diagonalize a non-normal matrix , without distinct eigenvalues I wonder how to diagonalize a matrix that is non-normal, and does not have distinct eigenvalues.

Let  $\lambda_i$ be the eigenvalue, and $v_i$ be the eigenvector with that eigenvalue. 

I think the process would go like this:


Determine if $\dim(\mathrm{span}(v_i)) = $ multiplicity of $\lambda_i$. If no, then it is not diagonolizable. If yes, go to 2
Is the eigenvectors linearly independent? If yes, we can diagonalize. If no, ... I don't know.
Non-normal matrices may or may not be diagonalizable.

Given $n\in \mathbb N$ and $A\in \mathcal M_{n\times n}(\mathbb C)$, it holds that
$A$ is diagonalizable (over) $\mathbb C$ if, and only if, there exists a basis $\{v_1, \ldots ,v_n\}$ of $\mathbb C^{n\times 1}$ such that $v_1, \ldots ,v_n$ are all eigenvectors of $A$.

If $A$ is diagonalizable, then to find a diagonalizing matrix $P$, you just have to find the vectors $v_1, \ldots ,v_n$ above and define $P$ by letting its columns be the eigenvectors found. This implies that $P^{-1}AP$ is a diagonal matrix.

This works whether $A$'s eigenvalues have all multiplicty $1$ or not.
Is this correct?


I'm given a matrix
I find out that it is non-normal
I find its eigenvalues. Let's say I have m of them.
I find out it does not have distinct eigenvalues
I find some set of eigenvectors to the eigenvalues, call this set S.
If and only if dim(span(the eigenvectors)) =  m , I can diagonalize A. (To put the equation in words, just to make sure "If and only if I have as many linearly independent vectors in S, as I have number of eigenvalues, the matrix is diagnoalizable").