MAT 211 Summer 2015 Homework 4
Solution Guide
Due in class: June 22nd .
Problem 1. Which of the following subsets of P2 are subspaces of P2 ? If it
is not a subspace, explain your reason; if it is a subspace, please find a basis
for it.
a) {p(t) : p(0) = 2};
Answer. No, because the zero polynomial is not in the set.
b) {p(t) : p(2) = 0}.
Answer. Yes. (You check the properties yourself.) Let p(t) = a + bt +
ct2 , then p(t) = a + 2b + 4c = 0. Hence, a = −(2b + 4c). Then
p(t) = −(2b + 4c) + bt + ct2 = b(t − 2) + c(t2 − 4).
The subspace is spanned by (t − 2, t2 − 4), and they are clearly linearly
independent, hence (t − 2, t2 − 4) form a basis of the subspace.
Problem 2. Which of the following subsets of Mat3×3 (R) are subspaces of
Mat3×3 (R)? If it is not a subspace, explain your reason; if it is a subspace,
please find a basis for it.
a) The set of all diagonal 3 × 3 matrices;
b) The set of all 3 × 3 matrices whose entries are all greater than or equal
to 0.
Answer. No, because it is not closed under scalar multiplication (take
a matrix in the set then time it by −1).
Problem 3. Find a basis for each of the following spaces and determine its
dimension.
1
a b
a) The space of all matrices A =
in Mat2×2 (R) such that
c d
a + d = 0;
Answer. This is Question 3 in the midterm 2. Check the solution for
midterm 2.
1 0
b) The space of all 2×2 matrices A such that commute with B =
.
0 2
c) The space of all 3 × 3 matrices A such

0 1
B = 0 0
0 0

a11

Proof. Let A = a21
a31
BA, i.e.,

0
0
0
that commute with

0
1 .
0

a12 a13
a22 a23  and A commutes with B. Then AB =
a32 a33

 
a21 a22 a23
a11 a12
a21 a22  = a31 a32 a33  .
0
0
0
a31 a32
Hence, we have


 ca11 = a22 = a33
a12 = a23
.

 a =a =a =0
21
31
32
Hence,








a11 a12 a13
1 0 0
0 1 0
0 0 1
A =  0 a11 a12  = a11 0 1 0 +a12 0 0 1 +a13 0 0 0 .
0
0 a11
0 0 1
0 0 0
0 0 0
The basis is

 
 

0 1 0
0 0 1 
 1 0 0
0 1 0 , 0 0 1 , 0 0 0 ,


0 0 1
0 0 0
0 0 0
and the dimension is 3.
2
Problem 4. Find out which of the following transformations are linear. For
those that are linear, determine whether they are isomorphisms. Explain
your reason.
a) T (M ) = M + I2 from Mat2×2 (R) to Mat2×2 (R), where I2 is the 2 × 2
identity matrix;
Answer. No, because
T (M1 + M2 ) = (M1 + M2 ) + I2
while
T (M1 ) + T (M2 ) =(M1 + I2 ) + (M2 + I2 )
=(M1 + M2 ) + 2I2 .
They are not equal.
a b
b) T
= ad − bc from Mat2×2 (R) to R;
c d
Answer. No, because
a b
ka kb
T k
=T
c d
kc kd
= (ka)(kd) − (kb)(kc)
= k 2 (ad − bc)
a b
2
=k T
.
c d
For any k 6= 0, 1, the linearity does not hold.
f (0) f (1)
c) T (f (t)) =
from P3 to Mat2×2 (R), where P3 is the space
f (2) f (3)
of all polynomials with degree smaller than or equal to 3.
Proof. Yes, it is a linear transformation. (Check the authenticity!) T
is an isomorphism because
A) dim(P3 ) = 4 = dim Mat2×2 (R);
3
B) ker(T ) = {0}. If T (f (t)) is the zero matrix in Mat2×2 (R), then
f (0) = f (1) = f (2) = f (3) = 0. Since ant nonzero polynomial in
P3 has at most 3 zeros, f can only be the zero function.
Problem 5. For which constants k is the linear transformation
2 3
3 0
T (M ) =
M −M
0 4
0 k
an isomorphism from Mat2×2 (R) to Mat2×2 (R)?
Answer. Given the standard basis B on Mat2×2 (R). Suppose
 
