Consider that for each $n \times n$ (possibly complex) matrix, $A_{k}$, $0 \leq k \leq m$, we have that \begin{align} A_{0}A_{k} &= kA_{k}, \qquad 1 \leq k \leq m \end{align} and suppose that \begin{align} \sum \limits_{k=1}^{m} \text{rank}(A_{k}) = \text{rank}(A_{0}). \end{align} Show that $A_{0}$ is diagonalizable.
I want to show that $\mathbb{C}^{n}$ can be written as a sum of (non-generalized) eigenspaces of $A_{0}$. I want to use the spectral theorem to try to do this. However, I can't seem to show it any way, so if the spectral theorem doesn't work, then I am open to other methods.
Hint: Note that each (non-zero) column of $A_k$ must be an eigenvector (corresponding to $k$) of $A_0$, since $$ A_0 \pmatrix{v_1&v_2&\cdots&v_n} = \pmatrix{A_0 v_1&A_0 v_2&\cdots&A_0 v_n} = \pmatrix{k v_1&k v_2&\cdots&k v_n} $$