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Math 222 - Selected Homework Solutions for Assignment 6 Instructor - Al Boggess Fall 1998 Section 3.5 11 There is a mistake in this problem. It should read, the transition matrix from F to E rather than the other way around. The transition matrix from F = [v1 ; : : : ; vn ] to E = [u1 ; : : : ; un ] is U ,1 V . On the other hand, the reduced row echelon form of (U jV ) involves executing row operations until U becomes the identity matrix. This is equivalent to multiplying both left and right hand sides by U ,1 , i.e. U ,1 U jU ,1 V = I jU ,1V Thus the transition matrix from F to E is the result (i.e the right hand side) of the reduced row echelon form of (U jV ). Section 3.6 7 We are given that A is m n with m > n and that N (A) = 0. a) Since N (A) = 0, the column vectors of A are linearly independent (Theorem 3.6.2). There are n column vectors and n < m, so the column vectors cannot span Rm . b) If b is not in the column space of A, there will be no solutions to Ax = b. If b is in the column space of A, there will be exactly one solution to Ax = b (again, by Theorem 3.6.2). 8 We are given that A is a 6 5 matrix. If the dimension of N (A) = 2, then the rank of A is 3 (using Theorem 3.6.4 since 3 + 2 = 5). If the rank of B is 4, then the dimension of N (B ) is 1. 1 10 If Ax = b, then the row echelon form of the augmented matrix (Ajb) has the same number of nonzero rows as the row echelon form of A. Otherwise, there would be a row of zeros for the row echelon form for A with something nonzero on the right side which would lead to an inconsistency (no solution). Since the number of nonzero rows equals the rank, we conclude that the rank of A and the rank of Ajb are the same. Conversely, if the rank of Ajb equals the rank of A then their row echelon forms have the same number of rows. In particular, the row echelon form of Ajb does NOT have a row of zeros on the left with something nonzero on the right. Therefore, the system Ax = b can be solved, which means the system Ax = b is consistent. 11 We are given that A is an m n matrix. a) We are to show that if B is a nonsingular m m matrix then BA and A have the same nullspace. If (BA)x = B (Ax) = 0 then Ax lies in the nullspace of B . Since B is nonsingular, its nullspace consists only of f0g. Thus Ax must be zero. This implies x belongs to the nullspace of A. We conclude that N (BA) N (A). Conversely, if Ax = 0, then (BA)(x) = B (Ax) = 0 and so x belongs to N (BA). Therefore N (A) N (BA). In the previous paragraph, we showed N (BA) N (A), so N (BA) = N (A). b) We are given that C is nonsingular. So C t is nonsingular. By part a), the nullity of C t At = (AC )t is the same as the nullity of At . Thus the rank of (AC )t is the same as the rank of At (Theorem 3.6.4). By Theorem 3.6.5, the rank of a matrix equals the rank of its transpose. So we conclude that the rank of A equals the rank of AC . 15 We are given that x0 is a particular solution to Ax = b. a) Suppose y is any solution to Ax = b, we must show that y = x0 + z where z 2 N (A). Consider the dierence z = y , x0 . Apply A to this vector and we obtain A(z ) = A(y , x0 ) = A(y) , A(x0 ) = b , b = 0 Therefore z belongs to N (A) and y = z + x0 . Conversely, if y = x0 + z with z 2 N (A), then Ay = A(x0 ) + A(z ) = b + 0 = b 2 b) If N (A) = 0, then the vector z in part a) must be zero. Therefore y = x0 is the only solution to Ax = b. 3