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I'm reading Chapter 6.3: Systems of Differential Equations from Gilbert Strang's Introduction to Linear Algebra.

The chapter mentions that $e^{At}$ always has the inverse $e^{-At}$ (assuming $A$ is a matrix). Later, it also mentions that sometimes $e^Ae^B$, $e^Be^A$, and $e^{A+B}$ can be all different.

My rookie question is: why $e^{At}$ is always invertible with inverse $e^{-At}$? Can I multiply them and add the exponents so that $e^{At}e^{-At}=e^{0t}=I+0+\cdots=I$? If yes, why would $e^A e^B$, $e^B e^A$, and $e^{A+B}$ sometimes be different?

I started reading Strang's textbook from the first page, and chapter 6.3 is so far the most confusing one. I'd really appreciate some help. Thanks in advance!

Rodrigo de Azevedo
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Skorpion
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    Because the product of two matrices is not commutative in general and that is the key for computing $e^{A+B}$. As in your example, $tA$ and $-tA$ always commutes so there is no problem. – Bebop Jun 01 '20 at 12:26
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    If $AB = BA$, then $e^{A+B} = e^A e^B$, as discussed at https://math.stackexchange.com/questions/3441619/ea-cdot-eb-eab-if-ab-ba-for-matrices-a-and-b or https://math.stackexchange.com/questions/762995/show-that-etab-etaetb-for-all-t-in-mathbbr-if-and-only-if for example. If $AB \ne BA$, then $e^{A+B}$ doesn't necessarily equal $e^Ae^B$ (see e.g. https://math.stackexchange.com/questions/652120/find-square-matrices-a-b-such-that-expa-b-neq-expa-expb or https://math.stackexchange.com/questions/880082/searching-two-matrix-a-and-b-such-that-expab-expaexpb-but-ab-is-not-equa). – Minus One-Twelfth Jun 01 '20 at 12:31

2 Answers2

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The product of matrices is not commutative and

$$e^Ae^B=e^Be^A$$ has a priori no reason to hold.

But if the sums converge,

$$e^{At}e^{-At}=\sum_{i=0}^\infty\sum_{j=0}^\infty(-1)^j\frac{A^it^iA^j t^j}{i!j!} =\sum_{j=0}^\infty\sum_{i=0}^\infty(-1)^j\frac{A^jt^jA^i t^i}{j!i!} =e^{-At}e^{At}. $$

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For your first question, if $\lambda$ is an eigenvalue of $A$, then $e^{\lambda t}$ is an eigenvalue of $e^{tA}$, and this accounts for all of the eigenvalues of $e^{tA}$. These eigenvalues are all nonzero, and as the determinant of a matrix is equal to the product of its eigenvalues, $\det(e^{tA})\ne0$, therefore it is invertible.

For the second, the matrices $A$ and $B$ must commute for the three expressions to be equal. There are several links to other questions that discuss this in a comment to your question, but I’ll repeat them in this answer since comments are more ephemeral than answers: see here, here, here and here, to start with.

amd
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