What algebra is generated by $\mathrm{O}(2)$? The unit complex numbers can be identified with the $2 \times 2$ special orthogonal matrices $\mathrm{SO}(2)$. The problem with $\mathrm{SO}(2)$, however, is that its not closed under $\mathbb{R}$-linear combinations. To rectify this, we can consider the $\mathbb{R}$-algebra generated by $\mathrm{SO}(2)$, namely $\mathbb{R}[\mathrm{SO}(2)]$, which turns out to be isomorphic to $\mathbb{C}$. It seems natural to play the same trick with $\mathrm{O}(2)$, and see what we get.

Question. What is $\mathbb{R}[\mathrm{O}(2)]$?

If its not too much to ask, I'd also like to know what $\mathbb{R}[\mathrm{SO}(3)]$ is. I'll ask another question if this isn't answered here.
 A: Well, as you said, any "complex number" $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ is a linear combinations of rotations. This implies that any matrix of the form $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} a & b \\ b & -a \end{pmatrix}$ is a linear combinations of products of $\mathrm{diag}(-1,1)$ and a rotation, so belongs to the vector space generated by $O(2)$ (which is the same thing as $\mathbb R[O(2)]$). With these two kinds of matrices, it's easy to get every $2 \times 2$-matrix, so $\mathbb R[O(2)] = M_2(\mathbb R)$.

EDIT: the question about $\mathbb R[SO(n)]$ has actually already be answered on MSE. It turns out that for $n \geq 3$, the answer is a bit disappointing (from your point of view): you get all of $M_n(\mathbb R)$.
A: You have a split short exact sequence
$$
0 \to \mathbb{R}[SO(2)] \to \mathbb{R}[O(2)] \to \mathbb{Z}/2 \to 0
$$
where the action of the generator of $\mathbb{Z}/2$ on $\mathbb{C}$ would be by complex conjugation. I'm not sure if there is a better description of this, or if this suffices to slake your curiosity.
(I'm not sure about $\mathbb{R}[SO(3)]$)
