Why is $e^{\int_0^t A(s)} \mathrm{d} s$ a solution of $x' = Ax$ iff all the entries of $A(s)$ are constant?

Question

I have seen this result in a few places on the internet and I am trying to prove it, but have found no way to start. I am trying to use the matrix $B = e^{\int_0^t A(s)} \mathrm{d} s$ and expand it into vectors, but this proved useless since I can't find a way to describe the matrix $B$ in terms of the vectors corresponding to the columns of $A$ or the entries of $A$. This was because when expanding the exponential in terms of the infinite series, the terms $A^k$ make using the entries of $A$ too complicated. I am assuming there is a simpler way.

Answer 1

This is not true. In the one-dimensional case the DE is satisfied for any differentiable function $A(s)$.

Answer 2

This is true if all entries of $A(s)$ are constant. Then $\int_0^t A(s)\;ds = tA$ of course.

More generally, it is true if we have a commutation condition: $A(s)A(t)=A(t)A(s)$ for all $s,t$.

So it is not "iff" as in the title.

Answer 3

$e^{\int_0^t A(s)ds}$ is a solution if $e^{\int_0^t A(s)ds}$ commutes with A(s). This can be found on Wikipedia. It is due to the fact that exp(A+B)=exp(A)exp(B)=exp(B)exp(A) if AB = BA. In fact, in many cases, such as taking the logarithm derivative of a matrix, one can get $A^-1 A'$ if $A^-1$ commutes with $A'$

https://en.wikipedia.org/wiki/Matrix_differential_equation

Explicit proof of the derivative of a matrix logarithm