Show that A is skew-symmetric if and only if $x^tAx = 0$

Question

I've tried by starting with setting $x^tAx = 0 = x^t(-A^t)x$ and checking it termwise, but I don't think this will show me anything.

Could you explain how to approach this problem please?

Answer 1

Hint: To prove the skew-symmetry of $\ A\ $ you need the equation to be true for all $\ x\ $. So what can you conclude about $\ x^t\left(A+A^t\right)y\ $ from the equation $\ 0 = \left(x+y\right)^tA\left(x+y\right)\ = x^tAx +x^tAy + y^tAx + y^tAy\ $?

Answer 2

Let p denotes A is skew-symmetric.
$\;\quad$q denotes $x^{T}Ax = 0$ such that $\forall x \in \mathbb{R}^n$.

1. Proof for $p \rightarrow q$
   Given A is skew-symmetric, that is $A^{T} = -A$.
   $x^{T}Ax = -x^{T}A^{T}x = [(-x^{T}A^{T}x)^{T}]^{T}=[-x^{T}Ax]^{T} = -x^{T}Ax$
   So, $x^{T}Ax = 0$ for all x in $\mathbb{R}^n$
2. Proof for $q \rightarrow p$ Prove by contradiction.
   Suppose A is not skew-symmetric and $x^{T}Ax = 0$ such that $\forall x \in \mathbb{R}^n$
   $$(x^{T}Ax)^{T} = x^{T}A^{T}x \neq -x^{T}Ax$$
   Due to $x^{T}Ax \in \mathbb{R}$
   $x^{T}Ax + (x^{T}Ax)^{T} = 0$
   $\Rightarrow x^{T}Ax \neq 0$, this statement contradicts with above assumptions.
   So, $x^{T}Ax = 0 \; ,\forall x \in \mathbb{R}^n \: \text{then A is skew-symmetric}$.
   $\therefore \: \text{A is skew-symmetric if and only if}\; x^TAx = 0 \enspace \forall x \in \mathbb{R}^n$.