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I'm working through Hale's book on ODE's

EDIT: my 4.1 question has been answered. 4.2 however still does not make sense to me.

He asks 4.1:

Prove that $Be^{At} = e^{At} B$ for all $t $ if and only if $BA=AB $

4.2:

Prove that $e^{At}e^{Bt} = e^{(A+B)t} $ for all $t $ if and only if $BA=AB $

This is all the question provides. We do not know anything about the determinants of the matrices.

My Question

1) the $t $ here is a scalar variable, correct? I need to ask this because...

2) although commutativity implies one direction in both cases, I can find examples for 4.2 for which the converse does not hold and

3) it seems like 4.1 can only be proven assuming that $B $ is an invertible matrix

Am I missing something? Is this problem perhaps not asking what I think?

Phillip Hamilton
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2 Answers2

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Note that $$ e^A = 1 + A + \frac {A^2} 2 + \frac{A^3} 6 + \cdots $$ This sum converges uniformly. If $AB = BA$ it should now be clear why $Be^A = e^AB$. The other inclusion follows by differentiation: $$ \frac d {dt} e^{tA} = A e^{tA}. $$ For $e^{A+B} = e^A e^B$ use the binomial theorem, which is valid for every commutative algebra.

  • The binomial theorem for a commutative algebra is not the problem. It's the other direction. Thanks for the 4.1 approach I see that avoids invertible properties. – Phillip Hamilton Feb 20 '16 at 17:35
  • http://math.stackexchange.com/questions/349180/if-ea-and-eb-commute-do-a-and-b-commute/349382#349382 see here. In general there are known examples of $e^{A}e^{B}=e^{(A+B)}$ not implying that A,B are commutative. – Phillip Hamilton Feb 20 '16 at 17:40
  • This is not a counterexample, since your equality holds for all $t \in \mathbb R$. So its a stronger assumption. Differentiation and evaluation at $t =0$ should work in your case. –  Feb 20 '16 at 17:44
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Yes, $t$ is a scalar.

You do not need $B$ to be invertible to prove 4.1. If you assume $Be^{At} = e^{At}B$ for all $t$, then using

$$e^{At} = \sum_{n=0}^{\infty} \frac{(At)^n}{n!}$$

and differentiating with respect to $t$, then setting $t = 0$ should yield the desired result.

GaussTheBauss
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