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Lets say $T$ is unitary operator and $T(W)\subset W$, where $W$ is a linear subspace of finite dimensional space $V$.

Prove that $T(W^{\bot}) \subset W^{\bot}$, where $W^{\bot}$ is a subspace orthogonal to subspace $W$.

I was trying to do it like this. Let $w^{\bot} \in W^{\bot}$ we have to show that $\langle w, Tw^{\bot} \rangle = 0$ where $w \in W$.

But I don't where do I have to use the fact that $T$ is unitary.

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Since $T$ is unitary, $\ker T=\{0\}$. Hence, $T|_W:W\to W$ is a linear isomorphism.

This means that every $w\in W$ can be written $w=Tw'$ for some $w'\in W$. What does this mean for $\langle w,Tu\rangle$ where $w\in W$ and $u\in W^\perp$?

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  • $\begingroup$ How do we know that ker$T = {0}$ $\endgroup$
    – janusz
    Jun 2 '16 at 19:09
  • $\begingroup$ Because $T$ is unitary, $\|Tx\|=\|x\|$ for all $x\in V$. Does this help? $\endgroup$
    – Michael M
    Jun 2 '16 at 19:10
  • $\begingroup$ Of course! Now it's easy. Would you take a look? Since $T$ is isomorphism we have $\langle w, Tu \rangle = \langle Tw^', Tu \rangle = \langle w^, u \rangle = 0$ $\endgroup$
    – janusz
    Jun 2 '16 at 19:12
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We know if $W$ is subspace of inner product space $V$, then if $$T(W) \subset W$$ then, $$T^*(W^{\bot}) \subset W^{\bot} $$ So, for any $w\in W$ and any $w^{\bot} \in W^{\bot}$, we have, $$ \langle w,T^*(w^{\bot} )\rangle=0$$ $$\langle Tw,w^{\bot}\rangle=0$$ $$\langle TTw,Tw^{\bot}\rangle=0$$ And we know $T$ is unitary operator and $T(W) \subset W$ , so $T$ must be an isomorphism from $V$ to $V$, so, $$TTw=w'$$ for some $w' \in W$ So, $$T(W^{\bot}) \subset W^{\bot}$$ Hence, the proof!

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