Diagonalizing a square matrix with degenerate eigenvalues

[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?

Thanks in advance,

Cheers
Vivek

 

[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.

 

[maverick280857]

Ok, let's 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 that's it, right?

 

[Defennder]

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

 

[maverick280857]

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

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.

Thanks for your help Defennder