Show that $\exp(C^{-1} AC ) = C^{-1} \exp(A C)$

Question

Show that $\exp(C^{-1} AC) = C^{-1} \exp(A C)$ for any matrices $A \in L_{n}(\mathbb{R})$ and $C \in GL_{n}(\mathbb{R})$.

The hint of the question is given below:

Consider the linear operator $\alpha$ given in some basis (e) by the matrix $A$, and find the matrix of $e^{\alpha}$ in the basis (e)$C$ in two ways.

But I just want any type of proof either by calculation or by the hint given above, because I will use this result in solving another problem, just to be convinced with it while using it.

Could anyone help me in doing so please?

Thank you!

Answer 1

Look at the power series defining $e^B$ for any $B \in L_n(\Bbb R)$:

$e^B = \displaystyle \sum_0^\infty \dfrac{B^n}{n!}; \tag 1$

now observe that for any $n \ge 0$,

$C^{-1}A^n C = (C^{-1}AC)^n; \tag 2$

then

$C^{-1}e^A C = C^{-1}\left ( \displaystyle \sum_0^\infty \dfrac{A^n}{n!} \right )C$ $= \displaystyle \sum_0^\infty \dfrac{C^{-1}A^nC}{n!} = \displaystyle \sum_0^\infty \dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. \tag 3$

That's about the shortest derivation of

$C^{-1}e^A C = e^{C^{-1}AC} \tag 4$

I know!

And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .