I have a rotation matrix represented as
$R(t)=e^{B(t)},\tag 1$
where $B(t)$ is a skew symmetric matrix (since any rotation matrix can be expressed as a matrix exponent of a skew symmetric matrix), and $t\in[0,1]$. We know the initial condition $R(0)$.
Specifications
- All matrices are of dimension $3 \times 3$. Rotation matrices are orthogonal, determinant 1
- There is a constraint condition imposed on these particular matrices, namely
$R'(t)=R(t)P(t), \tag 2 $
$P(t)=(1-t)P_0+t P_1,\tag 3$
where $P_0,P_1$ are given constant skew symmetric matrices and $R'(t)$ means derivative with respect to $t$. $P(t)$ is a linear interpolation of those two matrices, they do not commute with each other.
Question
Can we prove that $B(t)$ and $B'(t)$ commute at all $t$? I am not sure about how to prove it. Is it possible to prove that $B(t)=tA$?
NB: Remember matrix exponents are a bit different. Ref Link1-Matrix exponent, Ref Link2-Matrix exponent, Ref Link3-Rotation Matrix
If $P(t)=P$ is constant and $R(0)=I$ then $R(t)=e^{tP}$, more generally, if $P(t)$ commute with each other for different $t$ then $R(t)=e^{\int P(t)dt}$. In both cases different $R(t)$ commute with each other. But in general $R(t)$ is not of this form, it's what is called an ordered exponential.
The problem is that $e^{P(t)}$ do not commute with each other and you have to keep track of order in which you multiply them for the solution. They can't be collected into an integral in the exponent, and your $B(t)$ can only be computed after the fact, it's an artificial object. And if $R(0)\neq I$ then there is an additional issue of it commuting, or not, with $P(t)$ even if $P(t)$ commute with each other.