I am solving a ODE system with k = 3 (algebraic multiplicity) and s=2 (geometric multiplicity):
˙X(t)=AX(t), Where A=
[211121−2−2−1]
When I solve det(A−λ)= 0 I get 1 eigenvalue λ =1 with k=3, then I get 2 eigenvectors
[−101] [−110]
Now I am lost, I really dont know what to do next or how to find the general form of X(t).
EDIT:
I found what to do in the case of k=3 and s=1: you solve (A−λ)V1=0, (A−λ)V2=V1 and (A−λ)V3 = V2.
Then X(t)=C1V1eλ1+c2eλ1(V1t+V2)+C3eλ1(V1t22+V2t+V3).
I know that a part of the solution I'm looking is C1V1eλ1+C2V2eλ1, with V1 and V2 the eigenvetors that I already have found.
So I am looking for the last part of the solution.
