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Unformatted text preview: hat is, IG R λ i1 · · · λ ik . 1≤i1 <···<ik ≤n sk = For example, when n = 4, s1 = λ1 + λ2 + λ3 + λ4 , Y P s2 = λ1 λ2 + λ1 λ3 + λ1 λ4 + λ2 λ3 + λ2 λ4 + λ3 λ4 , s3 = λ1 λ2 λ3 + λ1 λ2 λ4 + λ1 λ3 λ4 + λ2 λ3 λ4 , s4 = λ1 λ2 λ3 λ4 . O C The connection between symmetric functions, principal minors, and the coeﬃcients in the characteristic polynomial is given in the following theorem. Coefﬁcients in the Characteristic Equation If λn + c1 λn−1 + c2 λn−2 + · · · + cn−1 λ + cn = 0 is the characteristic equation for An×n , and if sk is the k th symmetric function of the eigenvalues λ1 , λ2 , . . . , λn of A, then • • ck = (−1)k (all k × k principal minors), sk = (all k × k principal minors), • • trace (A) = λ1 + λ2 + · · · + λn = −c1 , det (A) = λ1 λ2 · · · λn = (−1)n cn . Copyright c 2000 SIAM (7.1.5) (7.1.6) (7.1.7) (7.1.8) Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.1 Elementary Properties of Eigensystems http://www.amazon.com/exec/obidos/ASIN/0898714540 495 Proof. At least two proofs of (7.1.5) are possible, and although they are conceptually straightforward, each is somewhat tedious. One approach is to successively use the result of Exercise 6.1.14 to expand det (A − λI). Another proof rests on the observation that if p(λ) = det(A − λI) = (−1)n λn + a1 λn−1 + a2 λn−2 + · · · + an−1 λ + an is the characteristic polynomial for A, then the characteristic equation is It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] λn + c1 λn−1 + c2 λn−2 + · · · + cn−1 λ + cn = 0, where ci = (−1)n ai . Taking the rth derivative of p(λ) yields p(r) (0) = r!an−r , and hence cn−r = D E (−1)n (r) p (0). r! (7.1.9) It’s now a matter of repeatedly applying the formula (6.1.19) for diﬀerentiating a determinant to p(λ) = det (A − λI). After r applications of (6.1.19), p(r) (λ) = T...
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## This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

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