Question on equivalence of inner product How to show that if $A, B \in M_{n}(\mathbb{C})$, and $x,y\in \mathbb{C}^{n}$ then the following two conditions are equivalent, 
1.) $\langle Ax,y\rangle =\langle Bx,y\rangle$ 
2.) $\langle Ax,x \rangle= \langle Bx,x \rangle$ 
for all $x,y$?

Both (1) and (2) are equivalent to $A=B$. 
By taking differences, it is equivalent to show that the following are equivalent:
(1) $A=0$.
(2) $\left<Ax,y\right>=0$ for all $x,y$. 
(3) $\left<Ax,x\right>=0$ for all $x$.
Obviously 1 implies 2 implies 3. Also 2 implies 1 by considering $y=Ax$. Assume 3 holds. Then compute, for arbitrary $\lambda\in\mathbb{C}$:
$0=\left<A(\lambda x+y),\lambda x+y\right>=\lambda \left<Ax,y\right>+\bar{\lambda}\left<Ay,x\right>.$
Hence for $\lambda\neq 0$ we get 
$\left<Ax,y\right>=-\frac{\bar{\lambda}}{\lambda}\left<Ay,x\right>$.
For fixed $x,y$, we can vary $\frac{\bar{\lambda}}{\lambda}$, so that $\left<Ax,y\right>=0$, proving 2.
Note that we really used that the field is $\mathbb{C}$! If the field was $\mathbb{R}$, then $\frac{\bar{\lambda}}{\lambda}$ is constant equal to 1 so there is nothing to vary; in $\mathbb{C}$ one could e.g. take $\lambda=1$ and $\lambda=i$. Indeed in the real case it is not true: take $n=2$ and $A$ a rotation of 90 degrees, then $A$ is not zero, but $\left<Ax,x\right>=0$ for all $x$.
However, if in the real case we add the assumption that $A=A^*$ (symmetric), then it remains true, as can be easily seen form the above proof noting that $\left<Ax,y\right>=\left<x,A^*y\right>$. Or alternatively, if $A=A*$ then $(x,y)\mapsto B(x,y):=\left<Ax,y\right>$ is a symmetric bilinear form, and such forms are uniquely determined by their associated quadratic form $x\mapsto q(x):=B(x,x)$ since $2B(x,y)=q(x)+q(y)+q(x+y)$.
