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.