# Matrix Exponential

## Introduction

Similar to exponential function $f(x) = \exp{x}$, matrix exponential $f(X) = \exp{X}$ also has different form of definitions but all of the definitions are mathematically equivalent.

Previously, we have discussed this topic for exponential function. In this blog post, I would like to discuss the same topic for matrix exponential.

## Prerequisites

### Matrix Norm

Given a field $K$ of either real or complex numbers, let $K^{m \times n}$ be the $K$-vector space of matrices with $m$ rows and $n$ columns and entries in the field $K$. A matrix norm is a norm on $K^{m \times n}$ and a function $\lVert \cdot \rVert : K^{m \times n} \rightarrow \mathbb{R}$ that must satisfy the properties.

For all scalars $\alpha \in K$ and matrices $A,B\in K^{m\times n}$,

• $\lVert A \rVert \geq 0$ (positive-valued)
• $\lVert A \rVert = 0 \iff A = 0$ (definite)
• $\lVert \alpha A^n \rVert \leq \lvert \alpha \rvert \lVert A \rVert^n$ (absolutely homogeneous)
• $\lVert A + B \rVert \leq \lVert A \rVert + \lVert B \rVert$ (sub-additive or satisfying the triangle inequality)

Matrix norms are particularly useful if they are also sub-multiplicative:

• $\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert$ (sub-multiplicative)

But it is not a necessary requirement for matrix norm.

An example of the matrix norm that also satisfies the sub-multiplicative property is Frobenius norm.

\begin{align} \lVert A \rVert_{F} &= \sqrt{\sum_{i=1}^{m} \sum_{i=1}^{n} \lvert a_{i,j}\rvert^2} \\ &= \sqrt{\text{trace}(A^{\dagger}A)} \\ \end{align}

### Power Matrix Series Convergence

If the analytic function $f$ has the Taylor expansion

\begin{align} f(x) &= \sum_{i=0}^{\infty} c_i x^i \\ &= c_0 + c_1 x + c_2 x^2 + \cdots \\ \end{align}

Then a matrix function $A \mapsto f(A)$ can be defined by substituting $x$ by a square matrix: powers become matrix powers, additions become matrix sums and multiplications by coefficients become scalar multiplications. If the series converges for $\lvert x \rvert < r$, then the corresponding matrix series

\begin{align} f(A) &= \sum_{i=0}^{\infty} c_i A^i \\ &= c_0 + c_1 A + c_2 A^2 + \cdots \\ \end{align}

converges for matrices $A$ such that $\lVert A \rVert < r$ for some matrix norm that satisfies the sub-multiplicative property $\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert$.

This can be proved by contradiction.

Proof

Because of the sub-multiplicative property $\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert$, we must have

$$\lVert A^n \rVert \leq \lVert A \rVert^n$$

We further take the limit on both side

$$\lim_{n \rightarrow \infty} \sup \lVert A^n \rVert \leq \lim_{n \rightarrow \infty} \sup \lVert A \rVert^n$$

$\lim_{n \rightarrow \infty} \sup$ was used for $\lVert A^n \rVert$ because we don’t know if $\lVert A^n \rVert$ has a limit. However, because $f(\lvert x \rvert)$ converges for $\lvert x \rvert < r$ and $\lVert A \rVert < r$, $f(\lVert A \rVert)$ must converge. So we must have

$$\lim_{n \rightarrow \infty} c_n \lVert A \rVert^n = 0$$

and

$$\lim_{n \rightarrow \infty} \sup \lVert A \rVert^n = \lim_{n \rightarrow \infty} \lVert A \rVert^n = 0$$

Because

$$\lim_{n \rightarrow \infty} \sup \lVert A^n \rVert \leq \lim_{n \rightarrow \infty} \sup \lVert A \rVert^n = 0$$

and with the basic property of matrix norm positive-valued

$$\lim_{n \rightarrow \infty} \inf \lVert A^n \rVert \geq 0$$

$\lVert A^n \rVert$ must have a limit and

$$\lim_{n \rightarrow \infty} \lVert A^n \rVert = 0$$

With the basic property of matrix norm definite, we must have

$$\lim_{n \rightarrow \infty} A^n = 0$$

Suppose the matrix series $\sum_{i=0}^{\infty} c_i A^i$ does not converge, we will have

$$\lim_{n \rightarrow \infty} c_n A^n \neq 0$$

and further

$$\lim_{n \rightarrow \infty} A^n \neq 0$$

This raised the contradiction.

