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Thanks for your answer Cedron! Taking your same assumption $N \leq M$, and by definition of PD, $x^* R_x^{-1} x > 0\quad \forall x \in \mathbb{R}^M \setminus \{0\}$ and since $C$ is full rank $\dim (\mathcal{R}(C)) = \min(M,N)= N$. By the rank-nulity theorem, $\dim (\mathcal{R}(C)) + \dim(\mathcal{N}(A))=N$, so we have that the null-space is trivial. This ...
First, this question is probably better for MATH.SE, but I'll give it a shot. It's been a long long time since I did this stuff. If $N > M$: 1) $C^*$ has rank M. 2) $R_x^{-1}$ has rank M. (or it wouldn't exist as an inverse) 3) Therefore $C^*R_x^{-1}$ has rank M, since $R_x^{-1}$ is a full rank square matrix. 4) $C$ has rank M 5) $C^*R_x^{-1}C$ ...