Let $A^{k+1} = A$ for some $k\in \mathbb{N}$. Prove that $A$ is diagonalisable.
Hint: $$(J_n(\lambda))^{k+1} = (\text{diag}_n(\lambda) + J_n(0))^{k+1}$$
So, I am not sure how to use the hint, and I'm just trying my own thing.
Since $A^{k+1} = A$, $A$ is a root of $p(x) = x(x^k-1)$. The minimal polynomial of $A$, say $m_A(x)$, divides $x(x^k-1)$. What's next? If I can show that $m_A(x)$ splits into distinct linear factors, I'll be done.
Let me know how I can proceed with this argument, and/or use the hint provided.
One thing that I found worth noticing is that $$\text{diag}_n(\lambda)^k = \text{diag}_n(\lambda^k)$$Does this help?
Thanks!
The statement is not true in general. E.g. $A=\pmatrix{0&1\\ 1&0}=A^3$ is not diagonalisable over $GF(2)$ and $A=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}=A^4$ is not diagonalisable over $\mathbb R$. It is true, however, over $\mathbb C$ because the minimal polynomial of $A$, which divides $x^{k+1}-x$, is a product of distinct linear factors.
Alternatively, you may use the given hint. Suppose the contrary that $A$ is not diagonalisable over $\mathbb C$. Then via a similarity transform, we may assume that $A$ is in Jordan form. Therefore it contains a nontrivial Jordan block $J_n(\lambda)$ of some size $n$ such that $J_n(\lambda)^{k+1}=J_n(\lambda)$. But this implies that $$ \sum_{r=0}^n\binom{n}{r}\lambda^{n-r}J_n(0)^r=\lambda I_n+J_n(0). $$ By comparing the coefficients of $I_n$ and $J_n(0)$ on both sides, we get $\lambda^n=\lambda$ and $n\lambda^{n-1}=1$. Therefore $|\lambda|=1$ and $n=1$, but this contradicts the assumption that $J_n(\lambda)$ is a nontrivial Jordan block. Hence $A$ must be diagonalisable.