Let be a field, a finite dimensional vector space over , and a linear transformation.
- Prove that if has a basis consisting of eigenvectors for , then the matrix is diagonal and moreover the diagonal entries are eigenvalues.
- If is a basis for , then is similar to a diagonal matrix (having nonzero diagonal entries) if and only if there is a basis for consisting of eigenvectors of .
- Let be a basis of eigenvectors for . Suppose that for each , is the eigenvalue associated to . That is, . In particular, we have , where if and 0 otherwise. So is diagonal, with eigenvalues as diagonal entries.
- Suppose is similar to a diagonal matrix ; say . Let be the diagonal entries of . We may take the columns of to be the coordinates (in terms of ) of a basis for . Then is the matrix of with respect to the basis . In particular, , so that each is an eigenvector of .
Conversely, if there is a basis of consisting of eigenvectors, then by part 1, is a diagonal matrix. For any other basis , we have an invertible change of basis matrix such that .