Math 2270 Spring 2004
Homework 12 Solutions
Section 4.3) 2, 4, 7, 8, 12, 20, 29, 34, 36, 42, 54, 61
(2) We will use the standard basis for 2 x 2 matrices (as given for problems 5 to 33).
"
1 1
1 1
#
B
1
1
1
1



=


"


,

1 2
3 4

#


=

B
1
2
3
4

"


,

2 3
5 7

#
B


=

2
3
5
7


"
1 4
6 8
0
1
0
0
0 −1
0 4 


1 −1 
0 0


,

#
B


=

We must determine if the four vectors in ℜ4 are linearly independent.





1
1
1
1
1
2
3
4
2
3
5
7
1
4
6
8








→

1
0
0
0
1
1
2
3
2
1
3
5
1
3
5
7








→

1
0
0
0
1
1 −2
 0
1 3 


→
 0
1 −1 
0
2 −2
0
1
0
0


1
4
6
8






The rref of the above matrix is not equal to the identity, so the column vectors are not
linearly independent. The linear transformation from ℜ2x2 into its coordinate vectors with
respect to the standard basis is an isomorphism. Therefore, since the four coordinate vectors
are not independent, the four matrices are not linearly independent.
(4) We will use the standard basis for P2 , {1, t, t2 }. Then,


1


[t + 1]B =  1  ,
0
h
i
t2 + t
B


0


=  1 ,
1
h
t2 + (2 + k)t + 2k
i
B
We must determine if the three vectors in ℜ3 are linearly independent.






2k


= 2+k 
1


1 0 2k
1 0 2k
1 0 2k






0
1
2
−
k
1
1
2
+
k
→
→

 0 1 2−k 



0 0 k−1
0 1
1
0 1
1
The rref of the above matrix equals the identity, only when k 6= 1. When k 6= 1, the three
vectors are linearly independent, and since there are three of them, they must span ℜ3 . Thus,
they form a basis for ℜ3 . Since the coordinate transformation with respect to the standard
basis is an isomorphism, it follows that the three polynomials are a basis of P2 only if k 6= 1.
(7, 8, 34, 42) We first find the matrix A of the linear transformation with respect to the
standard basis (U).
T
"
1 0
0 0
#!
=
"
1 0
0 0
#"
1 2
0 1
#
−
"
1 2
0 1
#"
1
1 0
0 0
#
=
"
1 2
0 0
#
−
"
1 0
0 0
#
=
"
0 2
0 0
#
Math 2270 Spring 2004
T
"
"
T
#!
0 1
0 0
#!
0 0
0 1
"
"
=
"
=
0 2
0 0
Homework 12 Solutions
#"
0 0
0 1
#
#"
0 1
0 0
U
1 2
0 1
1 2
0 1


0


=  2 ,
0
#
#
−
−
"
"
"
1 2
0 1
1 2
0 1
0 0
0 0
#
U

#"
#"
#
0 1
0 0
0 0
0 1

#
=
=

0


=  0 ,
0
"
"
"
#
0 1
0 0
−
#
−
0 −2
0 0
#
0 0
0 1
"
"
0 1
0 0
0 2
0 1
#
#
=
=
"
"
#
0 0
0 0
0 −2
0 0
#


0


=  −2 
0
U

0 0 0


A =  2 0 −2 
0 0 0
The matrix A is obviously not invertible (since rank(A)=1). Therefore, T is not an isomorphism. We will use the matrix A to find the image, rank, kernel and nullity of the linear
transformation T . The image of the matrix A is all linear multiples of the vector ~e2 , so the
rank of A is 1." Translating
this back into matrix form, the image of T is all scalar multiples
#
0 1
of the matrix
. This matrix will form a basis for the image of T .
0 0
Now, the kernel of the matrix A will be all vectors of the form




0
1




~x = r  0  + s  1 
0
1
Translating this back into matrix form, we see that the kernel consists of matrices of the
form
#
#
"
"
0 1
1 0
+s
r
0 1
0 0
Therefore, the nullity of T is two. We next find the matrix B of the linear transformation
with respect to the given basis (B).
T
T
"
1 0
0 1
#!
T
"
0 1
0 0
#!
"
1 0
0 −1
#!
=
"
1 0
0 1
#"
1 2
0 1
#
=
"
0 1
0 0
#"
1 2
0 1
#
=
"
1 0
0 −1
#"
−
"
1 2
0 1
#"
1 0
0 1
#
−
"
1 2
0 1
#"
0 1
0 0
#
1 2
1 2
−
0 1
0 1
#"
1 0
0 −1
# "
2
=
"
1 2
0 1
#
=
"
0 1
0 0
#
#
=
"
−
"
1 2
0 1
#
−
"
0 1
0 0
#
# "
=
"
0 0
0 0
#
=
"
0 0
0 0
#
#
"
0 4
0 0
1 2
1 −2
−
0 −1
0 −1
=
#
Math 2270 Spring 2004
"
#
0 0
0 0
Homework 12 Solutions


B
0


=  0 ,
0
"
0 0
0 0


#
B
0


=  0 ,
0
"
0 4
0 0

#
B


0 0 0


B= 0 0 4 
0 0 0

0


= 4 
0
We next find an invertible matrix S such that B = S −1 AS, or AS = SB.






