Prove that $\displaystyle\lim_{k \to \infty} \left( I + \frac{1}{k}A \right)^{k} = e^A$

Question

I'm having a little trouble here to prove the following statement:

Let $A$ be an $n \times n$ matrix (real or complex). Prove that

$$\lim_{k \to \infty} \left( I + \frac{1}{k} A \right)^{k} = e^{A}$$

Now I'm using matrix and possible non-commutative; I don't know where to begin. Can you give a spit? Thanks for your attention!

Answer 1

From the Binomial Theorem we have:

$$ \lim_{n \to \infty} \left(I + \left(\frac{A}{n} \right)\right)^{n} = \lim_{n \to \infty} \sum_{k=0}^{n} \binom{n}{k} I^{k} \left(\frac{A}{n} \right)^{n-k}$$

$$= \lim_{n \to \infty} \sum_{k=0}^{n} \binom{n}{k} \left(\frac{A}{n} \right)^{n-k}$$

Answer 2

Assuming that $\mathrm A$ is diagonalizable,

$$\begin{array}{rl} \left(\mathrm I_n + \frac 1k \mathrm A\right)^k &= \left(\mathrm I_n + \frac 1k \mathrm Q \Lambda \mathrm Q^{-1}\right)^k\\ &= \left(\mathrm Q \mathrm Q^{-1} + \frac 1k \mathrm Q \Lambda \mathrm Q^{-1}\right)^k\\ &= \mathrm Q \left( \mathrm I_n + \frac 1k \Lambda \right)^k \mathrm Q^{-1}\\ &= \mathrm Q \,\mbox{diag} ( (1 + \frac{\lambda_1}{k})^k, \dots, (1 + \frac{\lambda_n}{k})^k ) \mathrm Q^{-1}\\\end{array}$$

Since

$$\lim \left(1 + \frac{\lambda_i}{k}\right)^k = \exp{(\lambda_i)}$$

we have

$$\lim \left(\mathrm I_n + \frac 1k \mathrm A\right)^k = \mathrm Q \,\mbox{diag} ( \exp{(\lambda_1)}, \dots, \exp{(\lambda_n)} ) \mathrm Q^{-1} = \mathrm Q \exp(\Lambda) \mathrm Q^{-1} = \color{blue}{\exp (\mathrm A)}$$

Answer 3

Assume that $A\in M_n(\mathbb{C})$. Then it suffices to show the result when $A$ is a Jordan block $A=\lambda I+J$, where $J$ is the nilpotent Jordan block. When $k>n$:

\begin{align*}&(I+\dfrac{1}{k}A)^k\\=&((1+\lambda/k)I+J/k)^k\\=&(1+\lambda/k)^kI+\binom{k}{1}(1+\lambda/k)^{k-1}J/k+\cdots +\binom{k}{n-1}(1+\lambda/k)^{k-n+1}(J/k)^{n-1}\end{align*}

(a sum of $n$ matrices). Then \begin{align*}&\lim_{k\rightarrow +\infty}(I+\dfrac{1}{k}A)^k\\=&\lim_{k\rightarrow +\infty}(e^{\lambda}I+ke^{\lambda}1/kJ+\cdots+\dfrac{k^{n-1}}{(n-1)!}e^{\lambda}\dfrac{1}{k^{n-1}}J^{n-1})\\=&e^{\lambda}(I+J+\cdots+\cfrac{1}{(n-1)!}J^{n-1})\\=&e^A\end{align*}

and we are done.

Answer 4

Here is an answer without using diagonal arguments.

Let $\| \cdot \|$ be a sub-multiplicative norm over $M_n(\mathbb{C})$ meaning $$\forall (A,B) \in M_n(\mathbb{C})^2, \quad \|AB\|\leq\|A\|\space\|B\|$$ (The Schatten p norm is an exemple of such a norm)

We can easily prove: $$(\text {Lemma 1}): \forall A\in M_n(\mathbb{C}),\space \forall k \in \mathbb{N}\space\|A^k\|\leq\|A\|^k$$ Let $A \in M_n(\mathbb C)$ $$\eqalign{\left\| \left( I_n + \frac A p \right)^p - \exp A \right\| & = \left\| \sum\limits_{k=0}^p {\binom p k} {\frac {A^k} {p^k}} - \sum\limits_{k=0}^\infty \frac{A^k}{k!} \right\|\\ &= \left\| \sum\limits_{k=0}^p \frac {A^k} {k!} \left( \frac {p!} {(p-k)!p^k}-1\right) - \sum\limits_{k=p+1}^\infty \frac{A^k}{k!} \right\|\\ &\leq \sum\limits_{k=0}^p {\frac {\left\| {A^k} \right\|} {k!}} \left| \frac {p!} {(p-k)!p^k}-1 \right|+ \sum\limits_{k=p+1}^\infty {\frac {\left\| {A^k} \right\|}{k!}}&(\text{Triangular Inequality})\\ &\leq \sum\limits_{k=0}^p {\frac {\left\| {A} \right\|^k} {k!}} \left(1- \frac {p!} {(p-k)!p^k} \right)+ \sum\limits_{k=p+1}^\infty {\frac {\left\| {A} \right\|^k}{k!}} & (\text {Lemma 1})\\ &= \sum\limits_{k=0}^\infty {\frac {\left\| {A} \right\|^k} {k!}} - \sum\limits_{k=0}^p {\frac {\left\| {A} \right\|^k}{k!}}\frac {p!} {(p-k)!p^k}\\ &= \exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p }$$ We proved: $$0\leq\left\| \left( I_n + \frac A p \right)^p - \exp A \right\| \leq \exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p$$ But $$\exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p \underset {p\to {\infty}} {\longrightarrow} 0 $$ then by the squeeze theorem: $$\left\| \left( I_n + \frac A p \right)^p - \exp A \right\|\underset {p\to {\infty}} {\longrightarrow} 0 $$ which is equivalent to: $$ \left( I_n + \frac A p \right)^p \underset {p\to {\infty}} {\longrightarrow} \exp A $$