Is it possible for an order N square matrix to be diagonalizablebuthave less than 3 distinct eigenvalues? I've heard that it's possible, but can't think of an example.

Identity doesn't work, since it yields only the zero-vector for its e-value of 1, which is no good.

Edit:

It also doesn't imply that the matrix is invertible, right? Since you can have a singular matrix that is diagonalizable, right?

Edit: Here is a 3x3 matrix that is singular and has 2 distinct e-values, yet still has 3 linearly independent e-vectors, and hence is diagonalizable:

$\displaystyle A=\begin{bmatrix}229& -225 & -30 \\

225 & 229 & -30 \\

-30 & -30 & 450

\end{bmatrix}$

so I guess I have answered my own question.