Therefore, the matrix series $\sum_{i=0}^{\infty} c_i A^i$ must converge.

## Matrix Exponential Definitions

There are two common definitions for matrix exponential, including the series definition and the limit definition. Our goal is to prove the equivalence between the two definitions.

### Series Definition

For a real or complex matrix $X \in \mathbb{C}^{m \times n}$, the exponential function $\exp(X) = e^X$ is defined as follows.

\begin{align} \exp{X} &= \sum_{n=0}^{\infty} \frac{X^n}{n!} \\ &= I + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \cdots \\ \end{align}

### Limit Definition

For a real or complex matrix $X \in \mathbb{C}^{m \times n}$, the exponential function $\exp(X) = e^X$ is defined as follows.

\begin{align} \exp{X} &= \lim_{n \rightarrow \infty} \Big( I + \frac{X}{n} \Big)^n \end{align}

## Equivalence of Definitions

Proof

The majority of the proof is the same as the proof for exponential function that I derived previously. The subtle difference is at the convergence of the matrix exponential series.

Given real or complex number $x$, we could apply binomial expansion.

\begin{align} \Big( I + \frac{X}{n} \Big)^n &= \sum_{k=0}^{n} {n \choose k} \frac{X^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \frac{X^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{n (n-1) \cdots (n - k + 1)}{k!} \frac{X^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{X^k}{k!} \frac{n (n-1) \cdots (n - k + 1)}{n^k} \\ &= \sum_{k=0}^{n} \frac{X^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \\ \end{align}

We define

$$u_k(n) = \begin{cases} \frac{X^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) & \text{if k \leq n} \\ 0 & \text{if k > n} \\ \end{cases}$$

Let’s see if $\lim_{n \rightarrow \infty} u_k(n)$ exists for every possible $k$.

When $k > n$, it is obvious that $\lim_{n \rightarrow \infty} u_k(n) = 0$.

When $k \leq n$,

\begin{align} \lim_{n \rightarrow \infty} u_k(n) &= \lim_{n \rightarrow \infty} \bigg[ \frac{X^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \bigg] \\ &= \bigg(\lim_{n \rightarrow \infty} \frac{x^k}{k!} \bigg) \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{1}{n}\Big) \bigg) \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{2}{n}\Big) \bigg) \cdots \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{k-1}{n}\Big) \bigg)\\ &= \lim_{n \rightarrow \infty} \frac{X^k}{k!} \end{align}

So the question becomes if $\lim_{n \rightarrow \infty} \frac{X^k}{k!}$ exists for every possible $k \leq n$. For example, does $\lim_{n \rightarrow \infty} \frac{X^k}{k!}$ exist when $k = n, n - 1, n - 2, \cdots$?

Let $\lVert \cdot \rVert$ be the matrix norm that also satisfies the sub-multiplicative property, i.e., $\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert$. We already know the exponential function $\exp^{\lVert X \rVert} = \sum_{i=0}^{\infty}{\frac{\lVert X \rVert^k}{k!}}$ converges. The according to the power matrix series convergence property we showed in the prerequisites, the matrix exponential $\exp^{X} = \sum_{i=0}^{\infty}{\frac{X^k}{k!}}$ must also converge.

Therefore, the limit $\lim_{n \rightarrow \infty} \frac{x^k}{k!}$ must exist for every possible $k \leq n$.

\begin{align} \lim_{n \rightarrow \infty} u_k(n) &= \lim_{n \rightarrow \infty} \frac{X^k}{k!} \\ &= \frac{X^k}{k!} \\ \end{align}

Therefore,

\begin{align} \lim_{n \rightarrow \infty}\Big( I + \frac{X}{n} \Big)^n &= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{X^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \\ &= \lim_{n \rightarrow \infty} \sum_{k=0}^{\infty} u_k(n) \\ &= \sum_{k=0}^{\infty} \lim_{n \rightarrow \infty} u_k(n) \\ &= \sum_{k=0}^{\infty} \frac{X^k}{k!} \\ &= \sum_{n=0}^{\infty} \frac{X^n}{n!} \\ \end{align}

This concludes the proof.

Lei Mao

01-10-2022

01-10-2022