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Background Information

Previously I wrote a post about how to derive the limit definition of real exponential function

$$\exp(x) := \lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n, x \in \mathbb{R}$$

from the series definition

$$\exp(x) := \sum_{n=0}^{\infty} \frac{x^n}{n!}, x \in \mathbb{R}$$

And today when I was talking with one of my instructors about matrix functions, he said that the matrix exponential function for some arbitrary real square matrix $A = A_{k \times k}$ is defined as

$$\exp(A) := \sum_{n=0}^{\infty} \frac{A^n}{n!}, A \in M_k(\mathbb{R})$$

So one question naturally comes to my mind: Do we have an alternative definition through limit for $e^A$, that is to say

$$\exp(A) := \lim_{n \to \infty} \left(1 + \frac{A}{n} \right)^n, A \in M_k(\mathbb{R})$$

My Questions

So given the background information, my question are the following:

  1. Can we properly define

$$\exp(A) := \lim_{n \to \infty} \left(1 + \frac{A}{n} \right)^n, A \in M_k(\mathbb{R})$$

  1. If so, how to derive it from the series definition? My own idea is to find the matrix sequence/series version of Monotone Convergence and Dominated Convergence, and then cook up the argument. But this seems quite hard since I haven't done much in matrix function theory.

All ideas are welcome. Thanks for your help!

Yunxuan Zhang
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  • For any $x$ in some Banach algebra, $$\sum_{k=0}^n\frac{x^k}{k!}-{\left(1+\frac xn\right)}^n =\sum_{k=0}^n\left(1-\frac{n!}{(n-k)!,n^k}\right)\frac{x^k}{k!}$$ (this simply follows from the Binomial theorem and the expression of binomial coefficients). It is easy to see that the right-hand side tends to $0$ as $n\to\infty$. – nejimban Jan 24 '25 at 22:49
  • Thanks for your comment. But I don't have enough background for Banach algebra, and could you please consider giving answer with some details? Also, your equation is not directly connected with variables as matrices, so I don't know whether it is true for matrix functions. – Yunxuan Zhang Jan 24 '25 at 22:51
  • Look at the formula. It works when $x$ is a matrix, because $x$ and $I$ (replacing "1") commute. The rest is just combinatorics. Now everything you know about limits applies, and presumably your old proof goes through without any changes. – John Hughes Jan 24 '25 at 22:53
  • @JohnHughes Thanks for your comment! The point is whether the theorems in my old proof can be directly applied or not. I need to first see whether the monotone convergence and dominated convergence work for matrices. – Yunxuan Zhang Jan 24 '25 at 22:56

1 Answers1

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From the Binomial theorem, $${\left(I+\frac An\right)}^n=\sum_{k=0}^n\frac1{n^k}\binom nkA^k$$ (where $I$, the identity matrix, commutes with $A$). Therefore, \begin{align*}\sum_{k=0}^n\frac{A^k}{k!}-{\left(I+\frac An\right)}^n &=\sum_{k=0}^n\left(\frac1{k!}-\frac1{n^k}\,\frac{n!}{(n-k)!\,k!}\right)A^k\\[.4em] &=\sum_{k=0}^n\left(1-\frac{n!}{(n-k)!\,n^k}\right)\frac{A^k}{k!}. \end{align*} Thus, by triangle inequality (taking any matrix norm $\|\cdot\|$ for the finite-dimensional space of matrices), \begin{align*}\left\|\sum_{k=0}^n\frac{A^k}{k!}-{\left(I+\frac An\right)}^n \right\|&\le\sum_{k=0}^n\left(1-\frac{n!}{(n-k)!\,n^k}\right)\frac{\|A\|^k}{k!}. \end{align*} Fix $N$ and assume $n\ge N$. Then \begin{align*}\left\|\sum_{k=0}^n\frac{A^k}{k!}-{\left(I+\frac An\right)}^n \right\|&\le\sum_{k=0}^N\left(1-\frac{n!}{(n-k)!\,n^k}\right)\frac{\|A\|^k}{k!} +\sum_{k=N+1}^\infty\frac{\|A\|^k}{k!}, \end{align*} so (because $n!/(n-k)!\sim n^k$) \begin{align*}\limsup_{n\to\infty}\left\|\sum_{k=0}^n\frac{A^k}{k!}-{\left(I+\frac An\right)}^n \right\|&\le\sum_{k=N+1}^\infty\frac{\|A\|^k}{k!}, \end{align*} where the right-hand side, being the remainder of the convergent series of $\mathrm e^{\|A\|}$, tends to $0$ as $N\to\infty$.

nejimban
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