From https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Vector_space_structure
The real $n\times n$ matrix ${\textstyle A}$ is skew-symmetric if and
only if${\displaystyle \langle Ax,y\rangle =-\langle x,Ay\rangle \quad
\forall x,y\in \mathbb {R} ^{n}}$ This is also equivalent to
${\textstyle \langle x,Ax\rangle =0}$ for all ${\displaystyle x\in
\mathbb {R} ^{n}}$
I am trying to show $\langle x,Ax\rangle =0$ but I don't really see it. If I had something like $$\langle x,Ax\rangle = -\langle x,A x\rangle$$ then I can claim $\langle x,Ax\rangle = 0$, but instead I have $$\langle x,Ax\rangle = -\langle x,A^\top x\rangle = -\langle Ax, x\rangle \neq 0 \quad (?)$$
Best Answer
Note that $$ \langle x,Ax\rangle\stackrel{\text{skew symmetry}}{=}-\langle Ax,x\rangle= -\langle x,Ax\rangle\\ \implies 2\langle x,Ax\rangle=0\\ \implies \langle x,Ax\rangle=0 $$