$\forall n \geq 3, \text{Span}(SO_n(\mathbb{R})) = M_n(\mathbb{R})$ I would like to prove that the following holds for all $n \geq 3$ : 

$$\forall n \geq 3, \text{Span}(SO_n(\mathbb{R})) = M_n(\mathbb{R})$$

So far I have noticed the following : 


*

*For $n =2$ the result doesn't hold since $\text{dim(Span}(SO_2(\mathbb{R}))) = 2 \ne 4$

*For $n =2$ we can express every matrices of $SO_2(\mathbb{R})$ which help in finding a basis and thus the dimension. Yet I am enable to find a basis of $SO_n$ for $n \geq 3$
I don't have much idea of the way to go to prove the result since I am enable to find a basis of  $\text{Span}(SO_n(\mathbb{R}))$. 
 A: For odd $n \ge 3$ there is a straightforward answer:
Let $Q_1 = I, Q_2 = -I + 2 e_1 e_1^T$, where $e_1 = (1,0,...)^T$, note that $Q_1,Q_2 \in SO(n)$.
Then $Q_1+Q_2 = 2 e_1 e_1^T$. Now let $P_k$ be any even permutation matrix such that $P_k e_1 = e_k$, then
$P_i(Q_1+Q_2) P_j^T = 2 e_i e_j^T$, and $P_iQ_1 P_j^T, P_iQ_2 P_j^T$ are in $SO(n)$.
For even $n\ge 4$, let
$Q_1 = \operatorname{diag} \{ 1, \underbrace{ -1, \cdots, -1}_{{n \over 2}\text{ of these}}, \underbrace{ 1, \cdots, 1}_{{n \over 2}-1\text{ of these}} \}$,
$Q_2 = \operatorname{diag} \{ 1, 1, -1, \cdots, -1, 1,...,1\}$, $\cdots $,
$Q_{n-1} = \operatorname{diag} \{ 1, \underbrace{ -1, \cdots, -1}_{{n \over 2}-1\text{ of these}}, \underbrace{ 1, \cdots, 1}_{{n \over 2}-1\text{ of these}}, -1 \}$.
Note that $Q_1+\cdots + Q_{n-1} +I= 2 e_1 e_1^T$.
If we choose $P_k$ as above, get
$P_j(Q_1+\cdots + Q_{n-1} +I)P_j^T = 2 e_i e_j^T$ and we have
$P_j Q_j P_j^T \in SO(n)$.
A: We will construct a basis of $M_n(\mathbb R)$ consisting of special orthogonal matrices below. Denote by $E_{ij}$ the matrix with a $1$ at the $(i,j)$-th position and zeros elsewhere. For $i=1,2,\ldots,n-1$, define $D_i=I-2E_{ii}-2E_{nn}$. Define also $D_n=I$. Then the $D_i$s are linearly independent and they span the subspace of all diagonal matrices.
Now, for every pair of indices $(i,j)$ with $i<j$, pick an index $r\ne i,j$ and define
\begin{aligned}
R_{ij}&=E_{ij}-E_{ji}+\underbrace{\sum_{k\ne i,j}E_{kk}}_{\text{diagonal}},\\
S_{ij}&=E_{ij}+E_{ji}\,\underbrace{-\,E_{rr}+\sum_{k\ne i,j,r}E_{kk}}_{\text{diagonal}},\\
\end{aligned}
In other words, if $\mathcal I=\{i,j,r\}$ and $\mathcal J=\{1,2,\ldots,n\}\setminus\mathcal I$, then
$$
R_{ij}([\mathcal I,\mathcal J],[\mathcal I,\mathcal J])
=\pmatrix{0&1&0\\ -1&0&0\\ 0&0&1\\ &&&I_{n-3}},
\ S_{ij}([\mathcal I,\mathcal J],[\mathcal I,\mathcal J])
=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&-1\\ &&&I_{n-3}},
$$
Since the $D_i$s span all diagonal matrices, both $E_{ij}-E_{ji}$ and $E_{ij}+E_{ji}$ lie inside the matrix space spanned by the $D_i$s, $R_{ij}$s and $S_{ij}$s. It follows that the $D_i$s, $R_{ij}$s and $S_{ij}$s form a basis of $M_n(\mathbb R)$.
