We want to show that for every matrix $M \in S_n(\mathbf{R})$ we have $\pi(M) = d(M)$. By contradiction, assuming the existence of a vector subspace $G$ of $\mathcal{M}_{n,1}(\mathbf{R})$ of dimension $\dim G > \pi(M)$ satisfying condition $(\mathcal{C}_M)$, show $\dim(F_M^\perp \cap G) \geq 1$, deduce a contradiction and conclude.