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Let $V$ be a finite-dimensional real linear space, and let $K$ be a compact subgroup of $GL(V)$ (with the usual topology); then is there a basis of $V$ such that every $f\in K$ is an orthogonal matrix under this basis?

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Let $\mu$ the normalized Haar measure on $G$, and $\langle\cdot\mid\cdot\rangle$ the usual inner product on $\mathbb R^n$ (we assume that $V=\mathbb R^n$). Define an other inner product $\langle\cdot\mid \cdot\rangle_G$ by $\langle x\mid y\rangle_G:=\int_G \langle gx\mid gy\rangle\mu(dg).$ Since the Haar measure is invariant by translation, we have for all $g\in G$: $\langle gx\mid gy\rangle_G=\langle x\mid y\rangle$. Let M' such that \langle x\mid y\rangle_G=^tyM'x. We have ^tgM'g=M' for all $g\in G$, and since M' is positive definite, we can find an invertible matrix $M$ such that M'=^tMM. We have $^tg^tMMg=^tMM$ so $^t(MgM^{-1})\cdot (MgM^{-1})=I_n$, therefore $MGM^{-1}\subset O(n)$.

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    +1 A nice trick. Very much similar in spirit to a proof of Maschke's theorem, wherein you can use an orthogonal complement after you have made sure that the inner product is $G$-equivariant. In that case we can simply take the average, because the group is finite. I'm sure @joriki has seen that.2012-01-07