Can we claim that for an orthogonal matrix $A$ (satysfying $AA^T=I$),
its symmetric and skew-symetric parts are always commuting?
Symmetric part is calculated as $S=\frac{1}{2}(A+A^T)$ and skew symmetric part as $K=\frac{1}{2}(A-A^T)$.
It's easy to show the claim for dimension $2$.
Also for dimension $3$ we can see commutation for orthogonal matrices with determinant equal to 1.
For a rotation matrix $R$ it's evident from the Rodrigues formula,
when $R=\color{blue}{I+(1-\cos\theta) V^2 }+ \color{green}{\sin\theta V }$ for some skew-symmetric matrix $V $.
Here commutation follows because symmetric part and skew-symmetric part are expressed as polynomials of the same matrix $V $.
Is it possible to find similar explanation for higher dimensions of orthogonal matrices?
Best Answer
We have $AA^T=A^TA=I.$ Hence
$$(A+A^T)(A-A^T)=A^2-AA^T+A^TA-(A^T)^2=A^2-I+I+(A^T)^2=A^2-(A^T)^2.$$
A similar computation gives
$$(A-A^T)(A+A^T)=A^2-(A^T)^2.$$