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Let $V=\mathbb{R}^{n}$ and $T\,:V\to V$ be defined by $Tv=Av$ where $A\in M_{n}(\mathbb{R})$ is an orthogonal matrix.

My lecture wrote that if $W\subset V$ is a subspace of $V$ then if $W$ is $A$ invariant then $W^{\perp}$ is also $A$ invariant.

What I do know is that if $W$ is $A$ invariant then $W^{\perp}$ is also $A^*=A^{t}$ invariant, but I could not deduce from this that it is also $A$ invariant.

Is this 'fact' true ? I couldn't prove it (I tried writing a proof similar to the case I know, using inner products and failed), help is appreciated!

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Remember that $\langle Au, Av \rangle = \langle u,v \rangle$ whenever $u,v$ are column vectors in $\mathb{R}^n}$ and $\langle,\rangle$ is the usual inner (dot) product. – Geoff Robinson Jul 14 '12 at 12:05
up vote 3 down vote accepted

If $$\forall\,w\in W\,\,,\,Aw=w'\in W\Longrightarrow A^{-1}w'=A^{-1}Aw=w\in W$$

The above is enough since we know $\,A\,$ is bijective (why?), so for any $\,w'\in W\,$ there always exists $\,w\in W\,\,\,s.t.\,\,\,Aw=w'\,$

Thus, $\,W\,\,is\,\,A-\,$ invariant iff it is $\,A^{-1}-\,$ invariant, and now use Marc's answer.

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You need to use that $A$ is orthogonal, i.e., $A^t=A^{-1}$. Can you show that a subspace is $A$-invariant if and only if it is $A^{-1}$-invariant?

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I tried to show this as well and could not do it, I tried something like $A^{-1}w=A^{-1}(A^{-1}Aw)=A^{-1}A^{-1}w'$ and it doesn't help me... – Belgi Jul 14 '12 at 12:01
Try writing $V=A\cdot V$ and applying $A^{-1}$ to both sides... – Marc van Leeuwen Jul 14 '12 at 13:02

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