Let $A \in L(V)$ where $V$ is finite-dimensional inner-product space such that $\langle x,y \rangle = 0 $ implies $\langle Ax,Ay \rangle = 0 $. Show that there exist $U$ and $\lambda$ such that $A=\lambda U$ where $U$ is unitary operator on $V$ and $\lambda$ scalar.
I am given a Hint: If normed vectors $x$ and $y$ are orthogonal,$\langle x,y \rangle = 0 $ then $\langle x+y,x-y \rangle = 0 $.
