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We know for a skew-symmetric matrix A, $x^TAx = 0$. But is the converse statement true, i.e. does $x^TAx = 0$ imply A is skew-symmetric? If yes, then how to prove it?

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  • $\begingroup$ Are you saying that $x^TAx=0$ is true for all column vectors $x$ or just for one? $\endgroup$ – Rory Daulton Jan 23 '16 at 19:48
  • $\begingroup$ for all column vectors $x$. $\endgroup$ – SM10 Jan 23 '16 at 19:53
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It is true. We have: $$(x+y)^TA(x+y) = 0 \implies x^TAx + y^TAx + x^TAy + y^TAy = 0.$$But $x^TAx = y^TAy = 0$, so we have: $$x^TAy = -y^TAx.$$Take $x = e_i$ and $y = e_j$ to get $a_{ij} = -a_{ji}$.

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$x^T A x = 0\ \forall x$ can be written as: $\sum_i \sum_j x_j a_{ij} x_i = 0\ \forall x$

That is: $x_1^2 a_{1,1} + x_1 a_{1,2} x_2 + ... = 0$

The only way to have $x_i x_j$ cancel out $x_j x_i$ is by: $a_{ij} = -a_{ji}$, and the remaining terms $x_i^2 a_{ii}$ must be zero (for all $x$), so we must have $a_{ii}=0$. Therefore $A$ is antisymmetric.

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