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Question: Let $T : V → V$ be a unitary linear transformation on a finite-dimensional inner product space V . Let $W ⊂ V$ be a $T$-invariant subspace. Prove that $T(W) = W$ and $T(W^⊥) = W^⊥$

Having a lot of trouble with this question and don't know where to start. Can someone lend me a hand?

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You are given that $TW\subset W $; but $T $ is injective, so it preserves dimension and you get $TW=W $.

If $v\in W^\perp $ and $w\in W $, $$\langle T^*v,w\rangle=\langle v,Tw\rangle=0, $$ since $TW\subset W $. It follows that $T^*W^\perp\subset W^\perp$ and now you can deduce $T^*W^\perp=W^\perp $ since $T^*$ is injective.

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  • $\begingroup$ 1. Is $T$ always assumed to be injective iff $T$ is unitary? 2. Preserving dimension implies that $W ⊂ T(W)$, and thus $T(W) = W$, correct? $\endgroup$
    – The Monkey
    Oct 5, 2016 at 8:05
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    $\begingroup$ Assumed? A unitary map is always injective. You should be able to determine that. $\endgroup$ Oct 5, 2016 at 8:06
  • $\begingroup$ No. "Preserving dimension" means exactly that, that $\dim TW=\dim W $. $\endgroup$ Oct 5, 2016 at 8:23
  • $\begingroup$ Ohhh I see thank you for the clarification $\endgroup$
    – The Monkey
    Oct 5, 2016 at 8:25

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