Does this property of the anti-symmetric matrices hold in general?

Question

Consider an antisymmetric matrix $A=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$ and an arbitrary vector $v= \begin{pmatrix} a \\ b \end{pmatrix}$. It follows that $(A v)^T v = -ab+ab =0$.

My question: Can this be proved in general that $(Av)^Tv=0$, for arbitrary dimensions (and form) of $A$ (and hence $v$)?

Answer 1

Yes, since $A^T=-A$ for all $v\in V$ $\langle Av,v \rangle=0$

Proof:

Note that $\langle u,v\rangle=u^Tv$

$A^T=-A \Rightarrow \langle v,AV\rangle= v^TAv=-v^TA^Tv$

$\langle -Av,v\rangle=-v^TA^Tv \Rightarrow\langle -Av,v \rangle=\langle Av,v \rangle$

$\langle -Av,v \rangle=\langle Av,v \rangle \Rightarrow 2\langle Av,v\rangle =0$

$2\langle Av,v\rangle =0 \Rightarrow \langle Av,v\rangle=0 $

$\langle Av,v\rangle=0 \Rightarrow (Av)^Tv=0$

Answer 2

because $A^t = -A$, $\langle Av,v\rangle=\langle v,A^tv\rangle=\langle v,-Av\rangle=\langle -Av,v\rangle=-\langle Av,v\rangle$ and thus $\langle Av,v\rangle=0$