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Exercise 7.B.6

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Proof. Suppose $T$ is self-adjoint. Let $λ$ be an eigenvalue of $T$ and $v$ a corresponding eigenvector. Then

λ ⟨ v, v ⟩ = ⟨ Tv, v ⟩ = ⟨ v, Tv ⟩ = \bar{λ} ⟨ v, v ⟩ .

Therefore $λ = \bar{λ}$ , that is, $λ$ is real.

Conversely, suppose all eigenvalues of $T$ are real. Let $e_{1}, \dots, e_{n}$ denote an orthonormal basis consisting of eigenvalues of $T$ (which exists by 7.24) and let $λ_{1}, \dots, λ_{n}$ denote their corresponding eigenvalues (which are real). For any $v \in V$ , we can write

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

for some $a_{1}, \dots, a_{n} \in ℂ$ . Then

⟨ Tv, v ⟩ = ⟨ λ_{1} a_{1} e_{1} + \dots + λ_{n} a_{n} e_{n}, a_{1} e_{1} + \dots + a_{n} e_{n} ⟩ = λ_{1} | a_{1} |^{2} + \dots + λ_{n} | a_{n} |^{2} \in ℝ .

Thus, by 7.15, $T$ is self-adjoint. □

celiopassos

2017-10-06 00:00

Exercise 7.B.6

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