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Exercise 7.D.4

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Proof. Let $v \in V$ be an eigenvector of $\sqrt{T^{*} T}$ with $| | v | | = 1$ corresponding to $s$ and let $S \in L (V)$ be an isometry such that $T = S \sqrt{T^{*} T}$ . Then

| | Tv | | = | | S \sqrt{T^{*} T} v | | = | | \sqrt{T^{*} T} v | | = | s | | | v | | = | s | = s,

where the last equality follows because $\sqrt{T^{∗} T}$ is positive. □

celiopassos

2017-10-06 00:00

Exercise 7.D.4

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