I have seen it claimed multiple places (e.g. here and here) that a (not necessarily symmetric) matrix $M\in\mathbb{R}^{n\times n}$ is PD (i.e., $x^TMx>0$, $\forall x\in\mathbb{R}^n$) if and only if its symmetric part $\frac{1}{2}(M+M^T)$ is PD.
This is the case if $\frac{1}{2}x^T(M-M^T)x=0$, however I am only able to show this if $M$ is symmetric (in which case it is trivial).
Are the above linked claims true or false?
Note that $$\langle x, (M+M^{\mathrm{t}}) x\rangle = \langle x, M x\rangle + \langle x, M^{\mathrm{t}} x\rangle = \langle x, M x\rangle + \langle M x, x\rangle = 2\langle x, M x\rangle.$$