generalized eigenvector to solve system of ODE (matrix exponential) For a 4x4 matrix, A, with initial value problem 
x'=Ax, x(0) = ξ0
where ξ0 ∈ R4 is a vector such that (A − 3I)3
ξ0 = 0 
but ξ1 = (A − 3I)ξ0 $\neq$ 0 and ξ2 = (A − 3I)2ξ0 $\neq$ 0.
Make the change of variable x = e3ty, and derive an equation y'= By (find B). 
Then solve the initial value problem y'=By, y(0) = ξ0 representing y in terms of ξ0, ξ1, and ξ2. [Suggestion: use the matrix exponential.]
I have no idea how to get started but I found the following that looks relevant in my notes:
Let λ be an eigenvalue of A. Some vector η= 0 is a generalized eigenvector corresponding to λ if (A−λI)mη= 0 for some integer m≥1.  The smallest m for which this holds (so that(A−λI)m-1η $\neq$ 0 but (A−λI)mη = 0) is called the rank of η.
Any guidance and links to similar problems where I could understand the approach to solving this question would be greatly appreciated!

We have
$\mathbf x = e^{3t} \mathbf y, \tag 1$
whence
$\mathbf x' = 3e^{3t} \mathbf y + e^{3t}\mathbf y'; \tag 2$
we combine (2) with
$\mathbf x' = A\mathbf x, \tag 3$
and obtain
$3e^{3t} \mathbf y + e^{3t}\mathbf y' = A\mathbf x; \tag 4$
via (1),
$3e^{3t} \mathbf y + e^{3t}\mathbf y' = A e^{3t} \mathbf y = e^{3t} A\mathbf y; \tag 5$
thus,
$3 \mathbf y+ \mathbf y' = A\mathbf y, \tag 6$
or
$\mathbf y' = A\mathbf y - 3\mathbf y = A\mathbf y - 3 I \mathbf y = (A - 3I)\mathbf y; \tag 7$
set
$B = A - 3I; \tag 8$
then
$\mathbf y' = B\mathbf y; \tag 9$
the solution to this equation with
$\mathbf y(0) = \xi_0 \tag{10}$
is
$\mathbf y(t) = e^{Bt} \xi_0 = e^{(A - 3I)t}\xi_0; \tag{11}$
we use the given $(A - 3I)^i\xi_0$, $1 \le i \le 3$, to compute $e^{(A - 3I)t}\xi_0$:
$\mathbf y(t) = \displaystyle \sum_0^\infty \dfrac{(A - 3I)^j t^j}{j!} \xi_0$
$= \xi_0 + t (A - 3I) \xi_0 + \dfrac{1}{2} t^2 (A - 3I)^2 \xi,  \tag{12}$
where the higher-order terms vanish since
$(A - 3I)^3 \xi_0 = 0 \tag{13}$
implies
$(A - 3I)^j \xi_0 = (A - 3I)^{j - 3} (A - 3I)^3 \xi_0 = 0, \; \forall j \ge 3; \tag{14}$
thus in the light of
$(A - 3I)^j \xi_0 = \xi_j, \; j = 1, 2, \tag{15}$
(12) becomes
$\mathbf y(t) = \displaystyle \sum_0^\infty \dfrac{(A - 3I)^j t^j}{j!} \xi_0 = \xi_0 + t \xi_1 + \dfrac{1}{2} t^2 \xi_2, \tag{16}$
the sought-for solution.
