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

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- Nov 26th 2008, 03:37 PMvictor1487Computing 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 PMvincisonfire
$\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 PMchiph588@
$\displaystyle B(1,1) = B(2,2) $ should be $\displaystyle -1 $

- Nov 26th 2008, 05:06 PMvincisonfire
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 PMchiph588@
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 AMvincisonfire
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 AMawkward
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.)