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Show that if an $n\times n$ matrix $A$ satisfies $A^T=-A$, then $x^TAx=0$ for any $n\times 1$ vector $x$.

My attempt: Since matrix transpose won't affect the diagonal entries, so matrix $A$ has only zeros on its diagonal.

Then I tried to write $x$ in the form of $\begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{pmatrix}$ and $A$ in the form of $\begin{pmatrix} 0 & a_{11} & \cdots & a_{1n}\\ a_{21} & 0 & \cdots & \cdots \\ \vdots &\ddots &\ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & 0 \end{pmatrix}$ and multiply them in this form, but it doesn't make any sense to me.

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  • $\begingroup$ See what indices of $x$ multiply $a_{ij}$, and then see what indices of $x$ multiply $a_{ji}$. $\endgroup$ – NicNic8 Jun 5 '18 at 5:01
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    $\begingroup$ What's the transpose of $x^TAx$? $\endgroup$ – Lord Shark the Unknown Jun 5 '18 at 5:03
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    $\begingroup$ @LordSharktheUnknown $(x^TAx)^T=x^TA^Tx=-x^TAx$. will that help? $\endgroup$ – Thomas Jun 5 '18 at 5:05
  • $\begingroup$ @fcc $x^\top Ax$ is a scalar. $\endgroup$ – StubbornAtom Jun 5 '18 at 5:11
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What you can see is that $x^TAx$ is of type $(1\times1)$, hence symmetric, i.e.,

$$(x^TAx)^T = x^TAx.$$ On the other hand we have $$x^TA^Tx= x^TAx.$$

From $A^T=-A$ we get $$-x^TAx = x^TAx\iff 2x^TAx =0\iff x^TAx=0.$$

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We have that $$x^TA^Tx=\sum_{i,j}x_i(A^T)_{ij}x_j=\sum_{i,j}x_iA_{ji}x_j=\sum_{i,j}x_jA_{ji}x_i=x^TAx$$ But we also have that $$x^TA^Tx=\sum_{i,j}x_i(-A)_{ij}x_j=-x^TAx$$ So $$x^TAx=-x^TAx$$ So it must be $0$.

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  • $\begingroup$ for $x_j$ and $x_i$, do you mean entries of the vector? Why does $x_jA_{ji}x_i=xAx$? $\endgroup$ – Thomas Jun 5 '18 at 5:08
  • $\begingroup$ @fcc I was using einstein notation. But I will add the $\sum$ signs. $\endgroup$ – Botond Jun 5 '18 at 5:10

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