# Diagonalization of a matrix with repeated eigenvalues

1. Oct 1, 2006

### bemigh

Hey guys,
I know its possible to diagonalize a matrix that has repeated eigenvalues, but how is it done? Do you simply just have two identical eigenvectors??
Cheers
Brent

2. Oct 1, 2006

### Hurkyl

Staff Emeritus
Well, try an example -- the simplest you can think of. How about the identity matrix? Can you diagonalize it? What are its eigenvectors?

3. Oct 1, 2006

### bemigh

Ok, well... i would get an eigenvalue of 1, with a multiplicity of 3. Alright, so when solving for my eigenvalues, by plugging 1 into my matrix A-(lamda)I ; i would just be left with 0's. How can i get eigenvalues from this?? I appreciate your help by the way...

4. Oct 2, 2006

### HallsofIvy

Staff Emeritus
Am I misunderstanding something? If you know what eigenvalues are, then you must know how to find them in this simple case. If $\lambda$ is an eigenvalue of matrix A, then $A-\lambda I$ is NOT invertible (since the equation A- \lambda I= 0 does not have a unique solution) and so det(A- \lambda I)= 0. Since you are finding the eigenvalues, you do NOT yet know that $\lambda= 1$ so you are NOT "left with 0's. In this case, that equation is $(1- \lamba)^3= 0$ which obviously has 1 as a triple root.