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Exercise 6.5.15

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1.

We have

∥ U_{W} (x) ∥ = ∥ U (x) ∥ = ∥ x ∥

and so $U_{W}$ is an unitary operator on $W$ . Also, the equality above implies that $U_{W}$ is injection. Since $W$ is finite-dimensional, we get that $U_{W}$ is surjective and $U (W) = W$ .

2.

For each elment

w \in W

we have

U (y) = w

for some

y \in W

by the previous argument. Now let

x

be an element in

W^{⊥}

. We have

U (x) = w_{1} + w_{2}

for some

w_{1} \in W

and

w_{2} \in W^{⊥}

by Exercise 6.2.6. Since

U

is unitary, we have some equalities

∥ y ∥^{2} = ∥ w ∥^{2},

∥ x ∥^{2} = ∥ w_{1} + w_{2} ∥^{2} = ∥ w_{1} ∥^{2} + ∥ w_{2} ∥^{2}

by Exercise 6.1.10. However, we also have that $U (x + y) = 2 w_{1} + w_{2}$ . So we have that

0 = ∥ x + y ∥^{2} - ∥ 2 w_{1} + w_{2} ∥^{2}

= ∥ x ∥^{2} + ∥ y ∥^{2} - 4 ∥ w_{1} ∥^{2} - ∥ w_{2} ∥^{2} = - 2 ∥ w_{1} ∥^{2} .

This means that $w_{1} = 0$ and so $U (x) = w_{2} \in W^{⊥}$ .

jephianlin

2011-06-27 00:00

Exercise 6.5.15

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