Invertibility, eigenvalues and singular values

A square matrix is invertible if and only if it does not have a zero eigenvalue.

The same is true of singular values: a square matrix with a zero singular value is not invertible, and conversely.

The case of a square $n\times n$ matrix is the only one for which it makes sense to ask about invertibility. Its determinant is the product of all the $n$ algebraic eigenvalues (counted as to multiplicity). Since the determinant is nonzero if and only if the matrix is invertible, this is one way to recognize the equivalence of being invertible with not having a zero eigenvalue.

When $A$ is a square matrix, the eigenvalues of $A^TA$ are precisely the squares of the singular values of $A$. So if $A$ has a zero singular value, $A^TA$ has a zero eigenvalue, and conversely. Clearly $A^TA$ is invertible if $A$ is (the determinant of $A^T$ equals the determinant of $A$), and conversely if $A^TA$ is invertible, its determinant $|A^TA| = |A|^2$ is nonzero, and thus $|A|$ is nonzero, and $A$ is invertible.

Something can be said about the non-square case using singular values, but one needs a little care in discerning what it means for $A$ not to have a zero singular value. $A$ and $A^T$ will have equal rank, and at most one of $A^T A$ and $AA^T$ can be invertible in the nonsquare case. In any case one can discern that a non-square matrix is not "invertible" simply from its shape, and singular value computation is unnecessary.