= Solution
If $A$ is a <skew-symmetric matrix>, the scalar $x^\top Ax$ equals its own <transpose>, so
$$
x^\top Ax=x^\top A^\top x=-x^\top Ax,
$$
and it is zero over the <real numbers>.
Conversely, suppose the <quadratic form> $x^\top Ax$ vanishes for every real <vector>. Substituting $x=e_i$ gives $a_{ii}=0$. Substituting $x=e_i+e_j$ gives $a_{ij}+a_{ji}=0$ for $i\ne j$. Therefore $A^\top=-A$. This proves the <quadratic-form criterion for a skew-symmetric matrix>:
$$
\boxed{x^\top Ax=0\text{ for every }x\iff A^\top=-A.}
$$
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