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.)