prove that the maximal torus of $SO(3)$ is the maximal torus of $GL_3(\mathbb{R})$. I want to prove that the maximal torus of $SO(3)$ is the maximal torus of $GL_3(\mathbb{R})$. I want to use the theorem that every maximal torus of G equals $gTg^{-1}$ for some $g \in G$. But I am not sure how the argument can be made. I was trying to think of it in terms of how reflections can map on to the lines. And how for every rotation there can be a line associated with it, since every element of $SO(3)$ can be achieved by starting with the identity position and rotating by some angle.
I would really appreciate any guidance. 
Thanks
 A: So the two main ideas are that any compact, connected subgroup of $GL_{n}(\mathbb{R})$ is actually contained in a subgroup isomorphic to $SO(3)$ and that maximal tori in compact groups are conjugate.
Let $G$ be a compact subgroup of $GL_{3}(\mathbb{R})$. Take an inner product on $\mathbb{R}^{3}$. Then, we can make it $G$ invariant by averaging over the Haar measure of $G$. Thus, if we take $H$ to be the copy of $O(3)$ defined by this inner product, then $G$ is a subgroup of $H$. In addition, if $G$ is connected, then it is a subgroup of $H^{0}$ which is isomorphic to $SO(3)$.
Now, let $T$ be a maximal torus of $GL_{3}$ and, since it is compact and connected, pick some $H \cong SO(3)$ that contains $T$. Let $B$ be the standard inner product on $\mathbb{R}^{3}$ and let $B'$ be the inner product determining $H$ (so basically choose two bases $\{e_{1}, \ldots, e_{n}\}$ and $\{v_{1}, \ldots, v_{n}\}$ which are oriented, orthonormal bases for the respective inner product.) Let $M$ be the matrix (in the standard basis) that takes $e_{i}$ to $v_{i}$, and hence has $v_{i}$ for the $i$th column. The claim is then,
$$MSO(3)M^{-1} = H$$
where $SO(3)$ here is the standard $SO(3)$. For the proof, if $v, w \in \mathbb{R}^{3}$,and $N \in SO(3)$, we have
$$B'(Mv, Mw) = B(v, w)$$
and hence,
$$B'(MNM^{-1} v, MNM^{-1}w) = B(NM^{-1}v, NM^{-1}w) = B(M^{-1}v, M^{-1}w) = B'(v, w).$$
So, we see that $M^{-1}TM \subseteq SO(3)$. In particular, this means that $M^{-1}TM$ is a compact, connected, abelian subgroup of $SO(3)$, and must be maximal with respect to such properties, because if there were a bigger such subgroup of $SO(3)$, then this would also be a bigger such subgroup of $GL_{3}(\mathbb{R})$, which would contradict the fact that $T$ is a maximal torus of $GL_{3}$. Hence, $M^{-1}TM$ is a maximal torus of $SO(3)$. 
Now, if $S$ is an arbitrary maximal torus of $SO(3)$, it will be conjugate to $M^{-1}TM$ (because all maximal tori of compact groups are conjugate), and hence will be conjugate in $GL_{3}$ to $T$. Hence, $S$ will also be a maximal torus of $GL_{3}(\mathbb{R}$).
