For a real $n \times n$ matrix $A$ if $x^tA^tx \geq 0 \;\forall \;x,$ then $Au=0 $ iff $A^tu=0.$

Question

Let $A$ be a real $n \times n$ matrix such that $x^tA^tx \geq 0$ for all $x \in \mathbb{R}^n$. Then how can I show that $Au=0 $ if and only if $A^tu=0$ ?

Since $A$ and $A^t$ have the same rank enough to show that kernel of one of them is contained in the kernel of another. I need some help.