 2, 0) to the subspace W .
Solution: The required distance is d = ||perpW (~v)||.
From the Pithagorean Theorem ||perpW (~v)||2 + ||projW (~v)||2 = ||~v||2 .
4
2
4
2
4
2
||projW (~v)||2 = || √ ~q1 + √ ~q2 ||2 = || √ ~q1 ||2 +|| √ ~q2 ||2 +2 < √ ~q1 , √ ~q2 >
6
2
6
2
6
2
4
8
14
16
||~q1 ||2 + ||~q2 ||2 + 0 = + 2 = .
6
2
3
3
14
1
1
d2 = ||perpW (~v)||2 = ||~v||2 − ||projW (~v)||2 = 5 −
= , so d = √ .
3
3
3
=
(e) Describe W ⊥ (the orthogonal complement of W ) by giving a basis for
W ⊥ and describing what it is as a geometric object.
Solution: We know dim(W ) + dim(W ⊥ ) = 3, so dim(W ⊥ ) = 1 and
⊥
weget that
a line. A basis for W ⊥ is given by {perpW (~v)} =
√ W is



4
6
8/


√
√
√
√
4/ 6 + 2/ 2 or any multiple of it like 6 perpW (~v) = 2 + 3.
2
√
√
√


2− 3
4/ 6 − 2/ 2
(f) Write the linear equation for which the solution set is W .
Solution: Use the normal vector perpW (~v) to write the normal form of
the equation for the plane W .
√
√
√
√
8
√ x + (4/ 6 + 2/ 2)y + (4/ 6 − 2/ 2)z = 0
6
or
4x + (2 +
√
√
3)y + (2 − 3)z = 0
5. For this problem
R 1 we work in the vector space P2 with the inner product
hP (x), Q(x)i = 0 P (x)Q(x)dx.
(a) Show that the polynomials P (x) = 1 + x, Q(x) = 1 − x, R(x) = x2 are
linearly independent.
Solution:
c1 P (x) + c2 Q(x) + c3 R(x) = 0
4
c1 (1 + x) + c2 (1 − x) + c3 (x2 ) = 0
(c1 + c2 ) + (c1 − c2 )x + c3 x2 = 0


 c1 + c2 = 0
c 1 = 0
c1 − c2 = 0 =⇒ c2 = 0


c3
= 0
c3 = 0
Since the only solution to c1 P (x) + c2 Q(x) + c3 R(x) = 0 is the trivial
one, the polynomials P (x), Q(x), R(x) are linearly independent.
(b) Show that Span(P, Q, R) = P2 . (Use the same P, Q, R as in part (a)).
Solution:
c1 P (x) + c2 Q(x) + c3 R(x) = ax2 + bx + c
c1 (1 + x) + c2 (1 − x) + c3 (x2 ) = ax2 + bx + c
(c1 + c2 ) + (c1 − c2 )x + c3 x2 = ax2 + bx + c


 c1 + c2 = c
c1 = b+c
2
c1 − c2 = b =⇒ c2 = b−c
2


c3
= a
c3 = a
This shows that any element ax2 + bx + c of P2 can be written as a
linear combination b+c
P (x) + b−c
Q(x) + aR(x).
2
2
(c) Explain why {P, Q, R} is a basis of P2 .
Solution: From part (a) {P, Q, R} is a linearly independent set. From
part (b) Span{P, Q, R} = P2 , hence {P, Q, R} is a basis of P2 .
(d) Find an orthogonal basis of P2 with respect to the inner product described in the beginning of the problem.
Solution: Use Gramm-Schmidt applied to the basis P (x) = 1+x, Q(x) =
1 − x, R(x) = x2 .
~v1 = P (x) = 1 + x
R1
(1 + x)(1 − x)dx
< ~v1 , Q(x) >
~v1 = 1−x− 0 R 1
~v2 = perp~v1 Q(x) = Q(x)−
(1+x) =
< ~v1 , ~v1 >
(1 + x)2 dx
0
=1−x−
2 2
5 9
2/3
(1 − x) = 1 − x − − x = − x
7/3
7 7
7 7
5
Rescale to v~0 2 = 5 − 9x.
< ~v1 , R(x) >
< v~0 2 , R(x) > ~0
~v1 −
v2=
< ~v1 , ~v1 >
< v~0 2 , v~0 2 >
R1 2
R1 2
x
(1
+
x)dx
x (5 − 9x)dx
= x2 − R0 1
(1 + x) − R0 1
(5 − 9x) =
(1 + x)2 dx
(5 − 9x)2 dx
0
0
~v3 = R(x) −
= x2 −
7/12
−7/12
1
1
1
(1+x)−
(5−9x) = x2 − (1+x)+ (5−9x) = x2 −x+
7/3
7
4
12
6
(e) Find the coordinates of the polynomial T (x) = x2 + x + 1 with respect
to the orthogonal basis in (d).
Solution:

