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Exercise 7.D.16

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Proof. Let $S_{1}, S_{2} \in L (V)$ be isometries such that $T_{1} = S_{1} \sqrt{T_{1}^{*} T_{1}}$ and $T_{2} = S_{2} \sqrt{T_{2}^{*} T_{2}}$ and let $e_{1}, \dots, e_{n}$ and $f_{1}, \dots, f_{n}$ be orthonormal basis of $V$ consisting of eigenvectors of $T_{1}$ and $T_{2}$ , respectively, corresponding to the singular values $s_{1}, \dots, s_{n}$ . Defin $S \in L (V)$ by

S e_{j} = f_{j}

for each $j = 1, \dots, n$ . Then $S$ is also an isometry and we have

\begin{aligned} \sqrt{T_{1}^{∗} T_{1}} e_{j} & = s_{j} e_{j} \\ = S^{∗} (s_{j} f_{j}) \\ = S^{∗} \sqrt{T_{2}^{∗} T_{2}} f_{j} \\ = S^{∗} \sqrt{T_{2}^{∗} T_{2}} S e_{j} . \end{aligned}

So $\sqrt{T_{1}^{*} T_{1}} = S^{∗} \sqrt{T_{2}^{*} T_{2}} S$ . Therefore

T_{1} = S_{1} \sqrt{T_{1}^{*} T_{1}} = S_{1} S^{∗} \sqrt{T_{2}^{∗} T_{2}} S = S_{1} S^{∗} S_{2}^{∗} T_{2} S,

where the last equality follows by multiplying both sides of the equation $T_{2} = S_{2} \sqrt{T_{2}^{*} T_{2}}$ by $S_{2}^{∗}$ . This gives the desired result, because the product of isometries is also an isometry. □

celiopassos

2017-10-06 00:00

Exercise 7.D.16

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