Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue. I can't find the missing link between singularity and zero eigenvalues as is stated in the following proposition:
A matrix $A$ is singular if and only if $0$ is an eigenvalue.
Could anyone shed some light?

An $n \times n$ matrix, $\mathbf A$, is singular if and only if there is a non zero column vector $\mathbf x$ such that $\mathbf A \mathbf x = 0 = 0 \mathbf x$, i.e.,  $0$ is an eigenvalue.
