I try to solve it by KKT conditions.

The Lagrangian is $L=x^TAy-\lambda x^Ty$.

Its KKT conditions are given by $$ \begin{align} Ay-\lambda y=&0\quad (1)\\ A^Tx-\lambda x=&0\quad (2)\\ \lambda\geq &0\quad (3)\\ x^Ty\geq &0 \quad (4)\\ \lambda x^Ty=&0\quad (5)\\ \end{align} $$ From 1 and 2 we see that $\lambda$ is an eigenvalue of $A$, $\lambda\geq 0$, and $y=cx$ where $c\in\mathbf{R}$. From 3, 4 and 5, we know if $x^Ty>0$, $\lambda=0$, $\min x^TAy=\min \lambda x^Ty=0$. Otherwise, if $x^Ty=0$, we can also deduce that $\min x^TAy=\min \lambda x^Ty=0$. So $\min x^TAy=0$. Is it correct? Thx.