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?

I all depends on your starting definition. Here is one way, suppose $v$ is the eigen vector associated with $\lambda=0$ then $Av=0v=0$. Since $v\ne 0$ by definition then you have a nontrivial vector in the null space of $A$ that makes $A$ singular.
