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Context. I am currently reading a Quantum Mechanics book in which it is stated that it is "obvious" (without proof) that

$$ e^{(α Ω)} e^{(β Ω)} = e^{((α + β) Ω)} $$

where $\alpha, \beta \in \mathbb{C}$ and $\Omega$ is a square matrix of complex scalars, and $e^{\Omega}$ is defined as

$$ e^\Omega = \sum_{n=\color{red}{0}}^\infty \frac{\Omega^n}{n!} $$

whenever $\lim_{n \to \infty} \Omega^n$ converges.

Problem: I only know how to show this is true when $\Omega$ happens to be a diagonal matrix (in which case the matrix multiplication of $\Omega^n$ devolves to multiplication of the scalars $d_i^n$ along the diagonal of $\Omega$).

How does one show this result holds in general, when $\Omega$ isn't necessarily diagonal?

metamorphy
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user1770201
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    if $\Omega$ is diagonalizable you can write $\Omega = QDQ^{-1}$ where $D$ is diagonal, and check the formula in the same way. For the general case you can write $\Omega$ as a limit of a convergent sequence of diagonal matrices. – hunter Feb 09 '21 at 02:47
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    More generally, one has $e^A e^B =e^{A+B}$ if $AB=BA$. See https://math.stackexchange.com/questions/2459967/operator-exponential-ea-eb-eab for a rigorous proof. – Semiclassical Feb 09 '21 at 02:48
  • if not diagonalizable, the Jordan form writes $\Omega = D + N$ where $DN=ND$ and $D$ is diagonal and $N$ is nilpotent. Since they commute, $$ e^\Omega = e^D e^N,$$ where $e^N$ comes from a finite sum. Sometimes we can find a Jordan Chevalley decomposition, a generalization. – Will Jagy Feb 09 '21 at 03:11
  • Whose Quantum Mechanics book are you reading? –  Feb 09 '21 at 03:18
  • Are you asking to show that the definition of $e^{\omega}$ is correct, as your title says, or that the addition of exponents works? – Acccumulation Feb 09 '21 at 03:36

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