If A and B are similiar then are their exponentials equal or are they similiar? I would say the same since exponential is just $$\sum_{q=0}^\infty \frac{(1)}{q!}A^q$$ and q is just a constant. But I'm not sure so I'm asking you dear people of the internetz.
Also what would be the solution for x'(t)=Ax(t)? Is it: exp(tA).x(0) or exp(tA).x(t)???
Thank you.
HINT:$A\sim B \to A=PBP^{-1}$SO $$A^2=PBP^{-1}PBP^{-1}=PB^2P^{-1}\\A^3=PBP^{-1}PBP^{-1}PBP^{-1}=PB^3P^{-1}\\\vdots\\A^n=PB^nP^{-1}\to P^{-1}B^nP$$ now multiply $\large\sum_{q=0}^\infty \frac{(1)}{q!}A^q $by $P^{-1}P$ so we have $$P^{-1}\large\exp (A)P=\\P^{-1}\large\sum_{q=0}^\infty \frac{(1)}{q!}A^qP=\\ \large\sum_{q=0}^\infty \frac{(1)}{q!}P^{-1}A^qP=\\ \large\sum_{q=0}^\infty \frac{(1)}{q!}B^q=\exp(B)$$now you can do for $\exp(tA)$