I have difficulties in demonstrating the following theorem
For all $A\in \mathbb{R}^{n\times n}$, $P\in GL\left ( \mathbb{R}^{n} \right )$ . Then $e^{PAP^{-1}}=Pe^{A}P^{-1}$
I saw that the proof of this theorem begins
For $k\geq 0$, $P\left (\sum_{j=0}^{m} \frac{1}{j!}A^{j} \right )P^{-1}=\sum_{j=0}^{m}\frac{1}{j!}\left ( PA^{j}P^{-1} \right )=\sum_{j=0}^{m}\frac{1}{j!}\left ( PAP^{-1} \right )^{j}$
then taking the limit when $m\rightarrow \infty $ to both members of the equality, the theorem is tested
If you consider that $\left ( AB \right )^{j}=A^{j}B^{j}$,$\forall j\in \mathbb{N}$. Then $\left ( PAP^{-1} \right )^{j}=P^{j}A^{j}P^{-j}$,which implies that $PA^{j}P^{-1}\neq \left ( PAP^{-1} \right )^{j}$. Then $e^{PAP^{-1}}\neq Pe^{A}P^{-1}$.
I think it is not the demonstration process or missing data to the theorem. I would like you to clarify me please. Thank you for your help
In general, matrices don't commute.
On the other hand,\begin{align}(PAP^{-1})^2&=PA\overbrace{P^{-1}P}^{=\operatorname{Id}}AP^{-1}=PAAP^{-1}=PA^2P^{-1},\\(PAP^{-1})^3&=PA\overbrace{P^{-1}P}^{=\operatorname{Id}}A\overbrace{P^{-1}P}^{=\operatorname{Id}}AP^{-1}=PAAAP^{-1}=PA^3P^{-1},\end{align}and so on.