# Skew-symmetric matrices dot product condition

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.

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 $$(AB)^T=B^TA^T$$
• 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! – PJ Miller Aug 10 '13 at 2:29