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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.

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Hint: Note that each (non-zero) column of $A_k$ must be an eigenvector (corresponding to $k$) of $A_0$, since $$ A_0 \pmatrix{v_1&v_2&\cdots&v_n} = \pmatrix{A_0 v_1&A_0 v_2&\cdots&A_0 v_n} = \pmatrix{k v_1&k v_2&\cdots&k v_n} $$

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Let's phrase it this way: each $A_k$ gives you a $\text{rank}(A_k)$-dimensional eigenspace of $A_0$. Put them all together, and you find a basis for $\mathbb R^n$ made out of the eigenvalues of $A_0$ – Omnomnomnom Mar 17 '14 at 21:19
None of these eigenvalues have degree more than $1$. Where are you getting that from? An $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors, what more is there to this? – Omnomnomnom Mar 18 '14 at 2:05
None of these eigenvectors, the ones you have mentioned in your original comment have degree more than one, however these may not be all of the eigenvectors of the matrix $A_{0}$. I'm not saying you're wrong I'm just having trouble seeing why these eigenvectors represent all genuine and generalized eigenvectors of $A_{0}$, which is the necessary machinery (as far as I am aware) for the spectral theorem. – DRich Mar 18 '14 at 2:08
If the $m$ sum in the above problem was replaced with $n$, then I would agree. – DRich Mar 18 '14 at 2:09
An $n \times n$ matrix can have at most $n$ (linearly independent) eigenvectors. rank$(A_0)$ of those are in the columns of $A_k$, and there are null$(A_0)$ in the kernel of $A_0$. By the rank nullity theorem, we have $n$ total. – Omnomnomnom Mar 18 '14 at 2:09

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