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Unformatted text preview: rgument shows that it’s also impossible to have p < q, so p = q. Thus it is proved that if A ∼ B, then A and B have the same inertia. Conversely, if A and B have in= ertia (p, j, s), then the argument that produced (7.6.15) yields A ∼ E ∼ B. == 77 The fact that inertia is invariant under congruence is also a corollary of a deeper theorem stating that the eigenvalues of A vary continuously with the entries. The argument is as follows. Assume A is nonsingular (otherwise consider A + I for small ), and set X(t) = tQ + (1 − t)QR for t ∈ [0, 1], where C = QR is the QR factorization. Both X(t) and Y (t) = XT (t)AX(t) are nonsingular on [0, 1], so continuity of eigenvalues insures that no eigenvalue Y (t) can cross the origin as t goes from 0 to 1. Hence Y (0) = CT AC has the same number of positive (and negative) eigenvalues as Y (1) = QT AQ, which is similar to A. Thus CT AC and A have the same inertia. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 570 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Example 7.6.5 Taylor’s theorem in n says that if f is a smooth real-valued function defined on n , and if x0 ∈ n×1 , then the value of f at x ∈ n×1 is given by f (x) = f (x0 ) + (x − x0 )T g(x0 ) + (x − x0 )T H(x0 )(x − x0 ) 3 + O( x − x0 ), 2 where g(x0 ) = ∇f (x0 ) (the gradient of f evaluated at x0 ) has components gi = ∂f /∂xi , and where H(x0 ) is the Hessian matrix whose entries are It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu x0 given by hij = ∂ 2 f /∂xi ∂xj x0 . Just as in the case of one variable, the vector D E x0 is called a critical point when g(x0 ) = 0. If x0 is a critical point, then Taylor’s theorem shows that (x − x0 )T H(x0 )(x − x0 ) governs the behavior of f at points x near to x0 . This observation yields the following conclusions regarding local maxima or minima....
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