Is it possible for an order N square matrix to be diagonalizable but have 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.
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:
so I guess I have answered my own question.