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Matrix Functions, Definition of Matrix Exponential Functions 📂Functions

Matrix Functions, Definition of Matrix Exponential Functions

Definitions1

$$ \mathbf{x}(t) = \begin{pmatrix} x_{1}(t) \\ \vdots \\ x_{n}(t) \end{pmatrix},\quad \mathbf{A}(t) = \begin{pmatrix} a_{11}(t) & \cdots & a_{1m}(t) \\ \vdots & & \vdots \\ a_{n1}(t) & \cdots & a_{nm}(t) \end{pmatrix} $$

If each element of a matrix is a function of variable $t$, it is called a matrix function.

  • All elements of $\mathbf{A}(t)$, i.e., all $a_{ij}$ being continuous at a given point (or interval) means $\mathbf{A}(t)$ is continuous.

  • If all elements of $\mathbf{A}(t)$ are differentiable, then $\mathbf{A}(t)$ is considered differentiable. The derivative of $\mathbf{A}(t)$ is denoted as $\dfrac{d \mathbf{A}(t)}{dt}$ and defined as follows:

    $$ \frac{d \mathbf{A}}{dt} := \left[ \frac{d a_{ij}}{dt} \right] $$

  • The matrix consisting of the integral of each element of $\mathbf{A}(t)$ is called the integral of $\mathbf{A}(t)$.

    $$ \int_{a}^b\mathbf{A}(t) dt := \left[ \int_{a}^ba_{ij}(t)dt \right] $$

  • The exponential function of a matrix $e^{(\cdot)t} : M_{n\times n} \to M_{n\times n}$ is defined as follows:

    $$ e^{\mathbf{M} t}:=\sum \limits_{n=0}^\infty \frac{\mathbf{M}^n t^n}{n!}=\mathbf{I}+\mathbf{M}t + \frac{ \mathbf{M}^2t^2}{2!}+\frac{\mathbf{M}^3t^3}{3!}\cdots $$

Explanation

It is essentially no different from a vector-valued function. It’s just that the dimension of the index of the function values is 2.

It has the following properties regarding differentiation. When $\mathbf{A}(t)$, $\mathbf{B}(t)$ are matrix functions and $\mathbf{C}$ is a constant matrix,

$$ \frac{d}{dt}( \mathbf{CA})=\mathbf{C} \frac{ d \mathbf{A} }{dt} $$

$$ \frac{ d }{dt} ( \mathbf{A} + \mathbf{B})=\frac{ d \mathbf{A} }{dt} +\frac{d \mathbf{B}}{dt} $$

$$ \frac{d}{dt}(\mathbf{AB})=\mathbf{A}\frac{d \mathbf{B}}{dt}+\frac{d\mathbf{A}}{dt}\mathbf{B} $$

Example

Suppose matrix $\mathbf{A}(t)$ is as follows.

$$ \mathbf{A}(t) = \begin{pmatrix} \sin t & t \\ 1 & \cos t \end{pmatrix} $$

Then, $\mathbf{A}(t)$ is continuous across the entire interval, and its differentiation and integration are as follows.

$$ \mathbf{A}^{\prime}(t)=\begin{pmatrix} \cos t & 1 \\ 0 & -\sin t \end{pmatrix},\quad \int_{0}^\pi \mathbf{A}(t)dt=\begin{pmatrix} 2 & \pi^2/2 \\ \pi & 0 \end{pmatrix} $$

Matrix Exponential Function

As defined, a function with a constant matrix as the exponent’s coefficient is called a matrix exponential function. However, when trying to concretely conceptualize this definition, the following thought is natural.

$$ e^{\mathbf{A}t}:= \left( e^{a_{ij}t} \right)=\begin{pmatrix} e^{a_{11}t} &\cdots & e^{a_{1n}t} \\ \vdots & & \vdots \\ e^{a_{n1}t} & \cdots& e^{a_{nn}t} \end{pmatrix} $$

But defining it this way does not satisfy the definition of the exponential function, $\dfrac{d e^{t}}{dt} = e^{t}$. When defined in terms of series, it can be seen to satisfy it as follows.

$$ \begin{align*} \frac{d}{dt}\big( e^{\mathbf{A}t}\big) &=\frac{d}{dt}\left(\sum \limits_{n=0}^\infty \frac{\mathbf{A}^n t^{n}}{n!}\right) \\ &= \frac{d}{dt}\left( \mathbf{I}+\mathbf{A}t + \frac{ \mathbf{A}^2t^2}{2!}+\frac{\mathbf{A}^3t^3}{3!}\cdots \right) \\ &=\mathbf{A}+ \frac{\mathbf{A}^2t}{1!}+\frac{\mathbf{A}^3t^2}{2!}+\cdots \\ &= \mathbf{A} \left( \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!}+\cdots \right) \\ &= \mathbf{A} \sum \limits_{n=0}^\infty \frac{\mathbf{A}^n t^n}{n!} \\ &=\mathbf{A} e^{\mathbf{A}t} \end{align*} $$


  1. William E. Boyce, Boyce’s Elementary Differential Equations and Boundary Value Problems (11th Edition, 2017), p292-293, 332 ↩︎