C take m1 m2 m and apply the technique used in the

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Unformatted text preview: = 1, 2, . . . , n, or, by using the identity 1 − cos θ = 2 sin2 (θ/2), λij = 4 sin2 iπ 2(n + 1) + sin2 jπ 2(n + 1) , D E i, j = 1, 2, . . . , n. (7.6.8) Since each λij is positive, L must be positive definite. As a corollary, L is nonsingular, and hence Lu = g yields a unique solution for the steady-state temperatures on the square plate (otherwise something would be amiss). T H At first glance it’s tempting to think that statements about positive definite matrices translate to positive semidefinite matrices simply by replacing the word “positive” by “nonnegative,” but this is not always true. When A has zero eigenvalues (i.e., when A is singular) there is no LU factorization, and, unlike the positive definite case, having nonnegative leading principal minors doesn’t insure that A is positive semidefinite—e.g., consider A = 0 −0 . The positive 0 1 definite properties that have semidefinite analogues are listed below. IG R Y P Positive Semidefinite Matrices For real-symmetric matrices such that rank (An×n ) = r, the following statements are equivalent, so any one of them can serve as the definition of a positive semidefinite matrix. • xT Ax ≥ 0 for all x ∈ n×1 (the most common definition). (7.6.9) O C • • All eigenvalues of A are nonnegative. A = BT B for some B with rank (B) = r. • All principal minors of A are nonnegative. For hermitian matrices, replace ( )T by ( )∗ and (7.6.10) (7.6.11) (7.6.12) by C . Proof of (7.6.9) =⇒ (7.6.10). The hypothesis insures xT Ax ≥ 0 for eigenvectors 2 2 of A. If (λ, x) is an eigenpair, then λ = xT Ax/xT x = Bx 2 / x 2 ≥ 0. Proof of (7.6.10) =⇒ (7.6.11). Similar to the positive definite case, if each λi ≥ 0, write A = PD1/2 D1/2 PT = BT B, where B = D1/2 PT has rank r. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.6 Positive Definite Matrices http://www.amazon.com/exec/obidos/ASIN/0898714540 567 Proof of (7.6.11) =⇒ (7.6.12). If Pk is a principal submatrix of A, then Pk = QT AQ = QT BT BQ = FT F| =⇒ Pk = FT F It is ill...
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