1. 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

2. $e^A$ is defined as
$*\sum_{i=0}^{\infty} \frac{A^n}{n!}$
You can find that
if $B=e^A$
then $B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0$
$B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0$
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.

3. $B(1,1) = B(2,2)$ should be $-1$

4. I don't think so because it is
$\sum_{i=1}^{\infty} (-1)^{n}\cdot\frac{\pi^{2n}}{(2n)!}$ 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

5. well I typed e^A in my ti-89 and it gave me back $-I$ where I is the identity matrix

6. Yes you're right because we have to add the identity matrix at the beginning. SOrry.
$B=e^A$
then $B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0$
$B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0$
Somebody knows how to calculate these sum by hand?

7. 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 $A = P D P^{-1}$

where

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

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

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

(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.)