Real Distinct and 1 Double Eigenvalue with 1 Eigenvector

From the real eigenvalue/vector pair, l₁ and ?h₁, we find one solution,

e^l1t ?h₁

By our work in the 2x2 systems we know that from the double eigen-value l_2, along with single eigenvector, ?h₂, we find the following two solutions,

e^l2t ?h₂                                    t e^l2t ?x + e^l2t ?r

Here ?x and ?r should satisfy the following equations,

(A -l₂ I)  ?x = 0?                                                          (A -l₂ I)  ?r =  ?x

Remember that the first equation just tells us that ?x should be the single eigenvector for this eigenvalue, ?r, and we generally just say that the second solution we find from the double root case is,

t e^l2t ?h₂ + e^l2t ?r        here  ?r satisfies (A -l₂ I)  ?r =  ?h₂

1 Real Distinct and 1 Double Eigenvalue with 2 Linearly Independent Eigenvectors

We did not look at this case back while we were examining the 2x2 systems although this is easy adequate to deal along with. In this case we will have a particular real distinct eigenvalue/vector pair, l₁ and ?h_1, and also a double eigen-value l₂ and the double eigen-value has two linearly independent eigenvectors, ?h₂ and ?h₃.

Under this case all three eigenvectors are linearly independent and therefore we find the three linearly independent solutions as given,

e^l1t ?h_1,               e^l2t ?h_2,                       e^l2t ?h₃

We are here out of the cases which compare to those which we did with 2x2 systems and we now require to move in the brand new case which we pick up for 3x3 systems. That new case includes eigenvalues with multiplicity of 3. Since we noted above we can have 1, 2, or 3 linearly independent eigenvectors and thus we in fact have 3 sub cases to deal with here.  Therefore, let's going through these last 3 cases for a 3 x 3 system