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I have to calculate $e^{At}$ of the matrix $A$. We are learned to first compute $A^k$, by just computing $A$ for a few values of $k$, $k=\{0\ldots 4\}$, and then find a repetition. $A$ is defined as follows:

$ A = \begin{bmatrix} -2 & 2& 0 \\ 0 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix} $

Because I couldn't find any repetition I used Wolfram|Alpha which gave me the following, http://goo.gl/JxyIg:

$ \frac{1}{6} \begin{bmatrix} 3(-1)^k2^{k+1} & 2(2-(-1)^k2^{k+1}) & 0 \\ 0 & 6 & 0 \\ 3(-(-2)^k+0^k) & 2(-1+(-2)^k) & 6*0^k \end{bmatrix} $

Then $e^{At}$ is calculated as followed (note that $\sum_{k=0}^{\infty}\frac{0^kt^k}{k!} = e^{0t} = 1$, using that $0^0 = 1$): $ e^{At} = \begin{bmatrix} \frac{1}{6}\sum_{k=0}^{\infty}\frac{3(-1)^k2^{k+1}t^k}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{2(2-(-1)^k2^{k+1})t^k}{k!} & 0 \\ 0 & \frac{1}{6}\sum_{k=0}^{\infty}\frac{6t^k}{k!} & 0 \\ \frac{1}{6}\sum_{k=0}^{\infty}\frac{3(-(-2)^k+0^k)t^k}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{2(-1+(-2)^k)}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{6^k*0^k}{k!} \end{bmatrix} $

Now this matrix should give as a answer

$ \begin{bmatrix} e^{-2t} & e^{2t} & 0 \\ 0 & e^{t} & 0 \\ e^{t} & e^{-t} & 1 \end{bmatrix} $

Now when I compute this answer of $e^{At}$, I get different answers for some elements. Only the elements $A_{11} = e^{-2t}$, $A_{13} = A_{21} = A_{23} = A_{33} = 1$ and $A_{22} = e^t$. However when I calculate $A_{12}$ I get the following:

$ A_{12}=\frac{1}{6}\sum_{k=0}^{\infty}\frac{2(2-(-1)^k2^{k+1})t^k}{k!}=\frac{2}{6}\left(\sum_{k=0}^{\infty}\frac{2t^k}{k!}-\sum_{k=0}^{\infty}\frac{(-1)^k2^{k+1}t^k}{k!}\right)=\frac{4}{6}\left(\sum_{k=0}^{\infty}\frac{t^k}{k!}-\sum_{k=0}^{\infty}\frac{(-1)^k2^{k}t^k}{k!}\right)=\frac{4}{6}\left(e^t-e^{-2t}\right) $

Which is of course not equal to $e^{2t}$. Where do I make a mistake? Or does maybe Wolfram|Alpha make a mistake, I know it is correct for $0\ldots 4$.

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    Well to be fair I also thought they had to be 0, but both Wolfram|Alpha as well as Matlab, http://goo.gl/po8AC, quiet stupid of myself to just to take that over.2012-09-21

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Corrected : Your $A^k$ is right but for $e^{At}$ you should just have multiplied every term by $\frac {t^k}{k!}$ before computing the sum. So no $6^k$ in the central term for example.
Further (as explained by Robert Israel) for $k=0$ your $A^k$ expression is still valid with only diagonal $1$ terms (so no $1$ elsewhere).

Last you seem to be supposing that $e^{At}$ will simply be the exponential of each term : this is not true as shown by your $A_{1,2}$ term (i.e. the 'should give...' part is not right). I'll add that, in Mathematica, you must use MatrixExp to compute a matrix exponential and not Exp that returns the exponential of the individual terms! Result :

mma

Hoping this helped more,

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    @Wouter: you are welcome !2012-09-21
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In general, you would be better off by finding a diagonalization of $A$ and exponentiating the diagonal matrix, i.e.

$e^{At} = e^{VDV^Tt} = \sum_k \frac{(VDV^Tt)^k}{k!} = \sum_k V\frac{D^k t^k}{k!} V^T = Ve^{Dt}V^T$,

(since $V$ is orthonormal). Then, $e^D$ simply exponentiates the diagonal of $D$, premultiplying every diagonal entry with $t$, and you have to multiply things out...

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    See http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf2012-09-23