Lecture Notes in Differential Equations - Bruce E. Shapiro

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values, of the matrix. We find these from the determinant of A − λI,
0 =
∣ a − λ b
c d − λ∣ (34.39)
= (a − λ)(d − λ) − bc (34.40)
= λ 2 − (a + d)λ + ad − bc (34.41)
Writing
T = trace(A) = a + d (34.42)
∆ = det(A) = ad − bc (34.43)
the characteristic equation becomes
0 = λ 2 − T λ + ∆ (34.44)
There are two roots,
λ 1 = 1 (
T + √ )
T
2
2 − 4∆
λ 2 = 1 (
T − √ )
T
2
2 − 4∆
(34.45)
(34.46)
(34.47)
If there are two linearly independent eigenvectors v and w, then the general
solution of the linear system is
( x
= Ave
y)
λ1t + Bwe λ2t (34.48)
This result holds even if the eigenvalues are a complex conjugate pair, or if
there is a degenerate eigenvalue with multiplicity 2, so long as there are a
pair of linearly independent eigenvectors.
If the eigenvalue is repeated but has only one eigenvector, v then
( x
= [Av + B(tv + w)] e
y)
λt (34.49)
where w is the generalized eigenvector. In the following paragraphs we will
study the implications of equations (34.48) to (34.49) for various values of
the eigenvalues, as determined by the values of the trace and determinant.