a
b

[M ]B = 
c ,
d
a b
then M =
.
c d
2 3 a b
a b 3 0
T (M ) =
−
0 4 c d
c d 0 k
2a + 3c 2b + 3d
3a kb
=
−
4c
4d
0 kd
−a + 3c (2 − k)b + 3d
=
.
4c
(4 − k)d
4
Hence,


−a + 3c
(2 − k)b + 3d

[T (M )]B = 


4c
(4 − k)d
 


 


−1
0
3
0
0


 


 + b 2 − k  + c 0 + d  3 
= a
0
 0 
4
 0 
0
0
0
4−k

 
−1
0
3
0
a
 0 2−k 0
 b
3
 .
=
0
0
4
0  c
0
0
0 4−k
d
The matrix of T under the standard basis B is


−1
0
3
0
 0 2−k 0
3 
.
A=
0
0
4
0 
0
0
0 4−k
Then T is an isomorphism if and only if the matrix A is invertible, if and
only if the rank of the matrix is 4. Hence, k 6= 2 and k 6= 4.
Problem 6. Find the image and the kernel of the following linear transformations. Write your answer as a span of “vectors”. What is the rank and
nullity of each linear transformation? Use your result to test the Rank-Nullity
Theorem.
1 2
a) T (M ) = M
from Mat2×2 (R) to Mat2×2 (R);
3 6
b) T (f (t)) = f (7) from P2 to R;
Answer. Given P2 the basis B = (1, x, x2 ), then for any polynomial
a + bx + cx2 , then
 
a
T (f (X)) = f (7) = a + 7b + 49c = 1 7 49  b  .
c
5
Hence, the matrix of T under the basis is
A = 1 7 49 .
The image is span{1}, which is indeed all the real numbers. (Note that
T maps P2 to R.) The kernel of A is
  

−7
−49
ker(A) = span  1  ,  0  ,
0
1
namely,
ker(T ) = span −7 + x, −49 + x2 .
It is clear that dim(im(T )) = 1, dim(ker(T )) = 2 and
dim(im(T )) + dim(ker(T )) = 3 = dim(P2 ).
1 2
1 2
c) T (M ) = M
−
M from Mat2×2 (R) to Mat2×2 (R).
0 1
0 1
Problem 7. Determine whether the following “vectors” in the given linear
space are linearly independent or not. Explain your answer and show all your
work.
a) The polynomials f (t) = 7+3t+t2 , g(t) = 9+9t+4t2 and h(t) = 3+2t+t2
in the linear space P2 ;
Answer. Given P2 the standard basis B = (1, x, x2 ), then
 
 
 
7
9
3





[f ]B = 3 , [g]B = 9 , [h]B = 2 .
1
4
1
Since


7 9 3
rref 3 9 2 = I3 ,
1 4 1
The coordinate vectors are linearly independent. Hence, f, g and h are
linearly independent in P2 .
6
b) The matrices
1 1
1 1
,
1 2
3 4
2 3
1 4
,
,
5 7
6 8
in the linear space Mat2×2 (R).
(Hint: you can first give a basis to the linear space, then you just need to
show whether the coordinate vectors are linearly independent or not.)
Answer. The same idea as a). The matrices turn out to be linearly dependent.
Problem 8. Let T be a linear transformation from Mat2×2 (R) to Mat2×2 (R)
defined by
1 2
T (M ) = M
.
3 6
Find the matrix of T with respect to the following basis
1 0
0 1
1 0
0
B=
,
,
,
−1 0
0 −1
2 0
0
of Mat2×2 (R)
1
.
2
Answer. (Note that the basis hereisnot the standard basis.) given any
a
b

