= Solution
The entries of $M^TM$ are real, and the <transpose> rule gives $(M^TM)^T=M^TM$. Thus it is a real <symmetric matrix>, hence also a <Hermitian matrix>.
For any complex vector $v$,
$$
v^\dagger M^TMv=(Mv)^\dagger(Mv)=\|Mv\|^2\geq0.
$$
In particular, if $M^TMv=\lambda v$ with $v\ne0$,
$$
\boxed{\lambda=\frac{\|Mv\|^2}{\|v\|^2}\geq0.}
$$
Thus $M^TM$ is <positive semidefinite>. Its zero <eigenspace> is precisely the kernel of $M$, since $\|Mv\|^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative <eigenvalues>.
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