Eigenvalues/vectors diagonalization

[emyt]

Homework Statement

Suppose that A \in Mnxn(F) has two distinct eigenvalues \lambda_{1} and \lambda_{2} and that dim(E_{\lambda_{1}}) = n - 1. Prove that A is diagonalizable

Homework Equations

The Attempt at a Solution

hmm, I'm not sure.. how would I start this?

thanks

 

[HallsofIvy]

If the dimension of the eigenvector space corresponding to eigenvalue \lambda_1 is n- 1, then there are n-1 independent eigenvectors corresponding to \lambda_1

That, together with an eigenvector corresponding to eigenvalue \lambda_2 gives you n independent vectors (eigenvectors corresponding to distinct eigenvalues are always independent) and form a basis for the vector space. The matrix for the linear transformation, written in that basis, is a diagonal matrix having \lambda_1 n-1 times on the main diagonal and \lambda_2 once.

 

[emyt]

thanks, I was thinking about that too, but for some reason my mind strayed and thought that I would need 2n independent vectors..
I was also looking at a couple of the diagonal "if and only if" theorems. I saw a couple that might've been useful, like:
T is diagonalizable if and only if the multiplicity of\lambda is equal to dim(E_{\lambda}).
And now I have to ask: how can you deduce the multiplicity of an eigenvalue without working out the characteristic polynomial and looking at the multiplicity of the factors?
thanks :)