TITLE: Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?
QUESTION [15 upvotes]: (Disclaimer : I know very well that $SO(N)$ has a Lie algebra of dimension $N(N-1)/2$ etc. This absolutely not the point of my question.)
To make my problem more understandable, I start with the example of $SO(2)$. All $SO(2)$ matrices $M$ can be written as ($\theta\in [0,2\pi[$)
$$
M=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}.
$$
Using the basis of $2\times2$ real matrices
$\sigma_0=\begin{pmatrix}1 & 0\\0&1\end{pmatrix}$, $\sigma_1=\begin{pmatrix}0 & 1\\1&0\end{pmatrix}$,
$\sigma_2=\begin{pmatrix}0 & 1\\-1&0\end{pmatrix}$,
$\sigma_3=\begin{pmatrix}1 & 0\\0&-1\end{pmatrix}$,
one find that
$$M=\cos\theta\;\sigma_0+\sin\theta\;\sigma_2.$$
Clearly, $M$ does not have components along $\sigma_1$ and $\sigma_3$, so the dimension of the smallest linear subspace of $\mathrm{M}_2(\mathbb{R})$ that contains $SO(2)$ is $2$.
How to articulate the reasoning (for the cases $N>2$ in particular) is not completely clear. I guess that we can say that the component along $\sigma_0$ and $\sigma_2$ are independent because $\cos\theta$ and $\sin\theta$ are independent functions (in a functional analysis sense).
Assuming that made sense, we can try to increase $N$. For example, an $SO(3)$ matrix can be written as
$$
M=\left(\begin{matrix}
\cos\varphi\cos\psi - \cos\theta\sin\varphi\sin\psi & -\cos\varphi\sin\psi - \cos\theta\sin\varphi\cos\psi & \sin\varphi\sin\theta\\
\sin\varphi\cos\psi + \cos\theta\cos\varphi\sin\psi & -\sin\varphi\sin\psi + \cos\theta\cos\varphi\cos\psi & -\cos\varphi\sin\theta\\
\sin\psi\sin\theta                                  & \cos\psi\sin\theta                                   & \cos\theta
\end{matrix}\right)\,
$$
with $(\phi,\psi)\in [0,2\pi[^2$ and $\theta\in [0,\pi[$. Now, if I look at each matrix element one by one, they are all independent in a functional sense.$^*$ Does that mean that the ''dimension of the matrix space'' that $SO(3)$ matrices live in is $9$ ? Is there a way to generalize that to arbitrary $N$ ?
In the end, does any of what I wrote above make any sense ?

$^*$ It is slightly more subtle than that for the $\theta$ dependence, because in the end I am interested in doing integral over the Haar measure, which means that one should look at $x=\cos(\theta)\in[-1,1]$. But $x$ and $\sqrt{1-x^2}$ are orthogonal, so all should be fine.

REPLY [17 votes]: Let $S$ be the span of $SO(N)$ .
Then it's obvious that if $A \in S$ and $D_1, D_2 \in SO(N)$ then $D_1^{-1} A D_2 \in S$ .
Therefore it's enough to show that $B := diag(1,0,...0) \in S$ .
If $N$ is odd and $> 1$ then we can write $$B = \frac{1}{2} (\left[\begin{matrix}1&0\cr0&I_{N-1}\end{matrix}\right] + \left[\begin{matrix}1&0\cr0&-I_{N-1}\end{matrix}\right])$$ .
If $N$ is even and $> 2$ then we can write $$B = \frac{1}{4} (\left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&-I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&E\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&-E\end{matrix}\right])$$ ,
where we choose $E \in O(N-2)$ with $det E = -1$ .