# 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?

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

$$(A-A^T)(A+A^T)=A^2-(A^T)^2.$$

• Can we claim additionally that these two matrices are polynomials of some matrix $V$ and even more, are we capable to find this matrix? Commented Mar 16, 2021 at 9:58

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.

• I didn't expect that it is so easy to show the claim by direct calculation. We see that commutation is evident. But .. can we also find a matrix $V$ such that $S$ and $K$ are polynomials of this matrix as it's often happens in the case of commutation? Commented Mar 16, 2021 at 9:52
• @Widawensen: This is an interesting question which you should probably ask separately. The generalization of Rodrigues's formula (see emis.de/journals/BJGA/v18n2/B18-2-an.pdf) tells you that any orthogonal matrix with determinant one can be written as a polynomial in a skew-symmetric matrix $V$ and then the symmetric parts consists of the even powers while the skew-symmetric part consists of the odd powers and they clearly commute. However, finding $V$ is not trivial (it involves finding a logarithm of the original matrix). Commented Mar 16, 2021 at 12:42
• Very useful paper and inspiring. I was hoping in fact for some extension of Rodrigues formula for higher dimensions when I was writing "Is it possible to find similar explanation for higher dimensions of orthogonal matrices?" Thank you Levap.. Commented Mar 16, 2021 at 12:58

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.

• Excellent. Commutativity is visible here from polynomials. Only $4$ is not needed, it seems ?, in $4AA^T$ . We see additionally that we can identify these polynomials. Commented Mar 16, 2021 at 14:31