# Diagonalizing a square matrix with degenerate eigenvalues

1. Aug 2, 2008

### maverick280857

Hi

This is more of a math question but in the context of Quantum Mechanics, hence I posted it here. Suppose I have a matrix A of order 3x3 with three eigenvalues: 0, 0, 5. I am supposed to find the diagonalizing matrix for A.

I know that in general, if P denotes the matrix of eigenvectors of A, then $PAP^{-1}$ will be a diagonal matrix.

In my particular example, for the eigenvalue 0,

AX = 0

gives infinitely many solutions for X, so the eigenvector with eigenvalue 0 cannot be uniquely determined.

How do I diagonalize such a matrix?

Cheers
Vivek

2. Aug 3, 2008

### Defennder

There are infinitely many different eigenvectors for an eigenvalue, but are they all linearly independent? Remember that you only need 3 linearly independent eigenvectors for this. Just pick a simple eigenvector associated with the zero eigenvalue.

3. Aug 3, 2008

### maverick280857

Ok, lets say I take (1, 0, 0)' and (0, 1, 0)' as the eigenvectors for the zero eigenvalue. And then I just plug them along with the eigenvector for the eigenvalue 5...and thats it, right?

4. Aug 3, 2008

### Defennder

Well, that's provided you "plug them in" correctly, doesn't it?

5. Aug 3, 2008

### maverick280857

Yes, as the problem asks to find the diagonalizing matrix, I was wondering what to write as the eigenvectors are not uniquely determined here. By "plug them in", I meant that I construct a matrix P whose columns are precisely the Linearly independent eigenvectors of the matrix A...in this case I select two Linearly independent eigenvectors for the zero eigenvalue and the third column is the eigenvector for the eigenvalue 5.