# Matrix exponentials

1. Oct 3, 2008

### Somefantastik

$$A = $\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right)$$$

eigenvalues are $$\lambda_{1} = -1, \ \lambda_{2} = \lambda_{3} = 1$$

$$(A-\lambda_{1}I)u^{(1)} = 0 \ => \ u^{(1)} = $\left( \begin{array}{c} -1 \\ 0 \\ 1 \end{array} \right)$$$

$$(A-\lambda_{2}I)u^{(2)} = 0 \ => u^{(1)} = $\left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$$$

$$(A -\lambda_{3}I)u^{(3)} = u^{(2)} \ => \ \left( \begin{array}{ccc|c} 1&0&-1&-1\\ 0&0 &0&1\\ 0&0&0&0 \end{array} \right)$$

since we cannot have 0 = 1, we can say that there is only one eigenvector for

$$\lambda = 1$$

which means that the Jordan form will be

$$$\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right)$$$

Am I correct?

Now I need to find exp(Jt) and I'm not sure how.

If I only have 2 eigenvectors, how can I find the fund. matrix T such that

$$e^{At} = Te^{Jt}T^{-1}$$?

Last edited: Oct 3, 2008
2. Oct 3, 2008

### HallsofIvy

Staff Emeritus
In general, to find such a function of other than a "number", use the Taylor's series

eA= I+ A+ (1/2)A2+ (1/3!)A3+ ...

If A happens to be diagonal, that's easy: An is just the diagonal matrix with exponentials of the diagonal elements of A on its diagonal.

For Jordan form, its a little more complicated. Try calculating J2, J3, ... for this particular J yourself and see if you spot a pattern.

3. Oct 3, 2008

### Somefantastik

I noticed that A2k = I and A(2k+1)'s = A, but I don't know how to get eA from that.

4. Oct 3, 2008

### Hurkyl

Staff Emeritus
How exactly are you having trouble summing the series?

5. Oct 3, 2008

### Somefantastik

I'm just an idiot. I didn't realize that the Taylor expansions' limits came out to be cosh and sinh. I think I finished this problem. Thanks for the help so far.

6. Oct 3, 2008

### Hurkyl

Staff Emeritus
For the record, there's another method that can be useful. Note that you can separate that Jordan block into

$$\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] + \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$$

....

7. Oct 3, 2008

### Somefantastik

Ah, I see, and the second matrix is nilpotent.

But still, if I got that, how would I compute

$$e^{At} = Te^{Jt}T^{-1}$$

Not sure what T would be in this case.

8. Oct 3, 2008

### Hurkyl

Staff Emeritus
Right. And you'll find the results of doing arithmetic look a lot like derivatives -- a nilpotent matrix behaves a little bit like an infinitessimal.

(Again, look to the series to get started)