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Exercise 7.C.7

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Proof. Suppose $T$ is invertible. Let $R$ be the positive square root of $T$ . Then

⟨ Tv, v ⟩ = ⟨ R^{∗} Rv, v ⟩ = ⟨ Rv, Rv ⟩ > 0,

where the inequality follows because $R$ and $T$ have the same eigenvalues (as we saw in the proof of 7.36), but $0$ is not an eigenvalue of $T$ , thus $Rv \neq 0$ for all $v$ .

Conversely, suppose $⟨ Tv, v ⟩ > 0$ for every $v \in V$ with $v \neq 0$ . This directly implies that $null T = {0}$ . Thereofer $T$ is invertible. □

celiopassos

2017-10-06 00:00

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R and T don’t necessarily have the same eigenvalues. If b is an eval of R associated to v then b^2 is an eval of T associated to v. Then unless b = b^2, b cannot always be an eval of T, because either we’d have Tv = bv = b^2v which is impossible, or for w =/= v we’d have Tw = bw, which implies w is orthogonal to v, but w doesn’t necessarily exist, i.e in 1d.

Josh_ • 2023-12-08

Exercise 7.C.7

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