**Unformatted text preview: **y algebra states that if the coeﬃcients αi in
λn + αn−1 λn−1 + · · · + α1 λ + α0 = 0 T
H are integers, then every integer solution is a factor of α0 . For our problem, this
means that if there exist integer eigenvalues, then they must be contained in the
set S = {±1, ±2, ±3, ±6, ±9, ±18}. Evaluating p(λ) for each λ ∈ S reveals
that p(3) = 0 and p(−2) = 0, so λ = 3 and λ = −2 are eigenvalues for A.
To determine the other eigenvalue, deﬂate the problem by dividing IG
R λ3 − 4λ2 − 3λ + 18
= λ2 − λ − 6 = (λ − 3)(λ + 2).
λ−3
Thus the characteristic equation can be written in factored form as Y
P (λ − 3)2 (λ + 2) = 0, so the spectrum of A is σ (A) = {3, −2} in which λ = 3 is repeated—we say
that the algebraic multiplicity of λ = 3 is two. The eigenspaces are obtained
as follows.
For λ = 3,
⎧⎛
⎛
⎞
⎞⎫
1 0 1/2
⎨ −1 ⎬
A − 3I −→ ⎝ 0 1 0 ⎠ =⇒ N (A − 3I) = span ⎝ 0 ⎠ .
⎩
⎭
00 0
2 O
C For λ = −2, ⎛ 1
A + 2I −→ ⎝ 0
0 0
1
0 ⎧⎛
⎞
⎞⎫
1
⎨ −1 ⎬
−2 ⎠ =⇒ N (A + 2I) = span ⎝ 2 ⎠ .
⎩
⎭
0
1 Notice that although the algebraic multiplicity of λ = 3 is two, the dimension of the associated eigenspace is only one—we say that A is deﬁcient in
eigenvectors. As we will see later, deﬁcient matrices pose signiﬁcant diﬃculties. Copyright c 2000 SIAM Buy online from SIAM
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7.1 Elementary Properties of Eigensystems
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Please report violations to [email protected] Example 7.1.3 Continuity of Eigenvalues. A classical result (requiring complex analysis)
states that the roots of a polynomial vary continuously with the coeﬃcients. Since
the coeﬃcients of the characteristic polynomial p(λ) of A can be expressed
in terms of sums of principal minors, it follows that the coeﬃcients of p(λ)
vary continuously with the entr...

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