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I know how to prove this property with the matrix function $\pmb\Phi (t)=\exp((\textbf A+\textbf B)t)\exp(-\textbf Bt)\exp(-\textbf At)$ and its derivative $\pmb\Phi'(t)$, by proving that $\pmb\Phi'(t)=\textbf 0$ for all $t$.

It is also stated (Exercise 7.3.9, p. 539 of "Matrix Analysis and Applied Linear Algebra") that it could be proven with the function $\textbf F(t)=\exp((\textbf A+\textbf B)t)-\exp(\textbf At)\exp(\textbf Bt)$, proving that its derivative $\textbf F'(t)=\textbf 0$ so $\textbf F(t)$ is constant and since for $t=0$ it is zero so it is zero for all $t$.

I am trying to prove that $\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$ is zero for all $t$ but without success. Is my derivative function correct and if it is so how to prove it is identically zero?

Rodrigo de Azevedo
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Pekov
  • 1,045

3 Answers3

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Let's take as a starting point what you have calculated $\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$

Then by substituting $\textbf F(t)$ to the previous expression we get \begin{align*} \textbf F'(t)&=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\ &=(\textbf A+\textbf B)(\textbf F(t)+\exp(\textbf At)\exp(\textbf Bt))-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\ &=(\textbf A+\textbf B)\textbf F(t) \end{align*} Now take any vector $v$ and multiply the previous equality $$(\textbf F(t)v)'= (\textbf A+\textbf B)(\textbf F(t)v)$$ Which implies that $$\textbf F(t)v =e^{(\textbf A+\textbf B)t}\textbf F(0)v=0,\, \forall t \in \mathbb{R}$$ Therefore $$F(t)\equiv0$$

clark
  • 15,327
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\begin{align*} &(A+B)\exp((A+B)t) - A\exp(At)\exp(Bt)-B\exp(At)\exp(Bt)\\&=(A+B)\exp((A+B)t)-(A+B)\exp(At)\exp(Bt)\\ &=(A+B)\sum_{n=0}^{\infty}\frac{(A+B)^n t^n}{n!} - (A+B)\sum_{n=0}^\infty \frac{A^n t^n}{n!}\sum_{n=0}^\infty\frac{B^n t^n}{n!}\\ \end{align*}

But since $AB=BA$ then by standard multiplication of power series

$$\sum_{n=0}^\infty\frac{A^n t^n}{n!}\sum_{n=0}^\infty \frac{B^n t^n}{n!} = \sum_{n=0}^\infty\sum_{i=0}^n A^i B^{n-i} \frac{t^n}{i!(n-i)!} = \sum_{n=0}^\infty (A+B)^n\frac{t^n}{n!},$$

which you can prove by induction. Summarizing, the above is equivalent to

$$(A+B)\sum_{n=0}^\infty \frac{(A+B)^n t^n}{n!} - (A+B)\sum_{n=0}^\infty \frac{(A+B)^n t^n}{n!} = 0,$$

as desired.

William
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$f'(t)=(A+B)e(A+B)te(-At)e(-Bt)-Ae(A+B)te(-At)e(-Bt)-Be(A+B)te(-At)e(-Bt)$

Dozzy Cooper
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