*1 Triple Eigenvalue with 2 Linearly Independent Eigenvectors *

In this case we will have the eigen-value l with the two linearly independent eigenvectors ?h_{1} and ?h_{2} therefore we find the following two linearly independent solutions,

e^{l}^{t } ?h_{1} e^{l}^{t } ?h_{2}

We now need a third solution. The third solution will be in the form,

t e^{l2}^{t} ?x + e^{l2}^{t }?r

Here ?x and^{ }?r should satisfies the following equations,

(A -l I) ?x = 0? (A -l I) ?r = ?x

We have already verified that it will be a solution with such conditions in the double eigen-value case (which work only needed a repeated eigenvalue, not essentially a double one).

Though, unlike the previous times we've seen that we can't just say as ?x is an eigenvalue. In all the earlier cases in which we've noticed this condition, we had a particular eigenvalue and now we have two linearly independent eigen-values. It means that the most common possible solution to the initial condition is,

?x = c_{1} ?h_{1}+ c_{2} ?h_{2}

This makes problems in solving the second condition. The second conditions will not contain solutions for all choices of c_{1} and c_{2} and the choice which we use will be dependent on the eigenvectors. Therefore upon solving the first condition we would require plugging the general solution in the second condition and so proceed to find out conditions on c_{1} and c_{2} which would permit us to solve the second condition.