Eigenvectors and diagnolization

[Hallingrad]

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?

 

[VeeEight]

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

 

[HallsofIvy]

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?

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.)