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Let $V$ be a real inner product space of odd dimension and $S∈L(V,V)$ an orthogonal transformation. Prove that there is a vector $v$ such that $S^2(v)=v$.

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Hint: Since $V$ has odd dimension, you know $S$ must have at least one real eigenvalue ... – Neal Dec 2 '12 at 6:25
You might also consider posting what work you have done so far and a bit of motivation for recent spree of linear algebra question. – Holdsworth88 Dec 2 '12 at 6:30
Neal: and then? – i_a_n Dec 2 '12 at 7:00
"Then"? Then you're done! What are the possible eigenvalues of an orthonormal transformation? Ho are the possible eigenvalues of a power of any transformation related to the eignevalues of the transformation? – DonAntonio Dec 2 '12 at 11:26
DonAntonio: I need more explanation here. What are the possible eigenvalues of an orthogonal transformation? – i_a_n Dec 2 '12 at 11:46

Hint: Orthogonal transformation in inner product spaces satisfies the following relation

$$ <u,v>=<Su,Sv> = <u,S^TS v> ,$$

and have the property $S=S^T$. Note that, $<u,v>=u^T \, v$.

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Why $<Su,Sv>=<u,S^TSv>$, and why $S=S^T$? – i_a_n Dec 2 '12 at 7:02
@i_a_n: Check the inner product operations. – Mhenni Benghorbal Dec 2 '12 at 7:05
Sorry I don't think I get it. Can you explain them more clearly? – i_a_n Dec 2 '12 at 7:58

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