You're not always guaranteed that the algebraic multiplicity of an eigenvalue (multiplicity of the root in the characteristic equation) is equal to the geometric multiplicity of the eigenvalue (dimension of the space spanned by the corresponding eigenvectors). However, by the way the definition works, I believe you can prove that each distinct eigenvalue will give you at least one eigenvector....unless its guaranteed you get an eigenvector when you have an eigenvalue...

Question: How does this prove the needed result?