Solution
= Solution
If $A^\top=-A$, then the <dot product> satisfies
$$
(Ax)\cdot y=x^\top A^\top y=-x^\top Ay=-(Ay)\cdot x.
$$
Conversely, the assumed <dot product> identity gives $x^\top(A^\top+A)y=0$ for every $x,y$. Take $x=e_i$ and $y=e_j$ from the <standard basis>: every entry of $A^\top+A$ is zero. Hence \b[$A^\top=-A$], proving the equivalence.