Matrix exponential: Prove that $\mathrm{e}^A=\lim_{N\rightarrow\infty}(I+\frac{A}{N})^N.$

Asked Oct 27 '20 at 07:46

Active Oct 27 '20 at 07:46

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Let $A\in\mathbb{C}^{n\times n}$. Define the matrix exponential $$ \mathrm{e}^A=\lim_{N\rightarrow\infty}\sum_{j=0}^N \dfrac{A^j}{j!}. $$ Prove that $$\mathrm{e}^A=\lim_{N\rightarrow\infty}(I+\frac{A}{N})^N.$$

My ODE textbook uses the proposition directly. The proposition is obvious when $n=1$, but I still want to get I valid proof for the general case. It is easy to get $$ (I+\frac{A}{N})^N=I+A+\dfrac{1}{2!}(1-\dfrac{1}{N})A^2+\cdots+\dfrac{1}{N!}(1-\dfrac{1}{N})\cdots (1-\dfrac{N-1}{N})A^N $$ I want to compare the matrix norm then, but it seems do not work.

The main problem is we cannot directly use inequality here. Appreciate any help!

asked Oct 27 '20 at 07:46

user823011

1,335

You can use Santiago's answer for this question substituting $w=w_n=A$ and $|w|=|A|$. – Chrystomath Oct 27 '20 at 07:57
Replace $A$ with $At$ and differentiate both sides wrt $t$. The second identity amounts to using Euler's method to solve the ODE $X'(t) = AX(t)$ whose unique solution is the first. – Qiaochu Yuan Oct 27 '20 at 08:25

Matrix exponential: Prove that $\mathrm{e}^A=\lim_{N\rightarrow\infty}(I+\frac{A}{N})^N.$

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