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Is this statement true? It's easy to prove the forward direction: $$Av=\lambda v\Rightarrow\lambda (Av)=\lambda(\lambda v)\Rightarrow A(\lambda v)=\lambda^2 v\Rightarrow A(Av)=\lambda^2 v\Rightarrow A^2v=\lambda^2 v$$ But the backward direction is eluding me. I figured the contrapositive statement, $Av\neq\lambda v\Rightarrow A^2v\neq\lambda^2 v$, might help, but I don't see how. Thanks in advance.

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    $\begingroup$ What if $A=-I$ ? $\endgroup$ Jan 1, 2018 at 23:42
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    $\begingroup$ The reverse direction needs some interpretation. For example, $(-1)^2$ is an eigenvalue of $I^2$ but $-1$ is not an eigenvalue of $I$. $\endgroup$
    – copper.hat
    Jan 1, 2018 at 23:43

2 Answers 2

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The backward direction is false. Consider that $A^2v=\lambda^2 v$. Then $(A^2-\lambda^2I)v=0$, so $(A-\lambda I)(A+\lambda I)v=0$, implying that at least one of $\lambda$ or $-\lambda$ are eigenvalues of $A$, but not necessarily $\lambda$.

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Edit: @ErickWong 's comment shows this answer isn't one. I'll leave it here rather than deleting it since others may be tempted by the same thinko.

In the real plane rotation through a quarter of a circle has no (real) eigenvalues but its square is $-I$, which has eigenvalue $-1$ with multiplicity $2$.

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    $\begingroup$ I don't see what this has to do with the question. The phrasing suggests that you intend it to be a counterexample, but it doesn't appear to contradict anything, since there is no $\lambda \in \mathbb R$ for which $\lambda^2 = -1$. $\endgroup$
    – Erick Wong
    Jan 2, 2018 at 0:52
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    $\begingroup$ @ErickWong Thanks. See my edit. $\endgroup$ Jan 2, 2018 at 1:02

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