As in the title. Also, if anyone knows if all Hermitian-symmetric matrices with distinct diagonal elements are diagonalizable, that'd be great to know. Thanks.
Edit: Never mind about the Hermitian one, answered here: Are singular matrices diagonalizable?
The answer is no. Consider a Toeplitz matrix of this form:
$A = \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix}$
$A$ is a non-diagonalizable matrix with eigenvalues equal to $\lambda$, but it is not diagonalizable because the algebraic multiplicity (number of repeated eigenvectors) of $\lambda$ is 4, which is greater than its geometric multiplicity (number of linearly independent eigenvectors of $\lambda$), which is 1.
In fact, this is as diagonalized as this matrix is going to get, because this is the Jordan form of some defective matrix.