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Exercise 7.D.1

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Proof. A quick calculation shows that $T^{∗} Tv = | | x | |^{2} ⟨ v, u ⟩ u$ for every $v \in V$ . The map $R \in L (V)$ defined by

Rv = \frac{| | x | |}{| | u | |} ⟨ v, u ⟩ u

is a square root of $T^{∗} T$ . Moreover, it is easy to check that $⟨ Rv, v ⟩ \geq 0$ for all $v \in V$ . We but need to prove that $R$ is self-adjoint. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ . We can write

u = a_{1} e_{1} + \dots + a_{n} e_{n}

for some $a_{1}, \dots, a_{n} \in 𝔽$ . Note that

R e_{j} = \frac{| | x | |}{| | u | |} ⟨ e_{j}, u ⟩ u = \frac{| | x | |}{| | u | |} (a_{j} \bar{a_{1}} e_{1} + \dots + a_{j} \bar{a_{n}} e_{n}) .

Therefore, the matrix of $R$ with respect to the basis $e_{1}, \dots, e_{n}$ has entries defined by

M {(R)}_{j, k} = \frac{| | x | |}{| | u | |} a_{j} \bar{a_{k}} .

Thus $M {(R)}_{j, k} = \bar{M {(R)}_{k, j}}$ , that is, $M (R) = M (R^{∗})$ . Hence $R$ is self-adjoint. □

celiopassos

2017-10-06 00:00

Exercise 7.D.1

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