Calculating the exponential of a $4 \times 4$ matrix 
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
$$e^{N} = \frac{N^{0}}{0!} + \frac{N}{1!} + \frac{N^2}{2!} + \cdots = I_{4} + N + 0_{44} + \cdots = I_{4} + N$$
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.

A different, but rather specific, strategy would be to use the ring homomorphism
$${a+bi\in\mathbb C \mapsto \pmatrix{a&-b \\ b&a}\in\mathbb R^{2\times 2}}$$in the block decomposition. Then your problem is equivalent to finding
$$e^{t\pmatrix{1+i & 1\\ 0 & 1+i}}=e^{\pmatrix{t+ti & t\\ 0 & t+ti}}=e^{t+ti}e^{\pmatrix{0&t\\0&0}}=(e^{t+ti})\pmatrix{1&t\\0&1}$$
which unfolds to
$$\pmatrix{e^t\cos t & -e^t\sin t & t e^t \cos t & -t e^t \sin t \\ e^t \sin t & e^t \cos t & t e^t \sin t & t e^t \cos t \\ 0 & 0 & e^t\cos t & -e^t\sin t \\ 0&0& e^t\sin t & e^t\cos t }$$
