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Exercise 7.D.8

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Proof. Let $S^{'} \in L (V)$ be an isometry such that $T = S^{'} \sqrt{T^{*} T}$ . Then

\begin{aligned} 0 & = | | Tv | | - | | Tv | | \\ = | | S^{∗} Tv | | - | | {S^{'}}^{∗} Tv | | \\ = | | Rv | | - | | \sqrt{T^{*} T} v | | \\ = ⟨ R^{∗} Rv, v ⟩ - ⟨ T^{∗} Tv, v ⟩ \\ = ⟨ (R^{∗} R - T^{∗} T) v, v ⟩ \\ = ⟨ (R^{2} - T^{∗} T) v, v ⟩ \end{aligned}

where the last line follows because $R$ is self-adjoint. 7.16 now implies that $R^{2} = T^{∗} T$ . Since $R$ is positive and the positive square root of $T^{∗} T$ is unique (by 7.36), it follows that $R = \sqrt{T^{*} T}$ . □

celiopassos

2017-10-06 00:00

Exercise 7.D.8

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