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96) will have n roots; that is, A will have n eigenvalues l1 , l2 , . . , ln . The l’s will not necessarily all be distinct, or all nonzero, or even all real. ) After finding l1 , l2 , . . 96), the accompanying eigenvectors x1 , x2 , . . 95). If an eigenvalue is 0, the corresponding eigenvector is not 0. To see this, note that if l ¼ 0, then (A À lI)x ¼ 0 becomes Ax ¼ 0, which has solutions for x because A is singular, and the columns are therefore linearly dependent. 95) by a scalar k, we obtain k(A À lI)x ¼ k0 ¼ 0, which can be rewritten as (A À lI)kx ¼ 0 [by (2:12)]: Thus if x is an eigenvector of A, kx is also an eigenvector.

7a. Consider the system of equations x1 þ 2x2 ¼ 4 x1 À x2 ¼ 1 x1 þ x2 ¼ 3 or 0 1 @1 1 0 1 1 4 2   x À1 A 1 ¼ @ 1 A: x2 3 1 The augmented matrix is 0 1 (A, c) ¼ @ 1 1 1 2 4 À1 1 A, 1 3 which has rank ¼ 2 because the third column is equal to twice the first column plus the second: 0 1 0 1 0 1 1 2 4 2@ 1 A þ @ À1 A ¼ @ 1 A: 1 1 3 Since rank(A) ¼ rank(A, c) ¼ 2, there is at least one solution. If we add twice the first equation to the second, the result is a multiple of the third equation. Thus the third equation is redundant, and the first two can readily be solved to obtain the unique solution x ¼ (2, 1)0 .

6b. Let P be a nonsingular matrix. (i) If A is positive definite, then P0AP is positive definite. (ii) If A is positive semidefinite, then P0 AP is positive semidefinite. PROOF (i) To show that y0 P0 APy . 0 for y = 0, note that y0 (P0 AP)y ¼ (Py)0 A(Py). Since A is positive definite, (Py)0 A(Py) . 0 provided that Py = 0. 47), Py ¼ 0 only if y ¼ 0, since PÀ1 Py ¼ PÀ1 0 ¼ 0. Thus y0 P0 APy . 0 if y = 0. 36. A Corollary 1. Let A be a p  p positive definite matrix and let B be a k  p matrix of rank k p.