# Exponent of matrix/Diagonalization of matrix with repeated eigenvalue

1. Mar 25, 2012

### martizzle

Hello,

it's been a while since i did linear algebra. i need some help. I have this matrix:

1 1 0
0 1 0
0 0 0.

I know the eigenvalues are 1,1,0; and that the eigenvectors will be: (1,0,0), (0,0,0) and (0,0,1). But I cannot do the jordan decomposition on the matrix i.e. write it in the form: P M P^-1. Where P is the matrix made up of eigenvectors, M is a diagonal matrix containing the eigenvalues: 1,1,0.

My main interest however, is in finding the matrix exponent of
1-2i 1+3i 0
0 1-2i 0
0 0 0.

If I can diagonalize the first matrix, I should be able to use the same method to diagonalize/jordan decompose this matrix so that i can find the matrix exponent.

Thanks for any help.

Last edited: Mar 25, 2012
2. Mar 25, 2012

### HallsofIvy

Staff Emeritus
Obviously, you cannot find P-1 because, taking the 0 vector as a column, you are forcing the determinant to be 0. In order to "diagonalize" a matrix you need to find a non-zero eigenvector (some texts do not accept the 0 vector as an eigenvector at all).

The difficulty is that not every matrix is diagonalizable! In order that a matrix be diagonalizable, there must exist a "complete set" of eigenvalues. That is, a basis for the space consisting of eigenvectors. For a three by three matrix, there must be three independent eigenvectors and that is not true here.

What you can do, if you cannot diagonalize a matrix, is put it in "Jordan Normal" form. Here, the matrix you are given is already in Jordan Normal form!