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?
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$A(x)=0$ 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.
Tsemo Aristide
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Thanks! Do you have an counterexample for the infinite case? – Jan 30 '18 at 09:37
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2yes, consider the infinite dimension space ($l^2$) $(e_n,n\geq 1)$ and $A(e_n)=e_{n+1}$. $A$ is injective, but not surjective since $e_1$ is not in the image of $A$. – Tsemo Aristide Jan 30 '18 at 09:38
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Why isn't that map bijective? Wouldn't $B(e_n)=e_{n-1}$ be an inverse? – Jan 30 '18 at 09:40
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1@VML But what is $B(e_1) = e_{0}$ in that case? – Nigel Overmars Jan 30 '18 at 09:43