# 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:

1. Determine if $\dim(\mathrm{span}(v_i)) =$ multiplicity of $\lambda_i$. If no, then it is not diagonolizable. If yes, go to 2
2. Is the eigenvectors linearly independent? If yes, we can diagonalize. If no, ... I don't know.
• Eigenvectors relative to different eigenvalues are linearly independent. Mar 15, 2014 at 15:25

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.

• but distinct eigenvalues (i.e. all eigenvalues has multiplicity of 1) implies that the eigenvectors corresponding to those eigenvalues are in fact L.I. Thus I do not have check independency by calculating det≠0 or find out by gaussJordan. Mar 16, 2014 at 18:00
• This is true. Your point is? I thought you were asking how to diagonalize a matrix. Mar 16, 2014 at 18:04
• 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"). Mar 17, 2014 at 10:16
• @jacob If I understand you correctly, you got it wrong. The If and only if dim(span(the eigenvectors)) = m , I can diagonalize A part is correct if $A$ is a $m\times m$ matrix. But the If and only if I have as many linearly independent vectors in S, as I have number of eigenvalues, the matrix is diagonalizable part isn't right. You can have just one eigenvalue in a $3\times 3$ matrix for instance and the matrix can be diagonalizable and you can get three linearly independent eigenvectors. Mar 17, 2014 at 10:34
• So it would be easier to just skip this way of thinking and just stating: find some set of eigenvectors, if they are L.I. and the total # of vectors is p if the matrix is pxp, then you can diagnoalize the matrix. Mar 17, 2014 at 10:56

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").
• What is the order of the given matrix? Is it $m\times m$? Mar 17, 2014 at 10:35
• yes it is m x m. Mar 17, 2014 at 10:53
• The last bullet has two iff's. The first one is correct, the second is wrong. Mar 17, 2014 at 10:55
• @GitGud ok, got it. maybe. Mar 17, 2014 at 11:14