If $\sum_{i,j=1}^n p_{ij} A_{xi} \bar {A_{yj}}= \sum_{i,j=1}^n p_{ij} \bar{A_{xi}} A_{yj}$ for all $x$ and $y$, are each $A_{xy}$ real? Let $x,y \in \{ 1,2,...,n\}$ and $A_{xy} \in \mathbb C$ for each $x,y$. I have a matrix equation reduced to the equations: $$\sum_{i=1}^n \sum_{j=1}^n p_{ij} A_{xi} \bar {A_{yj}}= \sum_{i=1}^n \sum_{j=1}^n p_{ij} \bar{A_{xi}} A_{yj}\hspace{1cm} \forall x,y, p_{ij}$$
where $p_{ij}$ are real numbers. Is this enough to imply that $A_{xy} \in \mathbb R$ for all $x,y$? If not what are the conditions imposed on $A_{xy}.$
 A: No, this is not enough to imply that $A_{xy} \in \mathbb{R}$ for all $x, y$, a simple counterexample is the case when all $A_{xy} = i$ (or any complex number for that matter); then both sides are trivially equal.
However, we can say something interesting about the $A_{xy}$ values. Rewrite your equation as
$$\sum_{i=1}\sum_{j=1}p_{ij}(A_{xi}\overline{A_{yj}} - \overline{A_{xi}}A_{yj}) = 0$$
Choose any values $r$ and $s$ in $\{1, 2, \dots, n\}$, and choose $p_{ij}$ so that $p_{rs} = 1$ and $p_{ij} = 0$ for all other $(i, j) \neq (r,s)$. Then we find that
$$A_{xr}\overline{A_{ys}} = \overline{A_{xr}}A_{ys}$$
If both $A_{xr}$ and $A_{ys}$ are nonzero, then we can write this as:
$$\frac{A_{xr}}{\overline{A_{xr}}} = \frac{A_{ys}}{\overline{A_{ys}}}$$
But if $z = r\cdot e^{i\theta}$ (for $r \in \mathbb{R}$ and $\theta \in [0, 2\pi]$), then  $z/\overline{z} = e^{2i\theta}$. In particular, this implies that $A_{xr}/A_{ys}$ is real. It follows that the ratio of any two $A_{xy}$ values is real, so there exists some $\theta \in [0, 2\pi]$ such that you can write $A_{xy} = e^{i\theta}B_{xy}$, where all the $B_{xy}$ are real.  
