Characteristic Polynomial
نویسنده
چکیده
A [ An−1 + p1A n−2 + · · ·+ pn−1 In ] = −pn In . Since A is nonsingular, pn = (−1)n det(A) 6= 0; thus the result follows. Newton’s Identity. Let λ1, λ2, . . . , λn be the roots of the polynomial K(λ) = λ + p1λ n−1 + p2λ n−2 + · · · · · ·+ pn−1λ+ pn. If sk = λ k 1 + λ k 2 + · · ·+ λn, then pk = − 1 k (sk + sk−1 p1 + sk−2 p2 + · · ·+ s2 pk−2p1 + s1 pk−1) . Proof. From K(λ) = (λ − λ1)(λ − λ2) . . . . . . (λ − λn−1)(λ − λn) and the use of logarithmic differentiation, we obtain
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