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

Question

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 $\text{Span}(SO_n(\mathbb{R}))$.

Answer 1

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 $M_n(\mathbb R)$.

Answer 2

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 $P_j Q_j P_j^T \in SO(n)$.