Prove $e^AS = S e^B$ for $AS=SB$

Question

Let $A,B \in \mathbb{L}(\mathbb{R}^n,\mathbb{R}^n)$ be congruent, in the sense that there is an invertible matrix $S \in \mathbb{L}(\mathbb{R}^n,\mathbb{R}^n)$ such that $AS=SB$. Show that $e^AS = Se^B$, which implies that $e^A=e^B$ are congruent.

i) Show this by proving that for every $z_0 \in \mathbb{R}^n$ the solution $y(x)$ of $$\frac{dy}{dx} = Ay$$ with $y(0) = S z_0$equals $S z(x)$, where $z(x)$ is the solution to the differential equation $$\frac{dz}{dx} = Bz$$ with $z(0)=z_0$.

ii) Also show this in terms of the definition of a matrix exponential.

My work In terms of the definition of a matrix exponential, I've managed to prove ii) by saying $$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$ and rewriting $A = SAS^{-1}$. Then it follows quite clearly that $e^AS = Se^B$.

However, I'm not quite sure how to proceed on the first proof i). Any thoughts?

Answer 1

Consider the problem $\dot y=Ay,y(0)=Sz_0$. It has the solution $y(x)=e^{Ax}Sz_0$. Now make the change of variables $y(x)=Sz(x)$. You get $S\dot z=SBS^{-1}Sz,z(0)=z_0$, which implies that $z(x)=e^{Bx}z_0$. Hence you have $$ e^{Ax}Sz_0=Se^{Bx}z_0, $$ assuming that $z_0\neq 0$ and plugging $x=1$ you get the desired equality.

Answer 2

For invertible $S$ we have $$ e^{SAS^{-1}}=Se^AS^{-1}. $$ This is proved in all standard texts on the matrix exponential, e.g., here, Lemma $1$ on page $1$. It just uses the definition of the matrix exponential by its series. Now setting $B=SAS^{-1}$ gives the claim.