 Solution

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.