1
Math 110 Homework 11
Partial Solutions
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5.2.12
(a) Let
E
λ
be the eigenspace of
T
corresponding to
λ
, and let
F
λ

1
be the eigenspace of
T

1
corresponding to
λ

1
. If
x
∈
E
λ
, then
T
(
x
) =
λx
. Thus
x
=
T

1
(
λx
) =
λT

1
(
x
) and so
T

1
(
x
) =
λ

1
x
.
Thus
x
∈
F
λ

1
. By the same argument, we also get
F
λ

1
⊆
E
λ
and thus that they are equal.
(b) If
T
is diagonalizable, then there exists a basis
β
for
V
such that
[
T
]
fi
is diagonal. By (a), [
T

1
]
fi
must then also be diagonal.
5.2.20 Suppose that
V
=
W
1
⊕ · · · ⊕
W
k
.
Then, by Theorem 5.10, we can
find ordered bases
β
1
, . . . , β
k
for
W
1
, . . . , W
k
, respectively, so that
β
=
β
1
∪· · ·∪
β
k
is a basis for
V
. This then gives that dim(
V
) = dim(
W
1
)+
· · ·
+ dimW
k
by counting basis vectors.
On the other hand, suppose that the dimension equality holds. Picking
a basis
β
i
for each
W
i
, we have that
β
=
β
1
∪· · ·∪
β
k
has dim(
V
) vectors.
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 Spring '10
 FUCKHEAD
 Linear Algebra, WI, 0 K, 2 span, 1 Span

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