I am asked to find $e^{At}$, where
$A = \begin{bmatrix} 1 & -1 & 1 & 0\\ 1 & 1 & 0 & 1\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$.
So let me just find $e^{A}$ for now and I can generalize later.
I notice right away that I can write
$A = \begin{bmatrix} B & I_{2}\\ 0_{22} & B\\ \end{bmatrix}$, where
$B = \begin{bmatrix} 1 & -1\\ 1 & 1\\ \end{bmatrix}$.
I'm sort of making up a method here and I hope it works. Can someone tell me if this is correct?
I write:
$A = \mathrm{diag}(B,B) + \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$
Call $S = \mathrm{diag}(B,B)$, and $N = \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$
I note that $N^2$ is $0_{44}$, so $e^{N} = \frac{N^{0}}{0!} + \frac{N}{1!} + \frac{N^2}{2!} + \cdots = I_{4} + N + 0_{44} + ... = I_{4} + N$
and that $e^{S} = \mathrm{diag}(e^{B}, e^{B})$ and compute:
$e^{A} = e^{S + N} = e^{S}e^{N} = \mathrm{diag}(e^{B}, e^{B})\cdot[I_{4} + N]$
This reduces the problem to finding $e^B$, which is much easier.
Is my logic correct? I just started writing everything as a block matrix and proceeded as if nothing about the process of finding the exponential of a matrix would change. But I don't really know the theory behind this I'm just guessing how it would work.