# Diagonalizable Matrix

## Homework Statement

Are the following matrices diagonalizable:

[PLAIN]http://img693.imageshack.us/img693/4198/91350081.jpg [Broken]

## The Attempt at a Solution

I solved for the eigenvectors of both matrices, but only found one eigenvector for each. This means that neither of the two matrices have at least two linearly independent eigenvectors, which means neither are diagonalizable.

Am I correct?

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lanedance
Homework Helper
show your working & people can check it rather than having to do the whole problem (i haven't checked it)

however you are correct that for a 3x3 if the sum of the dimensions of the eigenspaces is <3, the matrix is not diagonalisable

vela
Staff Emeritus
Homework Helper
You must not be solving for the eigenvectors correctly, because they're both diagonalizable.

You must not be solving for the eigenvectors correctly, because they're both diagonalizable.

For a) did you get an eigenvalue of 3 and for b) 0, 1, and 3?

lanedance
Homework Helper
sonce again - show your working, and i'll try & help, but I won't do the whole problem for you ;)

a) I'm not so convinced this one is diagonalisable - as a start, what is the characteristic equation & what is the multiplicity for the eigenvalue $\lambda = 3$ (consider both algerabraic & geometric multiplicity) -

b) if you have 3 distinct real eigenvectors, you should be able to find 3 corresponding distinct real eigenvectors - what does this tell you ablout the diagonalisability of your matrix?

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Dick
Homework Helper
I got three distinct eigenvalues for the first one as well. Two are imaginary, but that's ok.

lanedance
Homework Helper
I assume the question means diagonalisable over the reals though, so that result would imply the matrix in a) is not diagonalisable...

Dick
Homework Helper
I assume the question means diagonalisable over the reals though, so that result would imply the matrix in a) is not diagonalisable...

That would be true, if that's what temaire means.

That would be true, if that's what temaire means.

Yes, we're only considering real numbers here. So for a) I found only one real eigenvalue, 3, which doesn't have at least 3 linearly independent eigenvectors, so it's not diagonalizable. For b) there are three real eigenvalues, 0, 1, and 3, and I was only able to come up with one linearly independent eigenvector, (-1, 1, 1) for the eignevector 0. For eigenvectors 1 and 3, I get no solutions, so therefore no eigenvectors. Did I go wrong here?

vela
Staff Emeritus