Decomposition of $SO(n)$ into $SO(2)$ and inversions

Question

I remember vaguely that any $R\in SO(n)$ is orthogonally equivalent to the direct sum of rotations in $SO(2)$, e.g., there exists some $Q\in O(n)$ such that $$ QRQ^T= \begin{bmatrix} \cos{\theta}&-\sin{\theta}&0&\cdots&0\\ \sin{\theta}&\cos{\theta}&0&\cdots&0\\ 0&0&\cdots&\cdots&0\\ \vdots&\vdots&0&\ddots&\vdots\\ 0&0&0&0&1 \end{bmatrix} $$ However, I don't quite remember the proof anymore, so it would be awesome if someone could prove some sketches or hints.

Answer 1

I think I now roughly remember the idea. Basically, you know that if were to extend $R\in SO(n)$ acting on $\mathbb{R}^n$ to $\mathbb{C}^n$, we know that the complex $R$ has at least one (possibly complex) eigenvalue in $\mathbb{C}^n$. Taking real and imaginary points, we see that there exists a 1-dim or 2-dim subspace $W$ in $\mathbb{R}^n$ which is invariant under $R$. We can then further show that $W^\perp$ is also $R$-invariant and thus use induction to decompose $\mathbb{R}^n$ into invariant subspaces of 1-dim or 2-dim. Restricting $R$ to each invariant subspace and the statement follows.