# $A$ skew-symmetric, then $x^T A^2 x \le 0$

Let $A$ be a skew-symmetric matrix.
Show that for any vector $x$ $$x^T A^2 x \le 0.$$

Since $A$ is skew-symmetric then $A^2$ is symmetric.
Then, I'm blocked.
I tried to consider $$a=x^T A^2 x,$$ without success.

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Hint: If $A$ is skew-symmetric, then $x^TA = -x^TA^T$. Therefore $$x^TA^2x = -x^TA^TAx = -(x^TA^T)(Ax) = \ldots\ ?$$
Is $B^T B \ge 0$ for any $B$ : $-\left( x^T A^T \right) \left(A x \right) = - \left(A x \right)^T \left(A x \right)$ –  Nicolas Essis-Breton Feb 7 '13 at 19:47
$= -\|Ax\|^2 \leq 0.$ Right. –  user7530 Feb 7 '13 at 19:56