Set of commuting Matrices $\Rightarrow $ Common Eigenvector

Question

I am trying to prove that if we have an arbitrary set of commuting matrices in $M_n(\mathbb C)$ then they have a common eigenvector.

Well, if we have only 2 matrices, the answer is easy and it has been already answered here: Matrices commute if and only if they share a common basis of eigenvectors? (acutally in this case we find a common eigenbasis, in case they are diagonalizable)

For the general case of an arbitrary set I am trying to find a subspace of $M_n(\mathbb C) $ which contains that arbitrary set of commuting matrices. Then Prove the statement for the basis ( is going to have $\leq n^2$ elements) by induction.

So if $Y $ is that arbitary set, take $span(Y)$ which has dimension $\leq n^2$ so find a basis $Y_1 ,Y_2,\dots Y_k,k\leq n^2$

Then prove by induction that we can find a common eigenvector $v\neq 0 $ for the vectors of the basis. So, if $A\in Y\Rightarrow $ $$A=\sum_{m=1}^k a_mY_m\Rightarrow Av=\sum_{m=1}^k a_mY_mv\Rightarrow$$ $$Av=\sum_{m=1}^k \left( a_m\lambda_m\right) v$$ i.e. $v$ is an eigenvector for $A$.

What do you think about that solution?

P.S. I asked if you could check my solution. I have already seen the post which is supposed to answer my question but I am not satisfied ( see comment).

I see... But I have an objection: We are taking $W_\infty=\bigcap\limits_{i=1}^\infty W_i$ i.e. we assume that we are having a countable set of matrices ... Can my case be reduced to this? What do you think about my proof? — Dimitris, Sep 08 '13 at 20:49
Dimitris Dallas: Your solution is correct. Voldemort's proof needs to be fixed in the case of uncountable families. — darij grinberg, Sep 08 '13 at 21:15
Then the fun thing is that my Correct solution is going to be closed, as a duplicate of the Wrong ONE ! Great community! — Dimitris, Sep 08 '13 at 21:18

Set of commuting Matrices $\Rightarrow $ Common Eigenvector

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