I have been studying the linear system of the form:

$$D_tX = AX + \textbf{b}$$

Where $A$ is not necessarily constant

Suppose we aim to find an integrating factor $M$ such that:

$$M[D_tX - AX] = D_t(MX)$$

This gives:

$$MD_tX - MAX = (D_tM)X + M(D_tX)$$

By equating coefficients we get:

$$D_tM = -MA$$

Solving this gives:

$$M = e^{-\int A \: dt}$$


$$D_t(e^{-\int{A} \: dt}) = -Ae^{-\int{A} \: dt} = -AM$$

So can we conclude that these two matrices commute?


I have proven that

$$AM = MA$$

if and only if

$$A\left(\int{A} \: dt \right ) = \left (\int{A} \: dt \right ) A$$

edit 2

After looking further into the question, it appears that for non-constant matrices

$$D_te^{A(t)} \neq \left ( D_tA(t) \right ) e^{A(t)}$$

more can be found here

  • 1
    $\begingroup$ I would need to look more carefully, but my suspicion is that existence of an $M$ satisfying your first condition leads to $A$ and $M$ commuting, which may be reason for nonexistence of $M.$ $\endgroup$
    – Will Jagy
    Feb 27, 2016 at 1:31
  • $\begingroup$ @WillJagy It's easy to prove commutativity for the case where A is constant, but my approach doesn't work so well when the entries of A are functions of t. $\endgroup$
    – bthmas
    Feb 27, 2016 at 1:40
  • $\begingroup$ Right. I do not think it works at all; there is a fairly elaborate theory where $A$ is not constant. It is just not like the one dimensional case, where there really is such a thing as an integrating factor, because one by one matrices commute with each other. $\endgroup$
    – Will Jagy
    Feb 27, 2016 at 1:45
  • 1
    $\begingroup$ @WillJagy Yea I just thought of a proof that $A$ and $M$ commute if and only if $A$ and $\int A \: dt$ commute which is very restrictive. (My intuition tells me that only constant matrices have this property, or at leas they must all be polynomials of the same degree) $\endgroup$
    – bthmas
    Feb 27, 2016 at 1:53
  • 1
    $\begingroup$ The solution to a linear system $\dot x=Ax$ can be written as $e^{tA}$ if and only if $A(t_1)A(t_2)=A(t_2)A(t_1)$ for any $t_1,t_2$, which is actually true for a constant $A$ but very seldom true for an arbitrary time-dependent $A$. $\endgroup$
    – Artem
    Feb 27, 2016 at 3:24

1 Answer 1


Let $$M = \left(e^{-\int A^T\, dt}\right)^T$$

Then $$D_t M = \left(D_t e^{-\int A^T\, dt}\right)^T = \left(-A^Te^{-\int A^T\, dt}\right)^T = \left(-A^TM^T\right)^T = -MA$$

  • $\begingroup$ but doesn't $(e^{- \int A^T \: dt})^T = e^{- \int A \: dt}$? $\endgroup$
    – bthmas
    Feb 28, 2016 at 6:10
  • $\begingroup$ Then there is your proof that $AM = MA$. $\endgroup$ Feb 28, 2016 at 19:05
  • $\begingroup$ But in general $AM \neq MA$. In fact $AM = MA \iff A$ commutes with $ \int A \: dt $ $\endgroup$
    – bthmas
    Feb 29, 2016 at 3:38
  • $\begingroup$ You can't have it both ways. My calculation above shows that defining $M$ as I did gives $D_tM = - MA$. You say this is the same as your definition of $M$, for which you showed that $D_tM = -AM$. Therefore either $AM = MA$, or you are mistaken in claiming that my $M$ and your $M$ are the same. I haven't investigated closely enough myself to decide. But regardless, by defining $M$ as I have, I get the formula you are after: $D_tM = -MA$, which is the point of the whole exercise. $\endgroup$ Feb 29, 2016 at 4:53
  • $\begingroup$ For non-constant matrices $A(t)$, $D_te^{A(t)} \neq \left ( D_tA(t) \right ) e^{A(t)}$ you can look at the link I posted in the 2nd edit of my question for more information if you like. $\endgroup$
    – bthmas
    Mar 2, 2016 at 2:45

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