If $\sum_{i,j=1}^n A_{xi} \bar A_{yj}=0$ does $A_{xy}=0$?

Question

I have a system of equations that I'm stuck with. Suppose $A$ is an $n\times n$ complex matrix with $xy^\text{th}$ component $A_{xy}$, if $$\sum_{i=1}^n \sum_{j=1}^n A_{xi} \bar A_{yj}=0$$ for all $x,y \in \{1,...,n\}$ must it be true that $A$ is the zero matrix? If not what are the conditions on $A$? The bar denotes the complex conjugate.

Answer 1

This is actually a comment but there is not enough room there. Note that $$ 0=\sum_{i=1}^n \sum_{j=1}^n A_{xi}\overline{A}_{y,j}=\left(\sum_{j=1}^n A_{xi}\right)\left(\sum_{j=1}^n \overline{A}_{yj}\right) $$ Given any complex $n\times n$ matrix $B$ let $R_x(B)$ be the sum of all entries of $B$ in the $x$th row. Then basically your condition is $R_x(A)R_y(\overline{A})=0$ for all $xy$. Note that $R_y(\overline{A})=0$ of and only if $R_y(A)=0$. So your condition reduces to $R_x(A)R_y(A)=0$ for all $x,y$. In particular for $x=y$, you have $(R_x(A))^2=0$, meaning $R_x(A)=0$, or the sum of each row is zero in $A$.

So you are basically asking if in a complex matrix the sum of each row is zero, is that matrix zero. The answer to which is actually a strong NO! Example: $$ A=\begin{pmatrix} 1 & -1\\ 1 & -1 \end{pmatrix} $$