= Matrix exponential respects similarity
{title2=$e^{PAP^{-1}}=Pe^AP^{-1}$}
For an invertible <matrix> $P$, induction gives $(PAP^{-1})^k=PA^kP^{-1}$, including $k=0$. The <power series> for the <matrix exponential> is absolutely convergent in every <matrix norm>, so multiplication by the fixed matrices $P,P^{-1}$ can be taken through the sum. This proves the displayed identity without any <diagonalizability> assumption. In particular, if $A=PDP^{-1}$ with $D$ diagonal, then $e^A=P\operatorname{diag}(e^{D_{11}},\ldots,e^{D_{nn}})P^{-1}$.
Back to article page
