If an n x n matix A is not invertible, then is 0 an eigenvalue of A? I know that if $0$ is an eigenvalue of a matrix $A$, then $A$ is not invertible, but is the opposite also true? If the matrix isn't invertible, is the eigenvalue always $0$?

Yes.
One way to think about it is if a square matrix is not invertible, it is linearly dependent. If the columns of a matrix are linearly dependent, there exists a linear combinations of the columns that sums to the $0$ vector.
$A \vec{b} = \vec0$
In this situation A is acting like a scalar multiple of $\vec{b}$. This is the definition of an eigenvalue. The factor it is scaling $\vec{b}$ by, $0$, is the eigenvalue.
$A\times0 = \vec0$
Thus, $0$ is an eigenvalue for all non-invertible square matrices.
