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 $\lambda=0?$
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.)