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**Unformatted text preview: **98714540 491 Let’s now face the problem of ﬁnding the eigenvalues and eigenvectors of
the matrix A = 7 −4 appearing in (7.1.1). As noted in (7.1.3), the eigen5 −2
values are the scalars λ for which det (A − λI) = 0. Expansion of det (A − λI)
produces the second-degree polynomial It is illegal to print, duplicate, or distribute this material
Please report violations to meyer@ncsu.edu p(λ) = det (A − λI) = 7−λ
5 −4
= λ2 − 5λ + 6 = (λ − 2)(λ − 3),
−2 − λ which is called the characteristic polynomial for A. Consequently, the eigenvalues for A are the solutions of the characteristic equation p(λ) = 0 (i.e.,
the roots of the characteristic polynomial), and they are λ = 2 and λ = 3.
The eigenvectors associated with λ = 2 and λ = 3 are simply the nonzero
vectors in the eigenspaces N (A − 2I) and N (A − 3I), respectively. But determining these eigenspaces amounts to nothing more than solving the two homogeneous systems, (A − 2I) x = 0 and (A − 3I) x = 0.
For λ = 2,
5
5 A − 2I = −4
−4 −→ 1
0 For λ = 3,
4
5 −4
−5 T
H =⇒ IG
R =⇒ N (A − 2I) = A − 3I = −4/5
0 D
E −→ Y
P x 1
0 =⇒ N (A − 3I) = x=α −1
0 x x1 = (4/5)x2
x2 is free 4/5
1 =⇒ x=β . x1 = x2
x2 is free
1
1 . In other words, the eigenvectors of A associated with λ = 2 are all nonzero
T
multiples of x = ( 4/5 1 ) , and the eigenvectors associated with λ = 3 are
T
all nonzero multiples of y = ( 1 1 ) . Although there are an inﬁnite number of
eigenvectors associated with each eigenvalue, each eigenspace is one dimensional,
so, for this example, there is only one independent eigenvector associated with
each eigenvalue.
Let’s complete the discussion concerning the system of diﬀerential equations
u = Au in (7.1.1). Coupling (7.1.2) with the eigenpairs (λ1 , x) and (λ2 , y) of
A computed above produces two solutions of u = Au, namely, O
C u1 = eλ1 t x = e2t 4/5
1 and u2 = eλ2 t y = e3t 1
1 . It turns out that all other solutions are linear combinations of these two particular
s...

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