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Exercise 6.4.10

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Directly calculate that

\begin{array}{l} ∥ T (x) \pm ix ∥^{2} & = ⟨ T (x) \pm ix, T (x) \pm ix ⟩ \\ = ∥ T (x) ∥^{2} \pm ⟨ T (x), ix ⟩ \pm ⟨ ix, T (x) ⟩ + ∥ x ∥^{2} \\ = ∥ T (x) ∥^{2} \mp i ⟨ T (x), x ⟩ \pm ⟨ T^{∗} (x), x ⟩ + ∥ x ∥^{2} \\ = ∥ T (x) ∥^{2} + ∥ x ∥^{2} . \end{array}

Also, $T \pm iI$ is injective since $∥ T (x) \pm x ∥ = 0$ if and only if $T (x) = 0$ and $x = 0$ . Now $T - iI$ is invertible by the fact that $V$ is finite-dimensional. Finally we may calculate that

\begin{array}{l} ⟨ x, {[{(T - iI)}^{- 1}]}^{∗} (T + iI) (y) ⟩ & = ⟨ {(T - iI)}^{- 1} (x), (T + iI) (y) ⟩ \\ = ⟨ {(T - iI)}^{- 1} (x), (T^{∗} + iI) (y) ⟩ \\ = ⟨ {(T - iI)}^{- 1} (x), {(T - iI)}^{∗} (y) ⟩ \\ = ⟨ (T - iI) {(T - iI)}^{- 1} (x), y ⟩ \\ = ⟨ x, y ⟩ \end{array}

for all $x$ and $y$ . So we get the desired equality.

jephianlin

2011-06-27 00:00

Exercise 6.4.10

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