There are two cases to the proof:
(we are in unitary spaces)
I) when we have n=k, we easily prove it as it follows:
$\Gamma(S) = \det (S) = \det ( M^*M)=|M^*||M|=|M|^2\geq 0$
II) when $ n < k$ :
it can be done it two ways. The first is geometrically; according to the Roché's theorem, vectors are linearly dependent if $|S| = 0$, or if they are linearly independent when $|S| \neq 0$. From here we know that a determinant represents a volume of a parallelotope. Therefore volume cannot be negative, hence we proved it.
My question is, how do we prove, the second case, if we do not look geometrically. If we say $\Gamma = \det(M^*M)\geq 0$ and we take an arbitrary vector $a\in U$, $a=(\alpha_{1},...,\alpha_{n})_{E}$, where $E$ is the basis.
How do we from this:
$a^*M^*Ma\geq 0$
prove that this
$\Gamma= |S| = \det(M^*M) \geq 0$ must be true?