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The converse is easy to show by multiplying $Ax$ on the left by $A$ (x is a eigenvector): $AAx=A(\lambda x)=\lambda Ax=\lambda^2x$ but I was wondering if the converse is true. The same approach doesn't work to prove what I want, I think this shouldn't be true but I'm not so sure, any hints?

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Not true as stated... $\omega^{k}$ is an eigenvalue of $I^{k}=I$ whenever $\omega$ is a $k$-th root of unity, but only $1$ is an eigenvalue of $I$. A better question would be: is $\lambda$ times some $k$-th root of unity an eigenvalue of $A$? – mjqxxxx Jan 7 at 4:52

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No. Consider the matrix $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. We have $A^2 = I$, and you can easily see that your statement no longer holds.

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@user55414 This is one reason why positive matrix is such a strong condition in your other problem; non-negativity isn't sufficient. – Calvin Lin Jan 7 at 4:51
I see... I asked this because there was another answer that was later deleted which stated something similar and I wondered if it was true. Thanks. – user55414 Jan 7 at 5:00
@user55414 Yes I know. I was about to comment on that answer, then it got deleted. It is a common misconception that $v$ must be an eigenvector of $A$, but sadly all that we know is there is some eigenvalue that is a root of $\lambda$ and corresponds to some eigenvector. You can easily recreate this scenario using complex roots of the form $\omega^k$ as your eigenvalues. – Calvin Lin Jan 7 at 5:09
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$\lambda$ is not the unique $k$-th root of $\lambda^k$, so the answer is no. If $m$ is an eigenvalue of $A^k$ then there is a $b$ with $b^k=m$ which is an eigenvalue of $A$.

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$(-1)^2$ is an eigenvalue of $1^2$, but $-1$ is not an eigenvalue of $A=1$ in $\mathbb{C}$.

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