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:


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

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:


 * $$A^n = -c_{n-1}A^{n-1} - c_{n-2}A^{n-2} - \cdots - c_1A - c_0I$$

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


 * $$A^{n+1} = -c_{n-1}A^n - c_{n-2}A^{n-1} - \cdots - c_1A^2 - c_0A$$

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 Am, 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:


 * $$e^A = \alpha_0I + \alpha_1A + \cdots + \alpha_{n-1}A^{n-1}$$

Where we can solve for the &alpha; 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:


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

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:


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

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


 * $$x'(t) = Ax(t) + f(t)$$

With initial conditions


 * $$x(t_0) = c$$

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


 * $$x(t) = e^{A(t - t_0)}c + \int_{t_0}^t e^{A(t - \tau)}f(\tau)d\tau$$

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:


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

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