Prove or disprove: If $x^T A x = 0 $ for all $x$, then $ A = 0 $.

If $x ^t Ax = 0$ for every $x \in \mathbb R^n $ then the diagonal entries of $A$ are all zero by taking $x$ to be the column vector with the $i$-th component as $1$ and $ 0$ elsewhere, $\forall i = 1,2,\ldots, n$.

then choose your $x$ as follows

the column vector with the $i$-th and the $j$-th components as $1$ and $0$ elsewhere. Then try to ahow $A^t=-A$

It is not true, $A=\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$ is a counterexample.

Take $x=e_i+e_j$, you can get $A_{ij}+A_{ji}=0$, which means $A^T+A=0$