Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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

share|improve this answer
    
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 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 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 at 2:08
    
If the $m$ sum in the above problem was replaced with $n$, then I would agree. –  DRich Mar 18 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 at 2:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.