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?

  • $\begingroup$ It's not true. Being similar to a skew-symmetric matrix doesn't mean skew-symmetric. $\endgroup$ – Lord Shark the Unknown Sep 13 '18 at 8:48
  • $\begingroup$ Lord Shark the Unknown ok $\endgroup$ – C Marius Sep 13 '18 at 8:52

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.

  • $\begingroup$ Siong Thye Goh ok $\endgroup$ – C Marius Sep 13 '18 at 8:52

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