# Solving Exponential of a Matrix

1. Oct 29, 2015

### Sherin

2. Oct 29, 2015

### Geofleur

When $a$ is just a real number, one can use a Taylor series to represent $e^{at}$ as

$e^{at} = \sum_{n=0}^{\infty} \frac{(at)^n}{n!} = 1 + at + \frac{1}{2}(at)^2 + \ldots$

By analogy, one can define the exponential of $\mathbf{A}t$, where $\mathbf{A}$ is now a matrix, as

$e^{\mathbf{A}t} = \sum_{n=0}^{\infty} \frac{(\mathbf{A}t)^n}{n!}$.

Because multiplying a matrix by itself is perfectly well defined, the above sum makes sense. Now if $t$ is not too large, we can truncate the series and write

$e^{\mathbf{A}t} \approx (\mathbf{A}t)^0 + \mathbf{A}t \equiv \mathbf{I} + \mathbf{A}t$,

where $\mathbf{I}$ is the identity matrix. Using this truncated sum, your formula can be derived, except for the time dependent functions $\alpha_1(t)$ and $\alpha_2(t)$. Perhaps someone else can shed some light on where those might be coming from. What is the context in which you are seeing this equation?

3. Nov 2, 2015

### jasonRF

Geofleur's infinite series is the place to start. The trick is to use the Cayley-Hamilton theorem, which tells you that a matrix satisfies it's own characteristic equation. Sine you have a 2x2 matrix this is a second order polynomial, so you can write $\mathbf{A}^2=\alpha_0 \mathbf{I} + \alpha_1 \mathbf{A}$ for some $\alpha_0$ and $\alpha_1$ . It follows that any power of your matrix can also be represented by a linear combination of $\mathbf{I}$ and $\mathbf{A}$. Hopefully that shows you where the result comes from.

This is a common approach used in electrical engineering.

Jason

4. Nov 2, 2015

### Sherin

Thank you so much for your help!