Solutions to PS 7 (Math 121)
J. Scholtz
November 18, 2006
Question 1: 5.2/12
1. Suppose that
v
∈
E
λ
, then
T
(
v
) =
λv
and hence by taking the inverse of both sides we get:
v
=
λT

1
(
v
) and so
v
∈
E
λ

1
. Hence the two eigenspaces are equal.
2. Suppose that
A
= [
T
] is diagonalizable then there exist
D
=
Q

1
AQ
such that
D
is diagonal.
Then by taking an inverse of each side we get
D

1
=
Q

1
A

1
Q
. Note, that since
D
is diagonal it
is invertible and our last steps is hence justified.
Question 2: 5.2/14a
Find a solution to the system of differential equations:
x
=
x
+
y
and
y
= 3
x

y
.
We need to diagonalize the matrix
A
=
1
1
3

1
, which gives us
D
=
2
0
0

2
and
Q
=

2
0
0
2
And so our new system looks like
p
=

2
p
and
q
= 2
q
, therefore
p
=
Ce

2
t
and
q
=
De
2
t
.
Now we only need to convert to the old basis where
p
=

x
+3
y
and
q
=
x
+
y
and so
y
=
1
4
{
Ce

2
t
+
De
2
t
}
,
whereas
x
=

1
4
{
Ce

2
t

3
De
2
t
}
.
Question 3: 5.2/17
Two operators T and U are simultaneously diagonalizable iff there exists a basis
β
such that both [
T
]
β
and [
U
]
β
are diagoanal. Similarly two matrices A, B are simultaneously diagonalizable if there exists an
invertible
Q
such that both
Q

1
AQ
and
Q

1
BQ
are diagonal matrices.
1. Show that if
T
and
U
are simultaneously diagonalizable then the matrices [
T
]
β
and [
U
]
β
are also
simultaneously diagonalizable.
If both
T
and
U
are both simultaneously diagonalizable then there exists a basis
γ
such that both
[
T
]
γ
and [
U
]
γ
are diagonal. But this means that
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Q−1 AQ, ∈ Eλ, simultaneously diagonalizable matrices

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