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I need to find the exponential of the following matrix:

$$ A= \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 1 & \cdots &1 \\ 0 & 0 & 1 & \cdots &1 \\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0&0&0 & \cdots&1 \end{pmatrix} $$

Where $A$ is a square matrix. I have already tried to find a general form for $A^{k}$, I found that again the diagonal is filled with 1, and for example in the case $n=4$ the super diagonal is filled with k, the next diagonal is the $\frac{k(k+1)}{2}$ and I thought the $A_{14}$ position was going to be something like $\frac{(k+1)(k+2)}{2}$ but it does not work as I expected. I'm trying to do this to find the exponential of a Jordan block and for $A$, I need to get this: $$ \begin{pmatrix} 0&e^{\lambda}& \frac{e^{\lambda}}{2!} & \cdots &\frac{e^{\lambda}}{(n-1)!}\\ 0&0&e^{\lambda}&\cdots &\frac{e^{\lambda}}{(n-2)!} \\ \vdots&\vdots&\vdots& \ddots& \vdots\\ 0&0&0& \cdots &0 \end{pmatrix} $$ Thanks in advance.

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  • $\begingroup$ I'm thinking this is going to be a roll up your sleeves kind of induction. Uniqueness of Taylor expansions tells us that the coefficients of $\lambda^n$ have to look like what we expect in each entry, so it should work out. But it probably won't be pleasant. $\endgroup$ Commented Sep 13, 2019 at 1:16
  • $\begingroup$ That last matrix doesn’t look right. If you’re looking for $e^A$, its diagonal should be filled with $e$, not $0$. $\endgroup$
    – amd
    Commented Sep 14, 2019 at 22:38

1 Answer 1

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The original matrix $A = I + N,$ where $I$ is the identity matrix, and $N$ is the strictly upper diagonal part. Since $IN = NI,$ we find that $$ e^{At} = e^{It} e^{N t} = e^t e^{Nt}. $$ You need to go through the details for small dimension, the point is that $N$ is nilpotent. In dimension $2,$ $N^2 = 0.$ In dimension $3,$ $N^3 = 0.$ So you can find $e^N$ and $e^{Nt},$ worth writing out.

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