Engineering Analysis/Matrix Exponentials

Matrix Exponentials

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

$e^A$

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:

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

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|Cayley-Hamilton Theorem]]. Solving the Theorem for Aⁿ, we get:

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

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

$A^{n+1}=-c_{n-1}{A}^n-c_{n-2}{A}^{n-1}-\cdots -c_1{A}^2-c_{0}A$

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:

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

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:

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

Derivative

The derivative of a matrix exponential is:

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

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:

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

Differential Equations

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

$x^{\prime }(t)=Ax(t)+f(t)$

With initial conditions

$x(t_0)=c$

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

$x(t)=e^{A(t-t_0)}c+\underset{t_0}{\overset{t}{\int }}{e}^{A(t-\tau )}f(\tau )d\tau$

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:

$ℒ[e^{At}]=(sI-A{)}^{-1}$

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