Show that for all $X \in \mathscr{M}_n(\mathbb{C})$ we have that: $\lim\limits_{n \rightarrow \infty}{\left(I + \frac{1}{m}X\right)}^{m} = \exp{X}$

Question

Show that for all $X \in \mathscr{M}_n(\mathbb{C})$ we have that:

$$\lim_{n \rightarrow \infty}{\left(I + \frac{1}{m}X\right)^{m}} = \exp{X}$$

Note that for matrices, the $e^X$ is defined by using the series expansion of the exponential function:

$$\exp{X} = I + X + \frac{1}{2!}X + \frac{1}{3!}X + \cdots$$

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I expanded out $\left(I + \frac{1}{m}X\right)^{m}$, which I figured I could do since powers of $X$ commute with each other:

$$\left(I + \frac{1}{m}X\right)^{m} = \sum_{k=0}^{m}{\binom{m}{k}\left(\frac{1}{m}X\right)^{k}} = \sum_{k=0}^{m}{\frac{m!}{k!\left(m-k\right)!}\left(\frac{1}{m^{k}}X^{k}\right)}$$

The $k$'th term in this expansion is can be simplified to:

$$\frac{m\left(m-1\right)\left(m-2\right)\cdots\left(m-k+1\right)}{m^{k}}\frac{1}{k!}X^{k}$$ (edited)

Thus, as $m \rightarrow \infty$, the identity will hold since $\frac{m\left(m-1\right)\left(m-2\right)\cdots\left(m-k-1\right)}{m^{k}} \underset{m\to \infty}{\longrightarrow} 1$

Answer 1

Here's a different approach based on the matrix logarithm. For $\|A\|<1$ we have the power series form $\log(I+A)=\sum_{k\geq1}(-1)^{k+1}A^k/k$, so for all sufficiently large $n$ we have $$\log\left((I+X/n)^n\right)=n\log(I+X/n)=n[X/n+o(1/n)]=X+o(1).$$ The desired result follows by the continuity of the matrix exponential.

Answer 2

Let $ m\in\mathbb{N} $, we have the following : \begin{aligned}\exp{X}-\left(I+\frac{1}{m}X\right)^{m}=\sum_{n=m+1}^{+\infty}{\frac{1}{n!}X^{n}}+\sum_{n=0}^{m}{\left(\frac{1}{n!}-\frac{1}{m^{n}}\binom{m}{n}\right)X^{n}}\end{aligned}

Let $ \left\Vert\cdot\right\Vert$, be a sub-multiplicative norm of $ \mathscr{M}_{n}\left(\mathbb{C}\right) $, then : $$ \left\Vert\exp{X}-\left(I+\frac{1}{m}X\right)^{m}\right\Vert\leq\left\Vert\sum_{n=m+1}^{+\infty}{\frac{1}{n!}X^{n}}\right\Vert+\sum_{n=0}^{m}{\left(1-\prod_{k=0}^{n-1}{\left(1-\frac{k}{m}\right)}\right)\frac{\left\Vert X\right\Vert^{n}}{n!}} $$

Notice that, for all $ n\in\mathbb{N} $,

$ 1-\prod\limits_{k=0}^{n-1}{\left(1-\frac{k}{m}\right)}=\sum\limits_{j=0}^{n-1}{\left(\prod\limits_{k=0}^{j-1}{\left(1-\frac{k}{m}\right)}-\prod\limits_{k=0}^{j}{\left(1-\frac{k}{m}\right)}\right)}=\frac{1}{m}\sum\limits_{j=0}^{n-1}{j\prod\limits_{k=0}^{j-1}{\left(1-\frac{k}{m}\right)}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tiny \ \ \ \ \normalsize\leq\frac{1}{m}\sum\limits_{j=0}^{n-1}{j}=\frac{1}{m}\binom{n}{2} $

Thus : $$ \left\Vert\exp{X}-\left(I+\frac{1}{m}X\right)^{m}\right\Vert\leq\left\Vert\sum_{n=m+1}^{+\infty}{\frac{1}{n!}X^{n}}\right\Vert+\frac{\left\Vert X\right\Vert^{2}}{2m}\sum_{n=0}^{m-2}{\frac{\left\Vert X\right\Vert^{n}}{n!}}\underset{m\to +\infty}{\longrightarrow}0 $$

Hence, $ \left(I+\frac{1}{m}X\right)^{m}\underset{m\to +\infty}{\longrightarrow}\exp{X} $.