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

analysislinear algebramatricesmatrix-calculussolution-verification

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:

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

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 $$\frac{m\left(m-1\right)\left(m-2\right)\cdots\left(m-k-1\right)}{m^{k}} \underset{m\to \infty}{\longrightarrow} 1$$

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.