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Let $H$ be a Hilbert space and $A(\neq \emptyset)\subset H$. If $T\in BL(H)$ is unitary, then $$T(A^\perp)=T(A)^\perp.$$

Let $y\in T(A^\perp)\implies \exists x\in A^\perp\ \text{s.t}\ T(x)=y. $ Since, $x\in A^\perp\implies \forall\ a\in A,\ \langle x,a\rangle=0$.

Since $T$ is unitary, implies $TT^*=I=T^*T$. \begin{align*} & T(x)=y\\ \implies& T^*T(x)=T^*(y)\\ \implies& T^*(y)=x \end{align*} Since, $\langle x,a\rangle=0\implies \langle T^*(y),a\rangle=0$.

Now I'm stuck, what to do. How to proceed further. Please help me.

Thanks.

Sachchidanand Prasad
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    For all $a \in A$, $$\langle T(x), T(a)\rangle = \langle x, T^T(a)\rangle = \langle x , a \rangle =0$$ since $x \in A^\perp$. So, $y = T(x) \in T(A)^\perp$. The converse can be proved similarly. If $y \in T(A)^\perp$, then for all $a \in A$, $$0 = \langle T(a),y \rangle = \langle a, T^(y) \rangle.$$ Hence $T^(y) \in A^\perp$ and $y = T(T^ (y)) \in T(A^\perp)$. – Sayantan Nov 22 '16 at 02:48

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