Triple Eigenvalue with 1 Eigenvector
The eigenvalue/vector pair in this case are l and ?h_{,}. Since the eigenvalue is real we know as the first solution we require is,
e^{l}^{t} ?h_{,}
We can utilize the work from the twice eigen-value with one eigenvector to find that a second solution is,
t e^{l}^{t} ?h + e^{l}^{t }?r here ?r satisfies (A -lI) ?r = ?h
For a third solution we can obtain a clue from how we dealt along with n^{th} order differential equations with roots multiplicity 3. In such cases we multiplied the original solution with a t 2. Though, just as with the double eigenvalue case which won't be adequate to find us a solution. Under this case the third solution will be,
t^{2} e^{l}^{t }?x +_{ }t_{ }e^{l}^{t }?r + e^{l}^{t }u?
Here^{ }?x^{ }?r and u? should satisfies,
(A -l I) ?x = 0? (A -l I) ?r = ?x (A -l I) ?u = ?r
You can verify that it is a solution and the conditions through taking a derivative and plugging in the system.
Here, the first condition simply tells us about the ?x = ?h since we only contain a single eigenvector here and therefore we can reduce that third solution to,
(1/2)t^{2} e^{l}^{t } ?h +_{ }t_{ }e^{l}^{t }?r + e^{l}^{t }u?
Here ?r, and ?u should satisfies,
(A -l I) ?r = ?h (A -l I) ?u = ?r
And at last notice that we would have solved the original first condition in finding the second solution above and therefore all we truly need to do here is solve the last condition.
As a last note in this case, the ½ is in the solution solely to maintain any extra constants from appearing in the conditions that in turn permits us to reuse previous results.