[proofplan]
We first prove that conjugation commutes with every matrix power: $(AXA^{-1})^m=AX^mA^{-1}$ for all $m\ge 0$. We then substitute this identity into the [power series](/page/Power%20Series) definition of the matrix exponential. [Absolute convergence of the matrix exponential series](/theorems/8777) justifies passing fixed left and right multiplication through the limiting series.
[/proofplan]
[step:Prove that conjugation commutes with all powers of $X$]
Let $I_n\in M(n,\mathbb C)$ denote the identity matrix. For each integer $m\ge 0$, define $P_m\in M(n,\mathbb C)$ by
\begin{align*}
P_m=(AXA^{-1})^m.
\end{align*}
We prove that
\begin{align*}
P_m=AX^mA^{-1}
\end{align*}
for every integer $m\ge 0$.
For $m=0$, the convention for zero powers gives $P_0=I_n$, and since $A\in GL(n,\mathbb C)$,
\begin{align*}
AI_nA^{-1}=AA^{-1}=I_n.
\end{align*}
Thus $P_0=AI_nA^{-1}$.
Assume now that $m\ge 0$ and $P_m=AX^mA^{-1}$. Multiplying by $AXA^{-1}$ on the right and using associativity of matrix multiplication gives
\begin{align*}
P_{m+1}=P_mAXA^{-1}.
\end{align*}
Substituting the inductive hypothesis,
\begin{align*}
P_{m+1}=AX^mA^{-1}AXA^{-1}.
\end{align*}
Since $A^{-1}A=I_n$, this becomes
\begin{align*}
P_{m+1}=AX^mI_nXA^{-1}=AX^{m+1}A^{-1}.
\end{align*}
By induction, $(AXA^{-1})^m=AX^mA^{-1}$ for every integer $m\ge 0$.
[guided]
The point of this step is to show that conjugating $X$ before taking powers is the same as taking the power of $X$ first and then conjugating once. Let $I_n\in M(n,\mathbb C)$ denote the identity matrix. For each integer $m\ge 0$, set
\begin{align*}
P_m=(AXA^{-1})^m\in M(n,\mathbb C).
\end{align*}
We claim that
\begin{align*}
P_m=AX^mA^{-1}
\end{align*}
for every integer $m\ge 0$.
The case $m=0$ must be handled explicitly because no factor $X$ appears. By the zero-power convention for matrices, $P_0=I_n$. Since $A\in GL(n,\mathbb C)$, the inverse $A^{-1}$ exists and satisfies $AA^{-1}=I_n$. Hence
\begin{align*}
AI_nA^{-1}=AA^{-1}=I_n=P_0.
\end{align*}
So the formula holds when $m=0$.
Now suppose the formula holds for some integer $m\ge 0$, meaning
\begin{align*}
P_m=AX^mA^{-1}.
\end{align*}
By definition of powers,
\begin{align*}
P_{m+1}=P_mAXA^{-1}.
\end{align*}
Substituting the inductive hypothesis gives
\begin{align*}
P_{m+1}=AX^mA^{-1}AXA^{-1}.
\end{align*}
Associativity of matrix multiplication lets us group the middle factors as $A^{-1}A$, and since $A^{-1}A=I_n$, we get
\begin{align*}
P_{m+1}=AX^mI_nXA^{-1}=AX^{m+1}A^{-1}.
\end{align*}
Thus the induction step holds, and induction proves the identity
\begin{align*}
(AXA^{-1})^m=AX^mA^{-1}
\end{align*}
for all integers $m\ge 0$.
[/guided]
[/step]
[step:Substitute the power identity into the exponential series]
Let $S_N\in M(n,\mathbb C)$ be the $N$th partial sum of the exponential series for $AXA^{-1}$, defined for each integer $N\ge 0$ by
\begin{align*}
S_N=\sum_{m=0}^{N}\frac{(AXA^{-1})^m}{m!}.
\end{align*}
Using the power identity from the previous step term by term,
\begin{align*}
S_N=\sum_{m=0}^{N}\frac{AX^mA^{-1}}{m!}.
\end{align*}
Since matrix multiplication distributes over finite sums and scalar multiplication commutes with matrix multiplication,
\begin{align*}
S_N=A\left(\sum_{m=0}^{N}\frac{X^m}{m!}\right)A^{-1}.
\end{align*}
[/step]
[step:Pass to the limit in the finite partial-sum identity]
Let $T_N\in M(n,\mathbb C)$ be defined for each integer $N\ge 0$ by
\begin{align*}
T_N=\sum_{m=0}^{N}\frac{X^m}{m!}.
\end{align*}
Since $X\in M(n,\mathbb C)$ and $A,A^{-1}\in M(n,\mathbb C)$, closure of $M(n,\mathbb C)$ under matrix multiplication gives $AXA^{-1}\in M(n,\mathbb C)$. By [Absolute Convergence of the Matrix Exponential]([citetheorem:8777]) applied to the two matrices $X$ and $AXA^{-1}$, the corresponding matrix exponential series converge absolutely, hence converge in the finite-dimensional [normed vector space](/page/Normed%20Vector%20Space) $M(n,\mathbb C)$. Therefore
\begin{align*}
T_N\to \exp(X)
\end{align*}
and
\begin{align*}
S_N\to \exp(AXA^{-1}).
\end{align*}
The map
\begin{align*}
L_A:M(n,\mathbb C)&\to M(n,\mathbb C)
\end{align*}
defined by
\begin{align*}
L_A(Y)=AYA^{-1}
\end{align*}
is linear, hence continuous because $M(n,\mathbb C)$ is finite-dimensional. Since $S_N=L_A(T_N)$ for every $N\ge 0$, taking limits gives
\begin{align*}
\exp(AXA^{-1})=L_A(\exp X).
\end{align*}
By the definition of $L_A$, this is exactly
\begin{align*}
\exp(AXA^{-1})=A\exp(X)A^{-1}.
\end{align*}
This proves the claimed conjugation invariance of the matrix exponential.
[/step]
