Matrix Exponentials

If we have a matrix A, we can raise that matrix to a power of e as follows:

${\displaystyle e^{A}}$

It is important to note that this is not necessarily (not usually) equal to each individual element of A being raised to a power of e. Using taylor-series expansion of exponentials, we can show that:

${\displaystyle e^{A}=I+A+{\frac {1}{2}}A^{2}+{\frac {1}{6}}A^{3}+...=\sum _{k=0}^{\infty }{1 \over k!}A^{k}}$.

In other words, the matrix exponential can be reduced to a sum of powers of the matrix. This follows from both the taylor series expansion of the exponential function, and the cayley-hamilton theorem discussed previously.

However, this infinite sum is expensive to compute, and because the sequence is infinite, there is no good cut-off point where we can stop computing terms and call the answer a "good approximation". To alleviate this point, we can turn to the Cayley-Hamilton Theorem. Solving the Theorem for Aⁿ, we get:

${\displaystyle A^{n}=-c_{n-1}A^{n-1}-c_{n-2}A^{n-2}-\cdots -c_{1}A-c_{0}I}$

Multiplying both sides of the equation by A, we get:

${\displaystyle A^{n+1}=-c_{n-1}A^{n}-c_{n-2}A^{n-1}-\cdots -c_{1}A^{2}-c_{0}A}$

We can substitute the first equation into the second equation, and the result will be Aⁿ⁺¹ in terms of the first n - 1 powers of A. In fact, we can repeat that process so that A^m, for any arbitrary high power of m can be expressed as a linear combination of the first n - 1 powers of A. Applying this result to our exponential problem:

${\displaystyle e^{A}=\alpha _{0}I+\alpha _{1}A+\cdots +\alpha _{n-1}A^{n-1}}$

Where we can solve for the α terms, and have a finite polynomial that expresses the exponential.

Inverse

The inverse of a matrix exponential is given by:

${\displaystyle (e^{A})^{-1}=e^{-A}}$

Derivative

The derivative of a matrix exponential is:

${\displaystyle {\frac {d}{dx}}e^{Ax}=Ae^{Ax}=e^{Ax}A}$

Notice that the exponential matrix is commutative with the matrix A. This is not the case with other functions, necessarily.

Sum of Matrices

If we have a sum of matrices in the exponent, we cannot separate them:

${\displaystyle e^{(A+B)x}\neq e^{Ax}e^{Bx}}$

Differential Equations

If we have a first-degree differential equation of the following form:

${\displaystyle x'(t)=Ax(t)+f(t)}$

With initial conditions

${\displaystyle x(t_{0})=c}$

Then the solution to that equation is given in terms of the matrix exponential:

${\displaystyle x(t)=e^{A(t-t_{0})}c+\int _{t_{0}}^{t}e^{A(t-\tau )}f(\tau )d\tau }$

This equation shows up frequently in control engineering.

Laplace Transform

As a matter of some interest, we will show the Laplace Transform of a matrix exponential function:

${\displaystyle {\mathcal {L}}[e^{At}]=(sI-A)^{-1}}$

We will not use this result any further in this book, although other books on engineering might make use of it.