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Exercise 5.A.16

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Proof. Suppose $λ$ is and eigenvalue of $T$ and $v$ a corresponding eigenvector. Let $v_{1}, \dots, v_{n}$ be a basis of $V$ with respect to which $(T)$ has only real entries. We have that $λ = a + bi$ for some $a, b \in ℝ$ . Thus

Tv = λv = av + biv .

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celiopassos

2017-10-06 00:00

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could you please elaborate ?

vedanshivaghela • 2025-09-06

Exercise 5.A.16

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