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From the Wikipedia page on skew-symmetric matrices:

Denote with $\langle\cdot,\cdot\rangle$ the standard inner product on $\mathbb{R}^n$. The real $n$-by-$n$ matrix $A$ is skew-symmetric if and only if $\langle Ax, y\rangle = -\langle x, Ay\rangle$ for all $x,y\in \mathbb{R}^n$.

I can't see how this follows from the definition $A^T=-A$ for skew-symmetric matrices.

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Hint: for two real vectors $v,w$, we note that $$ \langle v,w \rangle = v^Tw $$ Now, consider what this means when $$ v = Ax\\ w = y $$ Noting that for multiplicatively compatible matrices $A,B$, we have $$ (AB)^T=B^TA^T $$

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    $\begingroup$ Ah.. $\langle Ax,y\rangle+\langle x,Ay\rangle = 0$ iff $x^T(A^T+A)y=0$. This is obvious if $A^T+A=0$. On the other hand, we can choose $x=e_i$ and $y=e_j$ to get $A^T+A=0$. Done! $\endgroup$ – PJ Miller Aug 10 '13 at 2:29
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    $\begingroup$ That is even better-put than the proof I expected. Good job. $\endgroup$ – Omnomnomnom Aug 10 '13 at 2:32

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