Diagonalizability of singular matrices

Question

Can a matrix that has dependent columns have different eigen values and be diagonalizable?

Answer 1

$A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ has eigenvalues $\lambda_1=0,\,\lambda_2=1$, which are distinct (2 eigenvalues for a $2\times 2$ matrix means it's diagonalizable), and one of the columns is $\vec 0$, so the columns are linearly dependent. Having linearly dependent columns is equivalent to having zero as an eigenvalue. A matrix can have zero as an eigenvalue, and still be diagonalizable, as long as the geometric multiplicities equal the algebraic multiplicities for each eigenvalue.

Answer 2

To generelize Dave's answer:

An $n \times n$ matrix is diagonalizable if-and-only-if it has $n$ independent eigenvectors (which is not the same as $n$ independent columns in the matrix). Some of these eigenvectors can very well have eigenvalue 0 - which you definitely get if your matrix has linearly dependent columns.