Let $H$ be a Hilbert space, $A$ is unitary and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$?

Asked Dec 06 '14 at 23:42

Active Dec 06 '14 at 23:54

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Let $H$ be a Hilbert space, and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$?

What I have is if $x\in S^{\perp}$ then $x\perp A(A^*A^*Ax)$ then $(x,A^*Ax)=(Ax,Ax)=0$, so $x$ is in $\operatorname{Null}(A)$, is there a more obvious reason?

How about the other direction?

edited Dec 06 '14 at 23:54

asked Dec 06 '14 at 23:42

tepponia

In the title you write '$A$ is unitary'. Are you sure about this condition? Wouldn't it mean that $A$ is invertible (with $A^{-1}=A^*$)? But then $S=H$ and $S^\perp=0$. – Berci Dec 06 '14 at 23:58

Let $H$ be a Hilbert space, $A$ is unitary and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$?

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