Let $H_i$ be a $\mathbb C$-Hilbert space, $U$ be an unitary linear operator from $H_1$ to $H_2$ and $M\subseteq H_1$. How can we show that $U(M^\perp)=U(M)^\perp$?
Clearly, if $x\in M^\perp$ and $y\in M$, then $\langle Ux,Uy\rangle_{H_2}=\langle x,y\rangle_{H_1}=0$. This proves $U(M^\perp)\subseteq U(M)^\perp$. But how do we obtain the other inclusion?
Best Answer
Unitary operators are surjective.Suppose $y \in U(M)^{\perp}$. There exists $z$ such that $y=Uz$. Since $\langle y , Ux \rangle =0$ for all $x \in M$ we get $\langle Uz , Ux \rangle =0$ or $\langle z , x \rangle =0$ for all $x \in M$ so. $z \in M^{\perp}$ and $y =Uz \in U(M^{\perp})$.