[T ]B =

<T (x),~
v1 >
<~
v1 ,~
v1 >
 <T (x),v~0 2 > 


 <v~0 2 ,v~0 2 > 
<T (x),~
v3 >
<~
v3 ,~
v3 >
 35/12 
=
7/3
 −7/12 
 7 
1/180
1/180

5
4

1 
= − 12
1
6. True or false? If true give a proof if false say why it fails to be true.
a b
(a) W =
with a + c = b + d is a subspace of M2×2 with the
c d
usual addition and scalar multiplication.
Solution: True. We check that W is closed under addition and scalar
multiplication.
0 0
a b
a b
Let A =
, B = 0 0 be elements of W , so that a + c =
c d
c d
a + a0 b + b 0
0
0
0
0
b + d, a + c = b + d . Then A + B =
has the property
c + c0 d + d0
(a + a0 ) + (c + c0 ) = (b + b0 ) + (d + d0 ) obtained by adding together the
previous two equalities. Therefore A + B is in turn in W .
ka kb
Let k be a scalar. kA =
has the property ka + kc = kb + kd
kc kd
(obtained from a + c = b + d) which means kA is in W .
6
(b) W = {A ∈ M2×2 with det(A) = 1} is a subspace of M2×2 with the
usual addition and scalar multiplication.
Solution:
False.
W isnot closed under addition. For example A =
1 0
−1 0
,B =
are in W (since det(A) = 1, det(B) = 1), but
0 1
0 −1
A + B = 02×2 is not in W since det(02×2 ) = 0 6= 1.
(c) B = {1 + x, 2 − x + x2 , 3x − 2x2 , −1 + 3x + x2 } is a basis for P2 .
Solution: False. Since dim(P2 ) = 3 any basis of P2 must have 3 elements, but B has 4 elements.
(d) If V and W are subspaces of R3 then
V ∩ W = the set of vectors that are both in V and in W
is a subspace of R3 .
Solution: True. We check that V ∩ W is closed under addition and
scalar multiplication.
Let ~u, ~v be vectors in V ∩ W . Since ~u, ~v are in V , ~u + ~v is in V . Since
~u, ~v are in W , ~u + ~v is in W . Therefore ~u + ~v is in V ∩ W .
Let ~u be a vector in V ∩ W and c a scalar. Since ~u is in V , c~u is in V .
Since ~u is in W , c~u is in W . Therefore c~u is in V ∩ W .
(e) If V and W are subspaces of R3 then
~ with ~v ∈ V, w
~ ∈ W}
V + W = {~v + w
is a subspace of R3 .
Solution: True. We check that W is closed under addition and scalar
multiplication.
~ 0 be vectors in V + W . Then their sum (~v + w
~ , v~0 + w
~)+
Let ~v + w
~ 0 ) = (~v + v~0 ) + (~
~ 0 ) is in V + W because ~v + v~0 is in V and
(v~0 + w
w+w
~ 0 is in W .
~ +w
w
~ be a vector in V +W and c a scalar. Then c(~v + w
~ ) = c~v +c~
Let ~v + w
w
is in V + W since c~v is in V and c~
w is in W .
7