Answer
See below.
Work Step by Step
(Forward direction) First, let $\lambda=0$ be an eigenvalue of a matrix $A$. Therefore, $$\det(A-\lambda I) = \det(A-0I)=\det(A) = 0$$ This is the definition of singular. Therefore, $A$ is singular.
(Backwards direction) Next, let $A$ be a singular matrix. Then, $\det(A) = 0$. Since $\det(A)=\det(A-0I)=0$, $\lambda=0$ is an eigenvalue of $A$.
