2
$\begingroup$

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.
$\endgroup$
1
  • $\begingroup$ Eigenvectors relative to different eigenvalues are linearly independent. $\endgroup$
    – egreg
    Mar 15, 2014 at 15:25

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – jacob
    Mar 16, 2014 at 18:00
  • $\begingroup$ This is true. Your point is? I thought you were asking how to diagonalize a matrix. $\endgroup$
    – Git Gud
    Mar 16, 2014 at 18:04
  • $\begingroup$ 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"). $\endgroup$
    – jacob
    Mar 17, 2014 at 10:16
  • $\begingroup$ @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. $\endgroup$
    – Git Gud
    Mar 17, 2014 at 10:34
  • $\begingroup$ 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. $\endgroup$
    – jacob
    Mar 17, 2014 at 10:56
-1
$\begingroup$

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").
$\endgroup$
4
  • $\begingroup$ What is the order of the given matrix? Is it $m\times m$? $\endgroup$
    – Git Gud
    Mar 17, 2014 at 10:35
  • $\begingroup$ yes it is m x m. $\endgroup$
    – jacob
    Mar 17, 2014 at 10:53
  • $\begingroup$ The last bullet has two iff's. The first one is correct, the second is wrong. $\endgroup$
    – Git Gud
    Mar 17, 2014 at 10:55
  • $\begingroup$ @GitGud ok, got it. maybe. $\endgroup$
    – jacob
    Mar 17, 2014 at 11:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .