Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?

For $n>2$, the span of the matrices in $\mathrm{SO}(n)$ is the full space $M_n(\mathbb{R})$ of $n$-by-$n$ matrices with real entries.

One proof is using representation theory: If we let $S\subset M_n(\mathbb{R})$ denote the span of $\mathrm{SO}(n)$, then notice that this span is invariant under the action of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ defined by $$ (A,B)\cdot C = ACB^{-1}. $$ Thus, it suffices to show that this representation of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ on $M_n(\mathbb{R})$ is irreducible, since $S$ is clearly not the zero subspace.

When $n>2$, the center of $\mathrm{SO}(n)$ is discrete (it is either the identity (when $n$ is odd) or $\pm$ the identity (when $n$ is even), and the (irreducible) representation of $\mathrm{SO}(n)$ on $\mathbb{R^n}$ has commuting ring isomorphic to $\mathbb{R}$. Since the above representation of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ on $M_n(\mathbb{R})$ is clearly the tensor product of the irreducible $\mathbb{R}^n$-representations of the two $\mathrm{SO}(m)$-factors of $\mathrm{SO}(n)\times\mathrm{SO}(n)$, it is irreducible.

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 $det E = -1$ .

Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar decomposition, where $S\in Sym_n^+$ and $R\in O_n$. This decomposition is unique with $S\in SPD_n$ if $M$ is non-singular, but in general it exists and might be non-unique. If $R\in SO_n$, then ${\rm Tr}(SQ)=0$ for every $Q\in SO_n$ ; chosing $Q=I_n$, we have ${\rm tr}\,S=0$, which implies $S=0_n$. If on the contrary $R\in O_n^-$, we have ${\rm Tr}(SQ)=0$ for every $Q\in O_n^-$. Diagonalize $S$ in an orthogonal basis, the matrix of $D$ eigenvalues satisfies ${\rm Tr}(DQ)=0$ for every $Q\in O_n^-$. Chosing $Q$ the symmetry with respect to hyperplane $x_j=0$, we obtain ${\rm Tr}\,D=2d_j$ for every $j$. There follows $n\,{\rm Tr}\,D=2\,{\rm Tr}\,D$. Whence ${\rm Tr}\,D=0$, $d_j=0$ if $n\ge3$. This yields $M=0_n$, hence $E$ is the full space $M_n$.

When $n=2$, this gives only $d_1=d_2$ and we recover $M\in O_2^-$.