Showing $x^tAx=0$ implies $A=0$

Question

I want to show that " for a complex $n$ order matrix $A$ if $x^tAx=0$ for all column vectors $x$ in $\Bbb C^n$ then $A =0$".

My try: I know the fact that when we take base field $\Bbb R$ then result is not true because $x^tAx=0$ will imply $A$ is skew symmetric and that is not necessarily $0$ matrix. When base field is $\Bbb C$ then characteristic polynomial of $A$ will splits then we can apply schur's theorem that there exists an orthonormal basis such that $A$ is similar to upper triangular matrix.

I don't know how to proceed from here. Give me some hint and please tell if I am moving in right direction or not .

score 9 · Accepted Answer · answered May 31 '19 at 11:16

If you really mean $x ^t$ and not $x^*=\overline {x^t}$, we cannot conclude that $A=0$: $$\begin{pmatrix}z&w\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix} \begin{pmatrix}z\\w\end{pmatrix}=\begin{pmatrix}z&w\end{pmatrix}\begin{pmatrix}w\\-z\end{pmatrix}=zw-wz=0$$

Showing $x^tAx=0$ implies $A=0$

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