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Exercise 7.D.18

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Proof. (a) Use the same notation from the Singular Value Decomposition theorem (7.52). We have

\begin{aligned} | | Tv | |^{2} & = | s_{1} ⟨ v, e_{1} ⟩ |^{2} + \dots + | s_{n} ⟨ v, e_{n} ⟩ |^{2} \\ = | s_{1} |^{2} | ⟨ v, e_{1} ⟩ |^{2} + \dots + | s_{n} |^{2} | ⟨ v, e_{n} ⟩ |^{2} \\ \leq | s |^{2} | ⟨ v, e_{1} ⟩ |^{2} + \dots + | s |^{2} | ⟨ v, e_{n} ⟩ |^{2} \\ = | s |^{2} | ⟨ v, e_{1} ⟩ |^{2} + \dots + | ⟨ v, e_{n} ⟩ |^{2} \\ = s^{2} | | v | |^{2} \end{aligned}

for all $v \in V$ , where the first line follows from the Pythagorean Theorem (6.13) and last because $s$ is a positive real value. Taking the square root of both sides shows that $| | Tv | | \leq s | | v | |$ for all $v \in V$ . The proof that $ŝ | | v | | \leq | | Tv | |$ is almost the same.

(b) Let $v$ be an eigenvector of $T$ corresponding to $λ$ with $| | v | | = 1$ . Then

| λ | = | λ | | | v | | = | | Tv | | \leq s | | v | | = s .

Similarly, we have $ŝ \leq | λ |$ . □

celiopassos

2017-10-06 00:00

Exercise 7.D.18

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