$A\in M_n(C)$ such that $x^*Ax=0\; \forall x\in C^n,$ then $A=0$ 
$A\in M_n(C)$ such that $x^*Ax=0\; \forall x\in C^n,$ then $A=0$

I was able to show $A_{ii}=0 \; \forall 1\leq i \leq n$ by taking $x=e_i$
How can I proceed from here?
 A: Hint: Let $\langle Ax, y\rangle:=y^*Ax.$
Then $$0=\langle A(\alpha x+ y), \alpha x+ y\rangle=\alpha\langle A(x), y\rangle+\overline\alpha\langle A(y), x\rangle.$$
Consider the cases $\alpha=i$ and $\alpha=1$. Adding the two equations, we get $\langle A(x), y\rangle=0.$ Now take $y=Ax$.
A: To continue your idea you may now consider


*

*$e_k^{\star}Ae_l$ with $A=\left(a_{kl}\right)$
A quick calculation shows that
$$e_k^{\star}Ae_l = a_{kl}$$
Using this consider
$$(e_k + e_l)^{\star}A(e_k + e_l) = (e_k^{\star} + e_l^{\star})A(e_k + e_l)= \boxed{a_{kl} + a_{lk} = 0}$$
$$(e_k + ie_l)^{\star}A(e_k + e_l) = (e_k^{\star} -i e_l^{\star})A(e_k + ie_l)= ia_{kl} - ia_{lk} = \boxed{i(a_{kl} - a_{lk}) = 0}$$
From this follows immediately that $a_{kl} = a_{lk} = 0$.
A: A more general result:
Let $H$ be a complex Hilbert space and $A,B \in B(H)$, i.e. they are bounded linear operators. Then we have that
$$\forall x \in H: (Ax|x)=(Bx|x) \implies A=B$$
And the proof relies on the polarization formula for complex Hilbert spaces:
Let $H$ be a complex Hilbert space and $A \in B(H)$. Then $\forall x,y \in H$, we have that
$$(Ax|y)=\frac{1}{4} \sum_{n=0}^3 i^n (A(x+i^n y)|x+i^n y)$$
