# $Ae^{A} = e^{A}A$ for $A$ and operator and solving a differential equation

My first question is: For witch type of operator $A$ does we have the below equation satisfied:

$$Ae^{A} = e^{A}A$$

I think my question can be asked in the form : when an operator comute with its exponential?

But then my main goal is to set more rigorous background in this problem: For $A,B$ non-commutative operators prove that

$$e^ABe^{-A} = B + [A,B] + \frac{[A,[A,B]]}{2!} + \frac{[A,[A,[A,B]]] }{3!} + \dots$$

The author of the problem does not demand more things and I'm bit trying to understand where these operators can act, they need to be linear? They can act on Hilbert and Banach spaces that have infinite dimensions? This is true for any $A,B$ maps between two different spaces?

The hint given is the same as the standard way of proving this by using differential equations: prove that for an function (in this case a function that takes real values and gives operators) $F(t)$ that has the differential equation form $F'(t) = UF(t)$ where $U$ is an operator that does not depend on $t \in \mathbb{R}$ then the only solution to this for the condition $F(0)= \mathbb{1}$ is the exponential operator.So we took $F(t)$ for a fixed operator $A$ acting on another operator $B$ such that

$$F(t)B := e^{tA}Be^{-tA}$$

It is simple to see that the above is true for $UX := AX-XA$ the linear commutator operator. Then we can get that $F(t) = e^{Ut}$ is the solution and by uniqueness we solve the problem. Is this an correct aproach to deal with any kind of operators and not just matrix (linear) operators?

• The first question would hold for any bounded operator on a Banach space (this includes Hilbert spaces). Essentially you just use the Taylor series definition. – Joel Aug 12 '17 at 1:30
• @joel thank you this is a great part of the question since my most interest is in these spaces – Rafael Wagner Aug 12 '17 at 1:50