Let $M_n(\mathbb{R})$ be the algebra of $n\times n$ matrices with real coefficients, equipped with the operator norm, with respect to the $\ell^2$ norm $\displaystyle\|x\|_2=\left(\sum_{i=1}^n x_i^2\right)^{1/2}$, namely $\displaystyle\|A\|=\sup_{\|x\|_2=1}\|Ax\|_2$.
Let $\displaystyle\exp(A):=\sum_{n=0}^\infty \frac{A^n}{n!}$ be the exponential map on $M_n(\mathbb{R})$.
Let $A(n)$ be the subspace of skew-symmetric matrices $A=-A^T$.
Let $O(n)$ be the group of orthogonal matrices $A=(A^T)^{-1}$, and let $SO(n)$ be its subgroup made of determinant 1 matrices.
Now take $1\leq k\leq n-1$ and consider:
- $A(k,n)=\{[a_{ij}]\in A(n)\,:\, a_{ij}=0 \mbox{ for }i\leq k<j\mbox{ and }j\leq k<i\}$
- $X=\{[a_{ij}]\in A(n)\,:\, a_{ij}=0 \mbox{ for }i,j\leq k\mbox{ and }k<i,j\}$
In other terms $A(n)=A(k,n)\oplus X$ according to the block decomposition: $\begin{bmatrix}A_1&A_3\\A_2&A_4\end{bmatrix}=\begin{bmatrix}A_1&0_{k\times(n-k)}\\0_{(n-k)\times k}&A_4\end{bmatrix}+\begin{bmatrix}0_{k\times k}&A_3\\A_2&0_{(n-k)\times(n-k)}\end{bmatrix}$
Remark: It is clear that $\exp(X)\exp(A(k,n))\subset SO(n)$.
Claim: For every $T\in SO(n)$, there exist $B\in X$ and $A\in A(k,n)$ such that $\|B\|\leq \pi$ and $\exp(B)\exp(A)=T$.
Question: How would you establish this claim? If possible, I'd like to see different approaches (diagonalization, Lie algebra arguments, ...).