M ∈ Mat2×2 (R), suppose [M ]B = 
 c , then
d
1 0
0 1
1 0
0 1
a+c b+d
M =a
+b
+c
+d
=
.
−1 0
0 −1
2 0
0 2
2c − a 2d − b
Then
a+c b+d 1 2
T (M ) =
2c − a 2d − b 3 6
(a + c) + 3(b + d)
2(a + c) + 6(b + d)
=
(2c − a) + 3(2d − b) 2(2c − a) + 6(2d − b)
(a + 3b) + (c + 3d) (2a + 6b) + (2c + 6d)
=
2(c + 3d) − (a + 3b) 2(2c + 6d) − (2a + 6b)
1 0
1 0
0 1
0 1
= (a + 3b)
+ (c + 3d)
+ (2a + 6b)
+ (2c + 6d)
.
−1 0
2 0
0 −1
0 2
7
Hence,

 
a + 3b
1
2a + 6b 2
 
[T (M )]B = 
 c + 3d  = 0
2c + 3d
0
3
6
0
0
0
0
1
2
Therefore, the matrix of T with respect

1 3 0
2 6 0

0 0 1
0 0 2
  
0 a
1




0  b  2
=
3  c  0
6
d
0
3
6
0
0
0
0
1
2

0
0
 [M ]B .
3
6
to the given basis B is

0
0
.
3
6
Problem 9. Let V be the space of all upper triangular 2 ×
Consider the linear transformation
a b
T
= aI2 + bP + cP 2
0 c
1
from V to V , where I2 is the 2 × 2 identity matrix, P =
0
represents the product of P with itself.
a) Find the matrix A of T with respect
1 0
0
B=
,
0 0
0
to the basis
1
0 0
,
.
0
0 1
Answer. First, we have
 
 
1
1



[I2 ]B = 0 , [P ]B = 2 ,
1
3
1 8
and P =
hence
0 9
2
 
1
[P 2 ]B = 8 .
9
8
2 matrices.
2
and P 2
3
 
a
a b

Let M =
, then [M ]B = b .
0 c
c
[T (M )]B = a[I2 ]B + b[P ]B + c[P 2 ]B
 
 
 
1
1
1





= a 0 + b 2 + c 8
1
3
9


1 1 1

= 0 2 8 [M ]B .
1 3 9
Hence, the matrix of T under the basis B is


1 1 1
0 2 8 .
1 3 9
b) Find bases of the image and kernel of T , and thus determine the rank
of T .
Proof.
   
1
1
im(A) = span 0 , 2 ,
1
3
hence,
1 0
1 2
im(T ) = span
,
.
0 1
0 3
Rank(T ) = dim(im(T )) = 2.


3
ker(A) = span −4 ,
1
hence,
3 −4
ker(T ) = span
.
0 1
9
Problem 10. In the plane V defined by the equation 2x1 + x2 − 2x3 = 0,
consider the bases
   
1
2




2 , −2
U = (~a1 , ~a2 ) =
2
1
and
   
1
3
~
~




2 , 0 .
B = (b1 , b2 ) =
2
3
a) Find the change of basis matrix S from B to U;
Answer. Since ~b1 = ~a1 , ~b2 = ~a1 + ~a2 , we have
1
1
~
~
[ b1 ] U =
, [ b2 ] U =
.
0
1
Hence, the change of basis matrix S from B to U is
h
i 1 1 S = [~b1 ]U [~b2 ]U =
.
0 1
b) Find the change of basis matrix from U to B;
Answer. The matrix is
S
−1
1 −1
=
.
0 1
c) Write an equation relating the matrices [~a1 ~a2 ], [~b1 ~b2 ], and
S = SB→U .
Answer.
h
i ~b1 ~b2 = ~a1 ~a2 S.
10