# Exponential of an upper triangular matrix filled with 1.

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.

• 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. Commented Sep 13, 2019 at 1:16
• That last matrix doesn’t look right. If you’re looking for $e^A$, its diagonal should be filled with $e$, not $0$.
– amd
Commented Sep 14, 2019 at 22:38

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.