matrix transpose proof

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.

• See what indices of $x$ multiply $a_{ij}$, and then see what indices of $x$ multiply $a_{ji}$. Commented Jun 5, 2018 at 5:01
• What's the transpose of $x^TAx$? Commented Jun 5, 2018 at 5:03
• @LordSharktheUnknown $(x^TAx)^T=x^TA^Tx=-x^TAx$. will that help? Commented Jun 5, 2018 at 5:05
• @fcc $x^\top Ax$ is a scalar. Commented Jun 5, 2018 at 5:11

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 $$-x^TAx = x^TAx\iff 2x^TAx =0\iff x^TAx=0.$$
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 $0$.
• for $x_j$ and $x_i$, do you mean entries of the vector? Why does $x_jA_{ji}x_i=xAx$? Commented Jun 5, 2018 at 5:08
• @fcc I was using einstein notation. But I will add the $\sum$ signs. Commented Jun 5, 2018 at 5:10