Let $P$ be a real symmetric positive definite matrix and $S$ a real skew-symmetric matrix, both of dimension $n$. It is well known the fact that for any $x \in \mathbb{R}^n$ one has $$ x^T\cdot S \cdot x = 0$$ My question is if the product $P\cdot S$ has the same property? That is, is it possible to prove that for all $x \in \mathbb{R}^n$ one still has $$ x^T\cdot P \cdot S \cdot x = 0$$
My attempt
It is known that $$ P^{-\frac{1}{2}}\cdot P\cdot S = (P^{\frac{1}{2}}\cdot S \cdot P^{\frac{1}{2}}) \cdot P^{-\frac{1}{2}}$$ and $P^{\frac{1}{2}} \cdot S\cdot P^{\frac{1}{2}}$ is skew-symmetric ... Therefore $P\cdot S$ is similar to a skew symmetric ... is it skew symmetric?
Let $P= \begin{bmatrix} 1 & 0 \\ 0 & 2\end{bmatrix}$ and $S = \begin{bmatrix} 0 & 1 \\ -1 & 0\\ \end{bmatrix}$,
then we have $PS = \begin{bmatrix} 0 & 1 \\ -2 & 0\end{bmatrix} $ which is not skew symmetric.