# 0.13 Matrix exponential

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A formal description of the matrix exponential. The definition is given as well as examples of calculating it. The matrix exponential in thefrequency domain is given as well through treatment by the Laplace transform.

Since systems are often represented in terms of matrices and solutions of system equations often make use of the exponential, it makes senseto try and understand how these two concepts can be combined. In many previous applications, we've seen terms like $e^{at}$ come in handy for talking about system behavior. Here, awas always a scalar quantity. However, what would happen if the scalar $a$ was replaced by a matrix $A$ ? The result would be what is known as the matrix exponential .

## Definition

Recall the definition of the scalar exponential:

$e^{at}=1+a\frac{t}{1!}+a^{2}\frac{t^{2}}{2!}+a^{3}\frac{t^{3}}{3!}+$

The definition of the matrix exponential is almost identical:

$e^{At}={I}_{n}+A\frac{t}{1!}+A^{2}\frac{t^{2}}{2!}+A^{3}\frac{t^{3}}{3!}+$

Where $A$ is $n$ x $n$ and ${I}_{n}$ is the $n$ x $n$ identity matrix. While it is nice to see the resemblance between these two definitions, applying this infinite series does not turnout to be very efficient in practice. However, it can be useful in certain special cases.

Compute $e^{At}$ where $A=\begin{pmatrix}0 & 1\\ -1 & 0\\ \end{pmatrix}$ . We can start by taking powers of $A$ so that we can use the formal definition.

$A=\begin{pmatrix}0 & 1\\ -1 & 0\\ \end{pmatrix}$
$A^{2}=\begin{pmatrix}0 & 1\\ -1 & 0\\ \end{pmatrix}\begin{pmatrix}0 & 1\\ -1 & 0\\ \end{pmatrix}=\begin{pmatrix}-1 & 0\\ 0 & -1\\ \end{pmatrix}=-I$
$A^{3}=A^{2}A=-A$
$A^{4}=A^{2}A^{2}=I$
$A^{5}=AA^{2}=A$
$A^{6}=A^{2}A^{4}=-I$

And so the pattern goes, giving:

$A^{(4(n-1)+1)}=A$
$A^{(4(n-1)+2)}=-I$
$A^{(4(n-1)+3)}=-A$
$A^{(4(n-1)+4)}=I$

If we fill in the terms in the definition of $e^{at}$ , we'll get the following matrix:

$e^{At}=\begin{pmatrix}1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}- & t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\\ -t+\frac{t^{3}}{3!}-\frac{t^{5}}{5!}+ & 1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\\ \end{pmatrix}$

We notice that the sums in this matrix look familiar-in fact, they are the Taylor Series expansions of the sinusoids.Therefore, the solution further reduces to:

$e^{At}=\begin{pmatrix}\cos t & \sin t\\ -\sin t & \cos t\\ \end{pmatrix}$

## General method

The example above illustrates how the use of the true definition to simplify matrix exponentials might only be easily applied in caseswith inherent repetition. There is a more general method involving the Laplace Transform. In particular,

$(e^{At})=sI-A^{(-1)}$

We can verify that this is true by inserting the formal definition of the matrix exponential:

$(e^{At})=(I+A\frac{t}{1!}+A^{2}\frac{t^{2}}{2!}+)=\frac{1}{s}I+\frac{1}{s^{2}}A+\frac{1}{s^{3}}A^{2}+=sI-A^{(-1)}$

The jump between the third and fourth equations here may be a bit hard to believe, but this equality reduces to $I=I$ when both sides are multiplied by $sI-A$ . Taking an inverse Laplace of each side of Laplace Transform of the equation we find an expression for the matrix exponential:

$e^{At}=^{(-1)}(sI-A^{(-1)})$

We can do the same example as before, this time using the Laplace-based method.

$A=\begin{pmatrix}0 & 1\\ -1 & 0\\ \end{pmatrix}$
$sI-A^{(-1)}=\begin{pmatrix}s & -1\\ 1 & s\\ \end{pmatrix}^{(-1)}=\frac{1}{s^{2}+1}\begin{pmatrix}s & 1\\ -11 & s\\ \end{pmatrix}=\begin{pmatrix}\frac{s}{s^{2}+1} & \frac{1}{s^{2}+1}\\ \frac{-1}{s^{2}+1} & \frac{s}{s^{2}+1}\\ \end{pmatrix}$

Taking the inverse laplace of this gives us

$e^{At}=\begin{pmatrix}\cos t & \sin t\\ -\sin t & \cos t\\ \end{pmatrix}$

## Properties of the matrix exponential

In the scalar case, a product of exponentials $e^{a}e^{b}$ reduces to a single exponential whose power is the sum of the individual exponents' powers, $e^{a+b}$ . However, in the case of the matrix exponential, this is nottrue. If $A$ and $B$ are matrices,

$e^{A}e^{B}\neq e^{A+B}$

unless $A$ and $B$ are commutative (i.e. $AB=BA$ )

The derivative operates on the matrix exponential the same as it does on the scalar exponential.

$\frac{d e^{At}}{d t}}=0+A+A^{2}\frac{t}{1!}+A^{3}\frac{t^{2}}{2!}+=A(I+A\frac{t}{1!}+A^{2}\frac{t^{2}}{2!}+)=Ae^{At}$

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