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1 Real and 2 Complex Eigenvalues
From the real eigenvalue/vector pair, l1 and ?h1
el1t ?h1
We find the other two solutions in identical manner which we did with the 2x2 case. If the eigenvalues are l2,3 = a + bI with eigenvectors ?h2 and ?h3 We can find two real-valued solutions through using Euler's formula to expand;
el2t ?h2 = e(a + bi)t ?h2 = eat (cos(bt) + i sin(bt)) ?h2
in its real and imaginary parts, u? + i v? . The final two real valued solutions we need are then,
u? v?
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Real Distinct and 1 Double Eigenvalue with 1 Eigenvector From the real eigenvalue/vector pair, l 1 and ?h 1 , we find one solution, e l1 t ?h 1 By our work in the 2x
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