# What is wrong with this diagonalization problem.

1. Nov 26, 2008

### bluewhistled

1. The problem statement, all variables and given/known data
A is

[4 0 1
2 3 2
1 0 4]

Find an invertible P and a diagonal D so that D=P-1AP.

I keep getting two linearly dependent eigenvalues which means it's not diagonal but this problem doesn't state "If it can't be done explain why" or anything like that. So I just want to verify with some of you.

3. The attempt at a solution
I subtract with LI and take the determinant and get:

(L-3)((L-4)^2 - 1)
(L-3)(L^2-8L+16)-(L-3)
L^3-8L^2+16L-3L^2+24L-48-L+3
L^3-11L^2+39L-45

Which I then factor out to be 5, 3, 3.

Am I doing something wrong/missing something?

2. Nov 26, 2008

### VeeEight

The eigenvalues are correct. Now, did you plug each eigenvalue, one at a time, into your matrix to find the corresponding eigenvectors? There are two eigenvectors for eigenvalue 3, and one for eigenvalue 5.

3. Nov 26, 2008

### bluewhistled

Right but you can't find the diagonalization when the two eigenvalues are the same. They are linearly dependent. Right? I just wanted to verify that I wasn't going crazy.

Once you have two identical eigenvectors for P you can't invert it to eventually find D. Or am I supposed to just assume that D is the diagonal matrix of 5,3,3 even though you can't actually get to that point.

4. Nov 26, 2008

### e(ho0n3

The minimal polynomial of your matrix (x - 5)(x - 3), so it is diagonalizable.

5. Nov 26, 2008

### VeeEight

It is diagonalizable if and only if the characteristic polynomial of your matrix, A, splits and for each eigenvalue X of A, the multiplicity of X equals n - rank (A-XI). So check to see if these criteria apply here (you have almost done the first step)

6. Nov 26, 2008

### gabbagabbahey

When two or more of your eigenvalues are the same, you need to search for generalized eigenvectors. These generalized eigenvectors will be linearly independent, and you can construct your P-matrix from them.

7. Nov 26, 2008

### e(ho0n3

Or more simply, if and only if the minimal polynomial has the form (x - a1)(x - a2)...(x - an) where ai are the distinct eigenvalues of A. D will be the diagonal matrix having a1, a2, etc. down the diagonal and P will be the matrix whose columns are a linearly independent set of eigenvectors corresponding to the eigenvalues of A.

8. Nov 26, 2008

### bluewhistled

Can one of you help me find the generalized vectors I have no idea what to do.

9. Nov 26, 2008

### bluewhistled

Nm, I figured it out. when looking for the eigenvector for 3 it actually splits into 2. I think I did the math wrong the first time. Thanks you guys.