Let $A, B$ be $n\times n$ matrices over $\mathbb{C}$. Assume that $$[A,[A,B]]=[B,[A,B]] = 0$$
Show $$e^{\epsilon A}Be^{-\epsilon A} = B + \epsilon[A,B]$$
Then, I have to show the equation below. Here $[A, B]$ is equal to $AB-BA$ and $\epsilon$ is a real parameter. I can't find a way to show the equation below... Could anyone explain to me in detail?
On
$\textbf{Proof}$. Since $$ e^{\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(\varepsilon A)^k}{k!} = 1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + \frac{\varepsilon^3 A^3}{3!} + \ldots, $$ we have $$ e^{-\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(-\varepsilon A)^k}{k!} = 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} - \frac{\varepsilon^3 A^3}{3!} + \ldots. $$ So \begin{align*} e^{\varepsilon A} B e^{-\varepsilon A} &= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + \frac{\varepsilon^3 A^3}{3!} + \ldots\right)B \left( 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} - \frac{\varepsilon^3 A^3}{3!} + \ldots\right) \\ &= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + \frac{\varepsilon^3 A^3}{3!} + \ldots\right) \left( B - \varepsilon B A + \frac{\varepsilon^2 B A^2}{2!} - \frac{\varepsilon^3 B A^3}{3!} + \ldots \right) \\ &= B + \varepsilon (AB-BA) + 2! \varepsilon^2 (A^2B-2ABA + BA^2) +3! \varepsilon^3[A, [A,[A,B]] ] + \ldots \\ &= B + \varepsilon [A,B] + 2! \varepsilon^2 [A,[A,B]] +3! \varepsilon^3[A, [A,[A,B]] ] + \ldots \\ &= B + \varepsilon [A,B] \end{align*} since $[A,[A,B]]= A(AB-BA)-(AB-BA)A=A^2B-2ABA + BA^2=0$. $\hspace{2cm}\square$
Let $f(t)=\exp(t A)\, B \exp(-t A)$, an analytic matrix valued function of $t$ in a neighborhood of $t=0$. Clearly $f(0) = B.$ Verify $f'(t) = [A,f(t)]$, $f''(t)=[A,f'(t)]$, and so on: $f^{(k)}(t) = [A,f^{(k-1)}(t)]$.
Now evaluate all these derivatives at $t=0$, and assemble them into a Taylor series:$$f(t) = \sum_{n\ge0} \frac{t^n}{n!} f^{(k)}(0).$$
By the foregoing, $f(0) = B$ and $f'(0) = [A,B]$. Now $f''(0)=[A,[A,B]]$, which vanishes by hypothesis. By induction, $f^{(n)}(0) = 0$ for all $n>2$.