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Let $0 \neq A \in M_{3×3}(\mathbb{C})$ be a Hermitian matrix. Show that $A^k \neq 0$ for all positive integers $k$.

I'm stuck on this problem. I am not sure why the question is only about $ M_{3×3}(\mathbb{C})$. Isn't this problem true for $ M_{n×n}(\mathbb{C})$?

I also don't know how to solve it. I just know some theorems but my information doesn't seem useful. I appreciate any hint and suggestion.

And here is something that comes to my mind, but I'm pretty sure its wrong or incomplete.

Let $\alpha:v\to Av$ be an element of $\operatorname{End}(\mathbb{C}^3)$. Since $A$ is Hermitian, $\alpha$ is self-adjoint, so it's orthogonally diagonalizable. Let $0\neq \lambda \in \operatorname{spec}(\alpha)$. Hence, $\lambda^k\in \operatorname{spec}(A^k)$. Let $0\neq v$ be an eigenvector of $\alpha$ associated with eigenvalue $\lambda$, then $A^kv=\lambda^kv$ for all integers $k$. Suppose there exist an integer $k$ satisfying $A^k=0$, so $\lambda=0$ which means $\operatorname{spec}(\alpha)=\{0\}$ and this is a contradiction.

How about if the only eigenvalue is $\lambda=0?$

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    $\begingroup$ There is something wrong with the question. $k=1$ is a positive integer and $O=A$ is already given. $\endgroup$ – Anurag A Jan 3 '17 at 8:18
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    $\begingroup$ @Parisinia I've changed your question to $O \neq A$ because it doesn't make sense otherwise (see @ Anurag's comment). If this somehow wasn't what you meant, please edit it back. $\endgroup$ – TastyRomeo Jan 3 '17 at 8:31
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    $\begingroup$ The question presumably also assumes that $A \ne 0$. Hint: can you prove it for a diagonal matrix? $\endgroup$ – The Potter's Vessel Jan 3 '17 at 8:31
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    $\begingroup$ Your proof works: there is at least one nonzero eigenvalue (otherwise the diagonalization would be the zero matrix), and that's the one to use. $\endgroup$ – user357151 Jan 3 '17 at 15:00
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    $\begingroup$ @Parisina, it is obvious for a diagonal matrix, and hence it is also obvious for a diagonalizable matrix. Apparently the hint was too subtle. $\endgroup$ – The Potter's Vessel Jan 3 '17 at 16:20
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Hermitian matrices are unitarily diagonalizable: $A=Q^*DQ$ with unitary $Q$ and diagonal $D$. Hence $A^k = Q^*D^kQ$ and the conclusion follows: $$A\ne 0\implies D\ne 0\implies D^k\ne 0\implies A^k\ne 0$$ The size being $3\times 3$ does not matter.

(The same works for any diagonalizable matrix, not just Hermitian ones.)

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