Question

Does \(v^\top A v = 0\) for all vectors v imply A is skew symmetric?

Answer 1

Let $\mathbf v=\mathbf x+\mathbf y$. Then

$0=\left(\mathbf v^\top\mathbf{Av}\right)^\top$
$0=(\mathbf x+\mathbf y)^\top\mathbf A^\top(\mathbf x+\mathbf y)$
$0=\mathbf x^\top\mathbf A^\top\mathbf x+\mathbf x^\top\mathbf A^\top\mathbf y+\mathbf y^\top\mathbf A^\top\mathbf x+\mathbf y^\top\mathbf A^\top\mathbf y$
$0=\mathbf x^\top\mathbf A^\top\mathbf y+\mathbf y^\top\mathbf A^\top\mathbf x$
$\mathbf x^\top\mathbf A^\top\mathbf y=-\mathbf y^\top\mathbf A^\top\mathbf x$
$\mathbf x^\top\mathbf A^\top\mathbf y=-\left(\mathbf x^\top\mathbf{Ay}\right)^\top$

which requires $\mathbf A=-\mathbf A^\top$, so the answer is yes.