# Amazoncomexecobidosasin0898714540 characteristic

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Unformatted text preview: olutions—more is said in §7.4 on p. 541. Below is a summary of some general statements concerning features of the characteristic polynomial and the characteristic equation. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 492 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Characteristic Polynomial and Equation The characteristic polynomial of An×n is p(λ) = det (A − λI). The degree of p(λ) is n, and the leading term in p(λ) is (−1)n λn . • It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] • The characteristic equation for A is p(λ) = 0. • The eigenvalues of A are the solutions of the characteristic equation or, equivalently, the roots of the characteristic polynomial. • Altogether, A has n eigenvalues, but some may be complex numbers (even if the entries of A are real numbers), and some eigenvalues may be repeated. • If A contains only real numbers, then its complex eigenvalues must occur in conjugate pairs—i.e., if λ ∈ σ (A) , then λ ∈ σ (A) . D E T H Proof. The fact that det (A − λI) is a polynomial of degree n whose leading term is (−1)n λn follows from the deﬁnition of determinant given in (6.1.1). If IG R δij = then 1 0 if i = j, if i = j, Y P det (A − λI) = σ (p)(a1p1 − δ1p1 λ)(a2p2 − δ2p2 λ) · · · (anpn − δnpn λ) p is a polynomial in λ. The highest power of λ is produced by the term O C (a11 − λ)(a22 − λ) · · · (ann − λ), so the degree is n, and the leading term is (−1)n λn . The discussion given earlier contained the proof that the eigenvalues are precisely the solutions of the characteristic equation, but, for the sake of completeness, it’s repeated below: λ ∈ σ (A) ⇐⇒ Ax = λx for some x = 0 ⇐⇒ (A − λI) x = 0 for some x = 0 ⇐⇒ A − λI is singular ⇐⇒ det (A − λI) = 0. The fundamental theorem of algebra is a deep result that insures every polynomial...
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