# Computing e^A

• Nov 26th 2008, 03:37 PM
victor1487
Computing e^A
I need to compute e^A for the matrix
A= 0 ∏
....-∏ 0

where the diagonal are zeros and the other diagonal has pi on top and negative pi on the bottom.
I'm not quite sure where to start.

Thanks
• Nov 26th 2008, 03:54 PM
vincisonfire
$\displaystyle e^A$ is defined as
$\displaystyle *\sum_{i=0}^{\infty} \frac{A^n}{n!}$
You can find that
if $\displaystyle B=e^A$
then $\displaystyle B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$\displaystyle B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0$
$\displaystyle B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0$
If your prof wants you to evaluate those series then I don't know (I used Maple). Well I don't have the time... or both.
• Nov 26th 2008, 05:00 PM
chiph588@
$\displaystyle B(1,1) = B(2,2)$ should be $\displaystyle -1$
• Nov 26th 2008, 05:06 PM
vincisonfire
I don't think so because it is
$\displaystyle \sum_{i=1}^{\infty} (-1)^{n}\cdot\frac{\pi^{2n}}{(2n)!}$ because the first term is 0 not 1 we must integrate from 1 to infinity not 0 to infinity. But I may be mistaking.
Here are the 10 first terms
-4.934802202
-0.876090073
-2.211352843
-1.976022212
-2.001829103
-1.999899529
-2.000004167
-1.999999864
-2.000000003
-1.999999999
• Nov 26th 2008, 09:06 PM
chiph588@
well I typed e^A in my ti-89 and it gave me back $\displaystyle -I$ where I is the identity matrix
• Nov 27th 2008, 03:26 AM
vincisonfire
Yes you're right because we have to add the identity matrix at the beginning. SOrry.
$\displaystyle B=e^A$
then $\displaystyle B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$\displaystyle B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0$
$\displaystyle B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0$
Somebody knows how to calculate these sum by hand?
• Nov 27th 2008, 06:49 AM
awkward
Quote:

Originally Posted by victor1487
I need to compute e^A for the matrix
A= 0 ∏
....-∏ 0

where the diagonal are zeros and the other diagonal has pi on top and negative pi on the bottom.
I'm not quite sure where to start.

Thanks

You might consider starting by diagonalizing A--

Notice that $\displaystyle A = P D P^{-1}$

where

$\displaystyle P = \begin{pmatrix}1 &1 \\ i &-i \end{pmatrix}$

$\displaystyle D =\begin{pmatrix}\pi i &0 \\ 0 &-\pi i \end{pmatrix}$

$\displaystyle P^{-1} = \frac{1}{2} \begin{pmatrix}1 &-i \\ 1 &i \end{pmatrix}$

(I'm assuming you know how to go about diagonalizing a matrix; if not, there is an article on Wikipedia:

Diagonalizable matrix - Wikipedia, the free encyclopedia.)