Matrix exponential of an upper triangular matrix Let $a,b,c,d$ be real and nonzero. I am trying to find $e^{At}$ where
$$A = \begin{bmatrix}
a & b & c & d & e\\ 
 0& a & b& c &d \\ 
 0& 0 &a  &b&c \\ 
 0& 0 & 0 &a  &b \\ 
 0& 0& 0 & 0 & a
\end{bmatrix}.$$
I don't think diagonalizing would be a good approach here. I think there is some way I can split this up into different matrices added together and then go from there, but I am not sure what to do. Everything I try seems to be computationally heavy, but for some reason I feel like there is a simple way to do this.
Does anyone know how?

You can deal with this using two tricks. Firstly, if $AB = BA$ then $e^{A+B} = e^Ae^B$. Secondly, use the definition of exponential as
$$
e^A = \sum_{i=0}^\infty \frac{A^n}{n!}.
$$
Write out your matrix as
$$
\begin{bmatrix}
a & b & c & d & e\\ 
 0& a & b& c &d \\ 
 0& 0 &a  &b&c \\ 
 0& 0 & 0 &a  &b \\ 
 0& 0& 0 & 0 & a
\end{bmatrix} = \begin{bmatrix}
a & 0 & 0 & 0 & 0\\ 
 0& a & 0& 0 &0 \\ 
 0& 0 &a  &0&0 \\ 
 0& 0 & 0 &a  &0 \\ 
 0& 0& 0 & 0 & a
\end{bmatrix} + 
\begin{bmatrix}
0 & b & c & d & e\\ 
 0& 0 & b& c &d \\ 
 0& 0 &0  &b&c \\ 
 0& 0 & 0 &0  &b \\ 
 0& 0& 0 & 0 & 0
\end{bmatrix}
$$
and apply the above. It will be a bit computational heavy but you will see a nice pattern when making the powers of the part with zero diagonal. (Also note, that the condition of $AB = BA$  is crucial as the general case given by the Hamilton-Cayley Theorem is more tricky.)
