A map $A$ such that $(Ax,Ay)=(x,y)$ is bijective

Question

Having a finite dimensional Hilbert space $H$ and a $\mathbb{C}$-linear map $A:H\to H$ such that $(Ax,Ay)=(x,y)$ for all $x,y\in H$, why is $A$ automatically bijective?

Answer 1

$A(x)=0$ implies that $(A(x),A(x))=(x,x)=\|x\|^2=0$, this implies that $x=0$. So the kernel of $A$ is zero and $A$ injective. Since $H$ is finite dimensional and $A$ injective, $A$ is bijective.