TITLE: $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$
QUESTION [9 upvotes]: I am working on the following problem. Let $e^{Mt} = \sum\limits_{k=0}^{\infty} \frac{M^k t^k}{k!}$ where $M$ is an $n\times n$ matrix. Now prove that
$$e^{(M+N)}   = e^{M}e^N$$
given that $MN=NM$, ie $M$ and $N$ commute.
Now the left hand side of the desired equality is
$$e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots $$ 
On the right hand side of the equation we have
$$e^Me^N = \left(I + M + \frac{M^2}{2!}   + \frac{M^3}{3!}\ldots\right)      \left(I + N + \frac{N^2}{2!}   +   \frac{N^3}{3!} \ldots\right)   $$
Now basically this is as far as I got... I am unsure on how to work out the product of the two infinite sums. Possibly I need to expand the powers on the left hand side expression but I am unsure how to do this in an infinite sum... If anyone could give me an answer or a hint that can help me forward I would greatly appreciate it. Thanks

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
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\begin{align}
\color{#0000ff}{\large\expo{M}\expo{N}}
&=
\sum_{\ell = 0}^{\infty}{M^{\ell} \over \ell!}
\sum_{\ell' = 0}^{\infty}{N^{\ell'} \over \ell'!}
=
\sum_{\ell = 0}^{\infty}\sum_{\ell' = 0}^{\infty}{M^{\ell}N^{\ell'} \over \ell!\ell'!}
\sum_{n = 0}^{\infty}\delta_{n, \ell + \ell'}
\\[3mm]&=
\sum_{n = 0}^{\infty}\sum_{\ell = 0}^{\infty}{M^{\ell} \over \ell!}
\sum_{\ell' = 0}^{\infty}{N^{\ell'} \over \ell'!}\,\delta_{\ell',n - \ell}
=
\sum_{n = 0}^{\infty}
\sum_{\ell = 0 \atop {\vphantom{\LARGE A}n - \ell\ \geq\ 0}}^{\infty}
{M^{\ell} \over \ell!}\,{N^{n - \ell} \over \pars{n - \ell}!}
\\[3mm]&=
\sum_{n = 0}^{\infty}{1 \over n!}
\sum_{\ell = 0}^{n}
{n! \over \ell!\pars{n - \ell}!}\,M^{\ell}N^{n - \ell}
=
\sum_{n = 0}^{\infty}{1 \over n!}
\sum_{\ell = 0}^{n}{n \choose \ell}M^{\ell}N^{n - \ell}
\\[3mm]&
=\sum_{n = 0}^{\infty}{1 \over n!}\pars{M + N}^{n}
=
\color{#0000ff}{\large\expo{M + N}}\,,\qquad\mbox{since}\quad\bracks{M,N} = 0
\end{align}