Let $ad_{A}$ be the commutator operator on $nxn$ matrices $M_{n}(\mathbb{R})$, i.e. $ad_{A}(B)=AB-BA$.
Okay, for $s \in \mathbb{R}^{n}$ a real variable, let $\varphi(s)=\exp(sA)B\exp(-sA)\exp(A)$. I would like to show that this is equal to the function $\psi(s)=\exp(s ad_{A})(B)\exp(A)$, where $\exp$ is the usual exponential function for matrices.
I should be able to show this equality using the following theorem but am a little bit confused:
Theorem: For $A,C \in M_{n}(\mathbb{R})$, with $A$ nonzero and $a<0<b$, the differential equation $\alpha'(t)=\alpha(t)A$ has a unique solution $\alpha:(a,b)\rightarrow M_{n}(\mathbb{R})$ for which $\alpha(0)=0$.
I have shown that $\varphi(0)=\psi(0) = B \exp(A)$.
Now, presumably I need only show that $\varphi'(s)=\varphi(s)A = \psi(s)A$ for some matrix $A$ and conclude that $\varphi=\psi$ by the theorem? Assuming this is correct, I'm only stuck on the derivatives.
The result is valid in any matrix Lie group.
Here, $X\in M_n$ is fixed. When $s\in\mathbb{R}$, we consider the function $f(s):Y\in M_n\rightarrow e^{sX}Ye^{-sX}\in M_n$. We must prove that $f(s)(Y)=e^{s.ad_X}(Y)$, that is, $f(s)=e^{s.ad_X}$.
$\dfrac{d}{ds}(f(s)(Y))=Xe^{sX}Ye^{-sX}-e^{sX}Ye^{-sX}X=ad_X(e^{sX}Ye^{-sX})$, that is, $f'(s)=ad_X\circ f(s)$ (a composition of linear functions). Since $f(0)=id_{M_n}$, $f(s)=e^{s.ad_X}$ and we are done.