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I am trying to prepare for my exams and want to calculate some matrix exponents $\exp(A)$ and wanted to do this with $$A = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}.$$

Right now I have $\sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!}$

$$ e^{A} = \begin{bmatrix} 1/e & \sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!} & 1 \\ 1 & 1/e & 1 \\ 1 & 1 & 1/e \end{bmatrix}.$$

I would greatly appreciate any tips or tricks I could use to progress further or even understand were I might have made mistakes.

Further info: I arrived at my question through the following:

$A = (D+N)$

$$ (D+N)^n=\sum_{n=0}^{\infty}{n \choose k} D^{n-k} \cdot N^k = {n \choose 0} D^{n} \cdot N^0 + {n \choose 1} \cdot D^{n-1} \cdot N =D^n+n \cdot D^{n-1}\cdot N$$ Since $N^n = 0$ $\forall n\geq 2$ only the first two sum elements matter.

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klnstngr
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    Presumably each $i=0$ should be $j=0$? If so, please edit. – J.G. Jun 13 '21 at 12:36
  • Remember, $e^x = 1+x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$. – Minus One-Twelfth Jun 13 '21 at 12:39
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    @J.G. Yes, you are correct, sorry! – klnstngr Jun 13 '21 at 12:45
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    For $j=0$ you have $(-1)!$ which is undefined. You may want to check the indices... – jjagmath Jun 13 '21 at 12:47
  • @MinusOne-Twelfth hmm, so I am not quite sure how to progress from that. I did use $e^x=\sum_{j=0}^{\infty} \frac{x^j}{j!}$ to get all the other entries of the matrix exponent. So this is exactly my issue. – klnstngr Jun 13 '21 at 12:47
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    How did you arrive at your expression for $e^A$ actually? $A$ here is a Jordan matrix and its exponential should be upper triangular. There is a formula for exponential of Jordan matrix. See e.g. https://math.stackexchange.com/questions/1451276/matrix-exponential-for-jordan-canonical-form. – Minus One-Twelfth Jun 13 '21 at 12:51
  • I seperated $A = (D+N)$ so that D became the identity matrix times (-1) and N had the singular non zero element at $n_{12} = 1$. From there I went on with $$(D+N)^n=\sum_{k=0}^{\infty} {n \choose k} D^{n-k} \cdot N^k$$ Since $N^k=0 for k \geq 2$ I got $$D^n + n\cdot D^{n-1}\cdot N$$ should I add that to the original question? – klnstngr Jun 13 '21 at 12:59
  • @jjagmath that is a good point. And one of my issues. I will try to write up my whole thought process, because I can't find an error. – klnstngr Jun 13 '21 at 13:03
  • Bear in mind also the natural extension of $1/n!$ is $1/(-1)!=0$. – J.G. Jun 13 '21 at 13:05

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Rewrite the sum as $$\sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!}\overset{k=j-1\ }{\longrightarrow}\sum_{k=-1}^\infty \frac{(-1)^{k}}{k!}=-\frac{1}{(-1)!}+\sum_{k=0}^\infty \frac{(-1)^{k}}{k!}.$$

In order to understand what $\frac{1}{(-1)!}$ means, we have to look at the reciprocal Gamma function. We can express our term like this: $\frac{1}{(-1)!}=\frac{1}{\Gamma(0)}=0.$ If the starting point is actually $j=1$ and not $j=0$ we get the same result, since both outcomes are zero and do not change the value of the series.

Lastly, note that for all $x\in\mathbb R$, $\mathrm e^x = \sum_{n=0}^\infty \frac{x^k}{k!}$, so our final solution is $e^{-1}$.

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