📂Matrix Algebra

Properties of Matrix Exponentials

Definition¹

The matrix exponential function $\exp : M_{n \times n}(\mathbb{C}) \to M_{n \times n}(\mathbb{C})$ is defined as follows, and this is called the matrix exponential (function).

$$ \exp (X) = e^{X} := \sum\limits_{m=0}^{\infty} \dfrac{X^{m}}{m!} \tag{1} $$

Here the above limit is to be understood as the limit of matrices.

Explanation

The matrix exponential is a generalization of the exponential function to matrices, and it inherits the properties of the exponential function defined on the real numbers. $X^{0}$ is defined using the identity matrix $I$. Note that $e^{X}$ itself is also a $n \times n$ matrix.

Properties

Let $X, Y \in M_{n \times n}(\mathbb{C})$ and $\alpha, \beta \in \mathbb{C}$. Then the matrix exponential has the following properties.

(a) For every $X \in M_{n \times n}(\mathbb{C})$, the series $(1)$ converges and $e^{X}$ is a continuous function.

(b) $e^{O} = I$ ($O$ is the zero matrix)

(c) $(e^{X})^{\ast} = e^{X^{\ast}}$

(d) $e^{X}$ is an invertible matrix and $(e^{X})^{-1} = e^{-X}$.

(e) $e^{(\alpha + \beta) X} = e^{\alpha X} e^{\beta X}$

(f) If $XY = YX$, then $e^{X+Y} = e^{X} e^{Y} = e^{Y} e^{X}$.

(g) If $C \in \operatorname{GL}(n, \mathbb{C})$, then $e^{C X C^{-1}} = C e^{X} C^{-1}$. In this case $\operatorname{GL}(n, \mathbb{C})$ is the general linear group.

(h) For a diagonal matrix $D = [d_{ii}]$, we have $e^{D} = \begin{bmatrix} e^{d_{11}} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & e^{d_{nn}} \end{bmatrix}$.

Proof

(a)

From the properties of the matrix norm $\| AB \| \le \| A \| \| B \|$ and the triangle inequality $\| A + B \| \le \| A \| + \| B \|$ we obtain

$$ \| X^{m} \| \le \| X \|^{m} $$

$$ \implies \left\| \sum\limits_{m=0}^{\infty} \dfrac{X^{m}}{m!} \right\| \le \sum\limits_{m=0}^{\infty} \left\| \dfrac{X^{m}}{m!} \right\| \le \sum\limits_{m=0}^{\infty} \dfrac{\| X \|^{m}}{m!} \lt \infty $$

Since $\| X \|$ is scalar, the value on the right equals the value $e^{\| X \|}$ of the exponential function defined on the real numbers, and is bounded. Therefore $e^{X}$ is absolutely convergent, hence convergent.

Moreover, by the above inequality and the Weierstrass M-test, $e^{X}$ is uniformly convergent and continuous.

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(b)

Trivial.

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(c)

From the definition of the matrix exponential and properties of the limit of matrices we obtain

$$ (e^{X})^{\ast} = \left( \lim\limits_{M \to \infty} \sum_{m = 0}^{M} \dfrac{X^{m}}{m!}\right)^{\ast} = \lim\limits_{M \to \infty} \left( \sum_{m = 0}^{M} \dfrac{X^{m}}{m!}\right)^{\ast} = \lim\limits_{M \to \infty} \sum_{m = 0}^{M} \dfrac{(X^{\ast})^{m}}{m!} = e^{X^{\ast}} $$

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(d)

Method 1

The product of two convergent series can be expressed as follows.

$$ \begin{align*} e^{X} e^{-X} &= \left( \sum\limits_{m = 0}^{\infty} \dfrac{X^{m}}{m!} \right) \left( \sum\limits_{m = 0}^{\infty} \dfrac{(-1)^{m}X^{m}}{m!} \right) \\ &= \sum_{m = 0}^{\infty} \sum_{k = 0}^{m} \dfrac{X^{k}}{k!} \dfrac{(-1)^{m-k}X^{m-k}}{(m-k)!} \\ &= \sum_{m = 0}^{\infty} \dfrac{1}{m!} \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} (-1)^{m-k}X^{m} \\ &= \sum_{m = 0}^{\infty} \dfrac{X^{m}}{m!} \left( \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} (-1)^{m-k} \right) \end{align*} $$

The value inside the parentheses is given by the binomial theorem.

$$ \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} (-1)^{m-k} = \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} 1^{k}(-1)^{m-k} = (1 + (-1))^{m} $$

This value equals $1$ only when $m=0$, and equals $0$ otherwise; hence we obtain the following conclusion.

$$ e^{X}e^{-X} = \dfrac{X^{0}}{0!} = I $$

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Method 2

It follows by substituting $Y = -X$ into (f).

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(e)

It follows as a corollary of (f).

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(f)

The proof is analogous to the proof of (d) Method 1. $e^{X}e^{Y}$ is as follows.

$$ e^{X}e^{Y} = \sum_{m = 0}^{\infty} \dfrac{1}{m!} \left( \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} X^{m}Y^{m-k} \right) $$

Since $X$ and $Y$ commute, by the binomial theorem we obtain

$$ (X + Y)^{m} = \sum_{k = 0}^{m} \dfrac{m!}{k!(m-k)!} X^{m}Y^{m-k} $$

$$ \implies e^{X}e^{Y} = \sum\limits_{m = 0}^{\infty}\dfrac{(X + Y)^{m}}{m!} = e^{X+Y} = e^{Y}e^{X} $$

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(g)

From the definition of $e^{X}$ and properties of the limit of matrices we have

$$ \begin{align*} e^{CXC^{-1}} &= \sum\limits_{m = 0}^{\infty} \dfrac{(C X C^{-1})^{m}}{m!} \\ &= \sum\limits_{m = 0}^{\infty} \dfrac{C X^{m} C^{-1}}{m!} \\ &= \lim\limits_{M \to \infty}\sum\limits_{m = 0}^{M} \dfrac{C X^{m} C^{-1}}{m!} \\ &= \lim\limits_{M \to \infty} C \left( \sum\limits_{m = 0}^{M} \dfrac{X^{m}}{m!} \right) C^{-1} \\ &= C \left( \lim\limits_{M \to \infty} \sum\limits_{m = 0}^{M} \dfrac{X^{m}}{m!} \right) C^{-1} \\ &= C e^{X} C^{-1} \end{align*} $$

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