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Linear Algebra

Exercise 6.4.7

Answers

1.

We check

⟨ x, {(T_{W})}^{∗} (y) ⟩ = ⟨ T_{W} (x), y ⟩ = ⟨ T (x), y ⟩

= ⟨ T^{∗} (x), y ⟩ = ⟨ x, T (y) ⟩ = ⟨ x, T_{W} (y) ⟩

for all $x$ and $y$ in $W$ .

2.

Let

y

be an element in

W^{⊥}

. We check

⟨ x, T^{∗} (y) ⟩ = ⟨ T (x), y ⟩ = 0

for all $x \in W$ , since $T (x)$ is also an element in $W$ by the fact that $W$ is $T$ -invariant.

3.

We check

⟨ x, {(T_{W})}^{∗} (y) ⟩ = ⟨ T_{W} (x), y ⟩ = ⟨ T (x), y ⟩

⟨ x, T^{∗} (y) ⟩ = ⟨ x, {(T^{∗})}_{W} (y) ⟩ .

4.

Since

T

is normal, we have

T T^{∗} = T^{∗} T

. Also, since

W

is both

T

- and

T^{∗}

-invariant, we have

{(T_{W})}^{∗} = {(T^{∗})}_{W}

by the previous argument. This means that

T_{W} {(T_{W})}^{∗} = T_{W} {(T^{∗})}_{W} = {(T^{∗})}_{W} T_{W} = {(T_{W})}^{∗} T_{W} .

jephianlin

2011-06-27 00:00

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