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I am reading the book Naive Lie Theory

It proves that any isometry of $R^n$ that fixed the origin O is the product of at most n reflections in hyperplanes through O.

The proof is elementary and by induction. However, I cannot understand the arguments.

'Now suppose that $f$ is an isometry that fixes O and the result is true for $n=k-1$. If $f$ is not the identity, suppose $v \in R^k$ is such that $f(v)=w \neq v$.

Then the reflection $r_u$ in the hyperplane orthogonal to $u=v-w$ maps the subspace $Ru$ of real multiples of $u$ onto itself and the map $r_uf$ is the identity on the subspace $Ru$.'

Can anyone explain in detail to me why the map $r_uf$ is the identity on the subspace $Ru$?

Why do we have $$r_uf(u)=u$$ Can anyone explain why the last equality holds?

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1 Answer 1

There seems to be a little problem. $r_u$ maps $u\mapsto -u$, hence maps $\mathbb Ru$ to itself. It also maps $w\mapsto v$ and $v\mapsto w$, hence $r_uf$ maps $v\stackrel f\mapsto w\stackrel{r_u}\mapsto v$, that is instead of $\mathbb Ru$ the subspace $\mathbb Rv$ is left pointwise invariant under $r_uf$.

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should be $w \rightarrow v$ and $v \rightarrow w$ –  noot Aug 11 '13 at 22:05
    
@noot Yep, corrected –  Hagen von Eitzen Aug 11 '13 at 22:07

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