# Does $A$ commute with $e^{\int A \: dt}$

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:

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

But

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

So can we conclude that these two matrices commute?

edit

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

• 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 $M.$ Feb 27, 2016 at 1:31
• @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. Feb 27, 2016 at 1:40
• 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. Feb 27, 2016 at 1:45
• @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) Feb 27, 2016 at 1:53
• 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 $A$. Feb 27, 2016 at 3:24

Let $$M = \left(e^{-\int A^T\, dt}\right)^T$$
Then $$D_t M = \left(D_t e^{-\int A^T\, dt}\right)^T = \left(-A^Te^{-\int A^T\, dt}\right)^T = \left(-A^TM^T\right)^T = -MA$$
• but doesn't $(e^{- \int A^T \: dt})^T = e^{- \int A \: dt}$? Feb 28, 2016 at 6:10
• Then there is your proof that $AM = MA$. Feb 28, 2016 at 19:05
• But in general $AM \neq MA$. In fact $AM = MA \iff A$ commutes with $\int A \: dt$ Feb 29, 2016 at 3:38
• 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. Feb 29, 2016 at 4:53
• 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. Mar 2, 2016 at 2:45