If $P$ is a projection, $A$ is a matrix, and $AP=PA$, then $Pe^{tA} = e^{tPA}-I+P$. Proof:

\begin{align} Pe^{tA} & = P\sum_{i=0}^\infty \frac{A^it^i}{i!} \\ & = \sum_{i=0}^\infty \frac{PA^it^i}{i!} \\ & = P+\sum_{i=1}^\infty \frac{P^iA^it^i}{i!} \\ & = P+\sum_{i=1}^\infty \frac{(PA)^it^i}{i!} \\ & = e^{tPA}-I+P \end{align}

This seems like a simple enough result that someone has discovered it before, and (I think) interesting enough that someone might know who. So I'm interested if anybody has encountered it before, and if so, if they know a reference for it.

EDIT: Corrected something basic in response to a comment.

  • $\begingroup$ You're right, although that does not alter the proof I believe. $\endgroup$ Commented Jun 20, 2020 at 21:14


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