Compact subgroups of the general linear group

Question

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?

Answer 1

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 $MGM^{-1}\subset O(n)$.