Let $V$ be a finite-dimensional inner product space and let $x,y\in V$ be nonzero vectors. If there is a self-adjoint operator $A:V\rightarrow V$ such that $A(x)=y$ and $\langle A(v),v\rangle\geq0$ for all $v\in V$, then $\langle x,y\rangle>0$.
I think we can conclude the following inequality $$\langle x,y \rangle=\langle x,A(x) \rangle=\langle A^*(x),x \rangle=\langle A(x),x\rangle\geq 0$$
but I'm not able to show that strict inequality holds (or, in other words, that equality is not possible). Can someone give me a hint?