# Eigenvalues/vectors diagonalization

• emyt
In summary, if a matrix A has two distinct eigenvalues and the dimension of the eigenvector space for one of them is n-1, then A is diagonalizable. This is because the n-1 independent eigenvectors for the first eigenvalue, plus one eigenvector for the second eigenvalue, form a basis for the vector space. By writing the matrix in this basis, it can be shown to be diagonal with the first eigenvalue appearing n-1 times on the main diagonal and the second eigenvalue appearing once. This can also be deduced by looking at the multiplicity of the eigenvalue in the characteristic polynomial.

## 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

## The Attempt at a Solution

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

thanks

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.

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 :)