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**Unformatted text preview: **of degree n with real or complex coeﬃcients has n roots, but some
roots may be complex numbers (even if all the coeﬃcients are real), and some
roots may be repeated. Consequently, A has n eigenvalues, but some may be
complex, and some may be repeated. The fact that complex eigenvalues of real
matrices must occur in conjugate pairs is a consequence of the fact that the roots
of a polynomial with real coeﬃcients occur in conjugate pairs. Copyright c 2000 SIAM Buy online from SIAM
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7.1 Elementary Properties of Eigensystems
http://www.amazon.com/exec/obidos/ASIN/0898714540 493 Example 7.1.1 Problem: Determine the eigenvalues and eigenvectors of A = −1
1 1
1 . Solution: The characteristic polynomial is It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] det (A − λI) = 1−λ
1 −1
= (1 − λ)2 + 1 = λ2 − 2λ + 2,
1−λ so the characteristic equation is λ2 − 2λ + 2 = 0. Application of the quadratic
formula yields
√
√
2 ± −4
2 ± 2 −1
λ=
=
= 1 ± i,
2
2
so the spectrum of A is σ (A) = {1 + i, 1 − i}. Notice that the eigenvalues are
complex conjugates of each other—as they must be because complex eigenvalues
of real matrices must occur in conjugate pairs. Now ﬁnd the eigenspaces.
For λ = 1 + i, D
E A − λI = −i −1
1 −i −→ 1
0 i −1
1
i −→ 1
0 −i
0 IG
R For λ = 1 − i,
A − λI = T
H =⇒ N (A − λI) = span i
0 =⇒ N (A − λI) = span i
1 . −i
1 . Y
P In other words, the eigenvectors associated with λ1 = 1 + i are all nonzero
T
multiples of x1 = ( i 1 ) , and the eigenvectors associated with λ2 = 1 − i
T
are all nonzero multiples of x2 = ( −i 1 ) . In previous sections, you could
be successful by thinking only in terms of real numbers and by dancing around
those statements and issues involving complex numbers. But this example makes
it clear that avoiding complex numbers, even when dealing with real matrices,
is no longer possible—very innocent looking matrices, such as the one in this
example, can...

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