Matrix Commutativity - Integration

Question

Specifications and Data in question

1.We have a skew symmetric matrix $ A(t)_{3\times 3}= \begin{bmatrix}\,0&\!-a_3(t)&\,\,a_2(t)\\ \,\,a_3(t)&0&\!-a_1(t)\\-a_2(t)&\,\,a_1(t)&\,0\end{bmatrix} \tag 1$ and

$B(t)_{3 \times 3}= -\int_{0}^{t} A(s)\ ds \tag 2$

$a_1(t),a_2(t),a_3(t)$ are general non matrix functions. Just used here to show the form of $A(t)$

Question

1. Is it possible to prove $B(t)*B(t)^{'}=B(t)^{'}*B(t)$ (commutative)? if not why?

NB: * means multiplication and $B'(t) =\frac{\text{d}B(t) }{\text{d}t} $

Answer 1

It is sufficient to find any two non-commuting (skew-symmetric) matrix $P,Q$. Let $p_{i},q_{i}$ denote the entries of $P,Q$ respectively. If we then define $$ B(t) = \pmatrix{0&-p_{3}\;t^{q_3/p_3} & p_2\;t^{q_2/p_2}\\ p_{3}\;t^{q_3/p_3} & 0 & -p_{1}\;t^{q_1/p_1}\\ -p_2\;t^{q_2/p_2}&p_{1}\;t^{q_1/p_1}&0} $$ Then $B'(1)B(1) \neq B(1)B'(1)$. A similar trick using $\sin(t)$ or $e^t$ gives you such a $B(t)$ where the function is necessarily smooth at $0$.

Answer 2

In general the answer is negative as $B(t)$ and $B'(t)$ satisfy

$$B'(t)B(t)+B(t)B'(t)=\frac{d}{dt}\left(B^2(t)\right). $$

You can translate the commutativity condition

$$B'(t)B(t)=B(t)B'(t)~~(*)$$

into a system of equations for the elements of $B(t)$ and $B'(t)$. Explicitly, if

$$ B(t) = \pmatrix{ 0 &\alpha(t) & \beta(t) \\ -\alpha(t) & 0 & \gamma(t)\\ -\beta(t) & -\gamma(t) & 0} $$

then commutativity $(*)$ is equivalent to $\alpha(t)\gamma'(t)=\alpha'(t)\gamma(t),$ $\alpha(t)\beta'(t)=\alpha'(t)\beta(t),$ and $\beta(t)\gamma'(t)=\beta'(t)\gamma(t)$.

The triple

$$\alpha(t):=t,~~\beta(t):=-t,~~\gamma(t)=t^3$$

gives you a counterexample, for all $t$. (you can quickly derive the original skew symmetric matrix $A(t)$ for such a triple).