Math 235: Linear Algebra HW 7
Problem 3.4.6 [(d)]
(d)
Prove that if
B
is any ordered basis for
R
2
and [
T
]
B
=
A
, then
A
12
A
21
6
= 0.
Let
B
=
{
α
1
, α
2
}
where
α
1
= (
a, b
) and
α
2
= (
c, d
) be any basis for
R
2
.
Also let
B
s
denote the
standard basis.
Then the matrix
B
= [
I
]
B
B
s
=
a
c
b
d
is the change of basis matrix from the standard basis into
B
, and
B

1
= [
T
]
B
s
B
=
1
ad

bc
d

c

b
a
is its inverse. (note that
ad

bc
6
= 0 since change of basis
is invertible).
So the matrix
C
= [
T
]
B
B
is given by
BAB

1
=
1
ad

bc
a
c
b
d
·
0
1

1
0
·
d

c

b
a
So
C
=
m
n
r
s
where
m, n, r
and
s
are functions of
a, b, c
and
d
. (for example
m
=

ab

cd
, can
you figure out the rest?)
From part (c) we know that
T

cI
is invertible. So letting
c
=
m
we see that
C

mI
=
m
n
r
s

m
0
0
m
=
0
n
r
s

m
is invertible.
From previous work we know that this matrix is invertible if and only if 0 (
r

m
)

nr
6
= 0
⇒
nr
6
= 0.
For our matrix
n
=
A
2
,
1
and
r
=
A
1
,
2
, so we’re done.
(d)
Problem 3.4.7
Let
T
be the linear operator on
R
3
defined by
T
(
x, y, z
) = (3
x
+
z,

2
x
+
y,

x
+ 2
y
+ 4
z
)
(a)
What is the matrix of
T
in the standard ordered basis on
R
3
?
3
0
1

2
1
0

1
2
4
(a)
(b)
What is the matrix of
T
in the ordered basis
{
α
1
, α
2
, α
3
}
where
α
1
= (1
,
0
,
1),
α
2
= (

1
,
2
,
1), and
α
3
= (2
,
1
,
1)?
4

2
7

2
4

3
3
9
4
(b)
Problem 3.4.7 continued on next page. . .
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