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Exercise 7.B.14

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Proof. Suppose $U$ has a basis $e_{1}, \dots, e_{n}$ consisting of eigenvectors of $T$ . We can assume without loss of generality that this list is normalized (we can divide each vector by its norm if it isn’t). Define $⟨ \cdot, \cdot ⟩ : U \times U \to ℝ$ by

⟨ a_{1} e_{1} + \dots + a_{n} e_{n}, b_{1} e_{1} + \dots + b_{n} e_{n} ⟩ = a_{1} b_{1} + \dots + a_{n} b_{n} .

Since every vector in $U$ can be uniquely written as linear combination of $e_{1}, \dots, e_{n}$ , this function is well defined. Moreover, one easily checks that this function is an inner product. Note that

⟨ e_{j}, e_{k} ⟩ = ⟨ 0 e_{1} + \dots + 1 e_{j} + \dots + 0 e_{n}, 0 e_{1} + \dots + 1 e_{k} + \dots + 0 e_{n} ⟩ = {\begin{matrix} 1, if j = k \\ 0, if j \neq k \end{matrix} .

Therefore $e_{1}, \dots, e_{n}$ is also orthonormal. The Real Spectral Theorem (7.29) now implies that $T$ is self-adjoint.

The converse follows directly from 7.29. □

celiopassos

2017-10-06 00:00

Exercise 7.B.14

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