Commutation of symmetric and skew-symmetric part of orthogonal matrix 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?
 A: 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
$$(A-A^T)(A+A^T)=A^2-(A^T)^2.$$
A: One can verify it using a direct computation:
$$ SK = \frac{1}{2} \left( A + A^T \right) \cdot \frac{1}{2} \left(A - A^T \right) = \frac{1}{4} \left( A^2 - AA^T + A^TA - \left( A^T \right)^2 \right) = \frac{1}{4} \left( A^2 - \left( A^T \right)^2 \right)$$
while
$$ KS = \frac{1}{2} \left( A - A^T \right) \cdot \frac{1}{2} \left( A + A^T \right) = \frac{1}{4} \left( A^2 + AA^T - A^TA - \left( A^T \right)^2 \right) = \frac{1}{4} \left( A^2 - \left( A^T \right)^2 \right)$$
where we used the fact that $AA^T = A^TA$ (as they both are equal to $I_n$). More generally, the symmetric and and skew-symmetric part of a matrix commute iff the matrix is normal and an orthogonal matrix is normal.
A: Presumably the characteristic of the underlying field is not $2$.
The orthogonality condition implies that $(S+K)(S-K)=4AA^T=4A^TA=(S-K)(S+K)$. Hence $S$ commutes with $K$.
Alternatively, note that the inverse of a matrix $A$ is a polynomial in $A$. This is a consequence of Cayley-Hamilton theorem. It follows that when $A$ is orthogonal, $A^T$ is a polynomial in $A$. In turn, $S$ and $K$ are polynomials in $A$ as well. Hence they commute.
