If $M_{\alpha\beta}\ge0$ , how to show $M^\#\ge0$?

Question

We say $M_{\alpha\beta}\ge0$ if $M_{\alpha\beta}v^\alpha v^\beta\ge0$ (sum over $\alpha,\beta$) for all vectors $v=\{v^\alpha\}$.

If $M_{\alpha\beta}\ge0$ ,how to show $$ M^\#\ge0 ~? $$

Relative define is in the below picture. The picture blow is from the 186-188 and 212 of this paper.If there are some details be missed,please tell me, I will add it.

Answer 1

$M : \wedge^2 V\otimes_S \wedge^2 V\rightarrow {\bf R} $ so that there exists a basis $\{\phi_i\}$ which diagonalize $M$

So $$v=v^a\phi_a,\ M^\sharp (v,v)= v^a v^bC^{cd}_a C^{ef}_b M_{ce} M_{df} $$

$$ = v^a v^bC^{cd}_a C^{cd}_b M_{cc} M_{dd} =( v^a C^{cd}_a)^2 M_{cc}M_{dd} \geq 0 $$