Prove $e^{T^{-1}AT} = T^{-1}e^AT$.

Question

Let $T$ be a $n \times n$ invertible matrix, A $n \times n$ matrix. Prove that $e^{T^{-1}AT} = T^{-1}e^AT$. Also, if we know $T^{-1}AT$ and $T$, how can we calculate the matrix exponential $e^{At}$.

For now, I don't know where to start so I would really appreciate some hints, thank you!

Answer 1

Hint: first prove by induction $(T^{-1}AT)^n=T^{-1}A^nT$, then use $e^M=\sum_{n\ge0}\tfrac{M^n}{n!}$.

Answer 2

For any $n \times n $ matrix $A$, $e^A=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots$. From this it can be easily shown that $e^{T^{-1}AT}=T^{-1}e^A T$.