# $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$

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

• This doesn't hold in general, as far as I know. It works if $MN=NM$, though. – egreg Nov 15 '13 at 18:17
• Yes sorry, forgot to add that $M$ and $N$ commute. Will update and thanks for the spot! – Slugger Nov 15 '13 at 18:18
• Right, one needs something like $MN=NM$ for the equality. If one has that $M$ and $N$ commute, it's basically the same proof as the proof of $e^{z+w} = e^ze^w$ for complex numbers. – Daniel Fischer Nov 15 '13 at 18:18

## 4 Answers

Another take on it, which avoids the somewhat tedious term-by-term manipulation and term-by-term comparison of matrix power series:

Consider the ordinary, constant coefficient, matrix differential equation

$dX / dt = (M + N)X; \, \text{with} \, X(0) = I; \tag{1}$

the unique matrix solution is well-known to be

$X(t) = e^{(M + N)t}. \tag{2}$

Next, set

$Y(t) = e^{Mt}e^{Nt} \tag{3}$

and note that, by the Leibnitz rule for derivatives of products,

$dY / dt = (d(e^{Mt}) / dt) e^{Nt} + e^{Mt}(d(e^{Nt}) /dt) = Me^{Mt}e^{Nt} + e^{Mt}Ne^{Nt}, \tag{4}$

and since $MN - NM = [M, N] = 0$ we also have $[e^{Mt}, N] = 0$ so that (4) becomes

$dY / dt = Me^{Mt}e^{Nt} + Ne^{Mt}e^{Nt} = (M + N)e^{Mt}e^{Nt} = (M + N)Y, \tag{5}$

and evidently

$Y(0) = I, \tag{6}$

so that $X(t)$ and $Yt)$ satisfy the same differential equation with the same initial conditions; thus $X(t) = Y(t)$ for all $t$, or

$e^{(M + N)t} = e^{Mt}e^{Nt} \tag{7}$

for all $t$. Taking $t = 1$ yields the requisite result.QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• @Avitus: Thanks! – Robert Lewis Nov 15 '13 at 21:51
• @ Avitus: anything to avoid lots of eponents and indices! ;-) – Robert Lewis Nov 15 '13 at 22:57

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \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}

• @ Felix Marin: check your first equation in your first line, and$M$ needs to be replaced by an $N$. But thanks for the elegant grinding, +!! – Robert Lewis Nov 15 '13 at 21:20
• @RobertLewis This is the "cut and paste phantom". I already removed the typo. Thanks. – Felix Marin Nov 15 '13 at 23:53
• The use of the Kronecker delta is amazing! – Felix Crazzolara Mar 21 '18 at 18:53
• @Felix.C Yes, it's always helpful. Thanks. – Felix Marin Mar 21 '18 at 19:26

Hint.

$$(M+N)^2=M^2+MN+NM+N^2=(\text{if}~~[M,N]=0)=M^2+2MN+N^2.$$

Use this fact in the expansion of $\exp(M+N)$ to arrive at sum of monomials of the form $M^qN^r$ with $q,r\geq 0$ and rational coefficients (without $[M,N]=0$ you would have also monomials of the form $N^rM^q$!) . To finish the proof you need to collect the monomials of the same degree in $\exp(M)\exp(N)$, where the ordering is clear by definition.

• I will add more details if the OP asks for them. – Avitus Nov 15 '13 at 18:48

The product of two series (one of which is absolutely convergent) is $$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_{n-k}b_{k}\right).$$ Applying this to the series, $$e^Me^N = \left(I + M + \frac{M^2}{2!} + \frac{M^3}{3!}\ldots\right) \left(I + N + \frac{N^2}{2!} + \frac{N^3}{3!} \ldots\right)\\=I+ (MI+IN)+\left(\frac{M^2}{2}I+MN+I\frac{N^2}{2}\right)+\ldots$$

Now compare this to the other sum

$$e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots$$