0 0 0
a b c
a b c
0 0 0






 2 0 −2   d e f  =  d e f   0 0 4 
0 0 0
g h i
g h i
0 0 0




0 0 4b
0
0
0




 2a − 2g 2b − 2h 2c − 2i  =  0 0 4e 
0 0 4h
0
0
0
This gives us the following equations: a = g, b = h, b = 0, h = 0, c = 2e + i. The variables
d, e, f , g and i are free variables. Let’s take d = 0, e = 1, f = 0, g = 1 and i = −1. (Any
values can be chosen for these variables, as long as the resulting matrix is invertible. We’ll
see why these choices are nice in a minute.) It follows that a = g = 1 and c = 2 − 1 = 1.
Then,


1 0 1


S= 0 1 0 
1 0 −1
The columns of this matrix are linearly independent, so the matrix S must be invertible.
We could also have found a matrix S by defining the columns to be the coordinate vectors
of the B-basis with respect to the standard basis U (See the proof of Fact 4.3.4 part a).
"
1 0
0 1
#

U

1


=  0 ,
1
"
0 1
0 0
#

U

0


=  1 ,
0
"
1 0
0 −1
#


U
1


= 0 
−1
Notice that these three vectors are exactly the column vectors of the matrix S that we
defined previously.
(12) We use the standard basis for ℜ2x2 .
"
T
"
1 0
0 0
#!#
=
U
""
1 0
0 0
#"
3
2 0
0 3
##
U
=
"
2 0
0 0
#
U




=
2
0
0
0





Math 2270 Spring 2004
"
"
"
T
T
T
"
"
"
Homework 12 Solutions
0 1
0 0
#!#
0 0
1 0
#!#
0 0
0 1
#!#
U
""
U
""
U
""
=
=
=
0 1
0 0
#"
0 0
1 0
#"
0 0
0 1
#"
2 0
0 3
##
2 0
0 3
##
2 0
0 3
##
=
U
=
U
=
U
"
"
"
0 3
0 0
#
0 0
2 0
#
0 0
0 3
#
U
U
U

0
3
0
0


0
0
2
0


0
0
0
3



=



=



=













The matrix for the linear transformation with respect to the standard basis is




A=
2
0
0
0
0
3
0
0
0
0
2
0
0
0
0
3





This matrix is invertible, so the transformation T is an isomorphism.
(20, 36) T (f ) = f ′ from P2 to P2 . We use the standard basis U, {1, t, t2 }.


0


[T (1)]U = [0]U =  0  ,
0


1


[T (t)]U = [1]U =  0  ,
0
h
T (t2 )
i
U


0


= [2t]U =  2 
0
The matrix for the linear transformation with respect to the standard basis is


0 1 0

A= 0 0 2 

0 0 0
This matrix is not invertible, so the transformation T is not an isomorphism. The image of
A is spanned by the vectors ~e1 and ~e2 . Converting this back to polynomials, a basis for the
image of T is {1, t}. The rank of T is 2. The kernel of A is spanned by ~e1 . Converting this
back to polynomials, a basis for the kernel of T is {1}. The nullity of T is 1.
4
Math 2270 Spring 2004
(29) T (f (t)) =
R2
0
Homework 12 Solutions
f (t)dt from P2 to P2 . Again, we use the standard basis U, {1, t, t2 }.
[T (1)]U =
Z
[T (t)]U =
h
T (t2 )
i
U
=
2
1 · dt
0
Z
Z
2
tdt
0
2
U
t2 dt
0
U


2


= [2]U =  0 
0


2


= [2]U =  0 
0
=
U
8
3
U
The matrix for the linear transformation is


8/3


= 0 
0


2 2 8/3


A= 0 0 0 
0 0 0
This matrix is not invertible, so T is not an isomorphism.
(54) We need to calculate [T (~v1 )]B and [T (~v2 )]B where ~v1 and ~v2 are the given basis vectors.
The first vector will project to itself to [T (~v1 )]B = ~e1 . The second vector is perpendicular to
the first, so it will project to the zero vector.
B=
"
1 0
0 0
#
(61) T (f ) = f ′′ + f with basis U, {cos(t), sin(t), t cos(t), t sin(t)}.

0
0
0
0


0
0
0
0




[T (cos t)]U = [− cos t + cos t]U = [0]U = 



[T (sin t)]U = [− sin t + sin t]U = [0]U = 
5








Math 2270 Spring 2004
Homework 12 Solutions




[T (t cos t)]U = [−2 sin t − t cos t + t cos t]U = [−2 sin t]U = 



[T (t sin t)]U = [2 cos t − t sin t + t sin t]U = [2 cos t]U = 

The matrix of T with respect to the given basis U is




A=
0
0
0
0
0 0 2
0 −2 0
0 0 0
0 0 0
0
−2
0
0
2
0
0
0















We must solve the matrix equation A~x = ~e1 .





Solutions are of the form




~x = 
0
0
0
0
0
0
0
1/2
0 0 2|1
0 −2 0 | 0
0 0 0|0
0 0 0|0






+r


1
0
0
0











 + s


0
1
0
0





where r and s are real numbers. Converting this back to a function, we have general solutions
1
f (t) = t sin t + r cos t + s sin t
2
Let’s verify that these solutions solve the given equation.
1
1
sin t + t cos t − r sin t + s cos t
2
2
1
1
1
f ′′ (t) = cos t + cos t − t sin t − r cos t − s sin t
2
2
2
= cos t − f (t)
f ′ (t) =
6