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Exercise 7.A.2

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Proof. Clearly, it suffices to prove one direction. Suppose $λ$ is an eigenvalue of $T$ . Then

null (T^{∗} - \bar{λ} I) = null {(T - λI)}^{∗} = {(range (T - λI))}^{⊥},

where the last equality comes from 7.7. Now, since $\dim null (T - λI) > 0$ , the subspace on the right has non zero dimension and the result follows. □

celiopassos

2017-10-06 00:00

Exercise 7.A.2

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