Eigenvectors and diagnolization

1. Oct 23, 2009

Two questions. First, I'm given a 3x3 matrix with the last row all zeroes. I'm asked to diagonalizable it, but the determinant is 0, so there are no eigenvalues. Am I reasoning correctly here? It seems an odd question to ask.

Second, I'm asked to prove that if A n x n matrix in C space, then if A has n distinct eigenvalues, A is then diagonalizable. The way I proved it is to say that for A to be diagonalizable, it must have n distinct eigenvectors, which is only the case if it has n distinct eigenvalues. This seems far too easy, am I missing something?

2. Oct 23, 2009

VeeEight

If the determinant is 0 then 0 is an eigenvalue for the matrix

3. Oct 24, 2009

HallsofIvy

Pretty much nothing you have said is right. First of all, you are going the "wrong way". You were asked to prove that if A has n distinct eigenvalues, then it is diagonalizable.

What you are trying to prove is that "if A is diagonalizable then it has n distinct eigenvalues".

You cannot prove that because it is NOT true! Further, it is not true that a matirix having n distinct eigenvectors "is only the case if it has n distinct eigenvalues. The n by n identity matrix has n independent eigenvectors but all n of its eigenvalues is the same.

The other way is correct. Since eigenvectors corresponding to distinct eigenvalues are independent, if an n by n matrix has n distinct eigenvalues then it must have n independent ("distinct" is not enough) eigenvectors and so is diagonalizable. (Write the linear transformation corresponding in a basis consisting of those eigenvectors.)