$\operatorname{e}^{X}\operatorname{e}^{Y}= \operatorname{e}^{X+Y}$ for linear transformations.

Question

For $X\in \mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{n})$, define $\operatorname{e}^{X}= \sum_{k=0}^{\infty} \frac{X^{k}}{k!}$. I proved that $\operatorname{e}^{X}$ is well-defined. A question ask to prove that for two elements $X,Y \in \mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{n})$ such that $XY=YX$, $\operatorname{e}^{X}\operatorname{e}^{Y}= \operatorname{e}^{X+Y}$ .

At this point, we don't know about continuity and differentiability. The proof that I found uses differentiability.

This question is from the book of Elon Lages, curso de análise vol. 2.

Do you know how to prove $e^x e^y =e^{x+y}$ for real numbers $x,y$? Because you can just use the same proof, since $XY=YX$ you can treat them as numbers. — Demophilus, Nov 05 '17 at 18:27

score 1 · Accepted Answer · answered Nov 05 '17 at 21:04

$$ e^X e^Y = \left( \sum_{k=0}^{\infty} \frac{X^{k}}{k!} \right) \left( \sum_{l=0}^{\infty} \frac{Y^{l}}{l!} \right) = \sum_{k=0}^{\infty} \sum_{l=0}^{\infty} \frac{X^{k}}{k!} \frac{Y^{l}}{l!} = \{ n = k+l \} \\ = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{X^{k}}{k!} \frac{Y^{n-k}}{(n-k)!} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} X^{k} Y^{n-k} \\ = \sum_{n=0}^{\infty} \frac{1}{n!} (X+Y)^n = e^{X+Y} $$

$\operatorname{e}^{X}\operatorname{e}^{Y}= \operatorname{e}^{X+Y}$ for linear transformations.

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