Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 4.1 Chapter 4 4.1 1. Not a subspace since it does not contain the neutral element, that is, the function f ( t ) = 0, for all t . 3. This subset V is a subspace of P 2 : The neutral element f ( t ) = 0 (for all t ) is in V since f 0 (1) = f (2) = 0. If f and g are in V (so that f 0 (1) = f (2) and g 0 (1) = g (2)), then ( f + g ) 0 (1) = ( f 0 + g 0 )(1) = f 0 (1) + g 0 (1) = f (2) + g (2) = ( f + g )(2), so that f + g is in V . If f is in V (so that f 0 (1) = f (2)) and k is any constant, then ( kf ) 0 (1) = ( kf 0 )(1) = kf 0 (1) = kf (2) = ( kf )(2), so that kf is in V . If f ( t ) = a + bt + ct 2 then f 0 ( t ) = b + 2 ct , and f is in V if f 0 (1) = b + 2 c = a + 2 b + 4 c = f (2), or a + b + 2 c = 0. The general element of V is of the form f ( t ) = ( - b - 2 c ) + bt + ct 2 = b ( t - 1) + c ( t 2 - 2), so that t - 1 , t 2 - 2 is a basis of V . 5. If p ( t ) = a + bt + ct 2 then p ( - t ) = a - bt + ct 2 and - p ( - t ) = - a + bt - ct 2 . Comparing coefficients we see that p ( t ) = - p ( - t ) for all t if (and only if) a = c = 0. The general element of the subset is of the form p ( t ) = bt . These polynomials form a subspace of P 2 , with basis t . 7. The set V of diagonal 3 × 3 matrices is a subspace of R 3 × 3 : a. The zero matrix 0 0 0 0 0 0 0 0 0 is in V , b. If A = a 0 0 0 b 0 0 0 c and B = p 0 0 0 q 0 0 0 r are in V , then so is their sum A + B = a + p 0 0 0 b + q 0 0 0 c + r . 91
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Chapter 4 SSM: Linear Algebra c. If A = a 0 0 0 b 0 0 0 c is in V , then so is kA = ka 0 0 0 kb 0 0 0 kc , for all constants k . 9. Not a subspace; consider multiplication with a negative scalar. I 3 belongs to the set, but - I 3 doesn’t. 11. Not a subspace: I 3 is in rref, but the scalar multiple 2 I 3 isn’t. 13. Not a subspace: (1 , 2 , 4 , 8 , . . . ) and (1 , 1 , 1 , 1 , . . . ) are both geometric sequences, but their sum (2 , 3 , 5 , 9 , . . . ) is not, since the ratios of consecutive terms fail to be equal, for example, 3 2 6 = 5 3 . 15. The set W of all square-summable sequences is a subspace of V : The sequence (0 , 0 , 0 , . . . ) is in W . Suppose ( x n ) and ( y n ) are in W . Note that the inequality ( x n + y n ) 2 2 x 2 n + 2 y 2 n holds for all n , since 2 x 2 n + 2 y 2 n - ( x n + y n ) 2 = x 2 n + y 2 n - 2 x n y n = ( x n - y n ) 2 0. Thus n =1 ( x n + y n ) 2 2 n =1 x 2 n + 2 n =1 y 2 n converges, so that the sequence ( x n + y n ) is in W as well. If ( x n ) is in W ( so that n =1 x 2 n converges ) , then ( kx n ) is in W as well, for any constant k , since n =1 ( kx n ) 2 = k 2 n =1 x 2 n will converge. 17. Let E ij be the n × m matrix with a 1 as its ij th entry, and zeros everywhere else. Any A in R n × m can be written as the sum of all a ij E ij , and the E ij are linearly independent, so that they form a basis of R n × m . Thus dim( R n × m ) = nm . 19. a + bi c + di = a 1 0 + b i 0 + c 0 1 + d 0 i The vectors 1 0 , i 0 , 0 1 , 0 i form a basis of C 2 as a real linear space, so that dim( C 2 ) = 4. 21. Use Summary 4.1.6. The general element of the subspace is a 0 0 d = a 1 0 0 0 + d 0 0 0 1 . Thus 1 0 0 0 , 0 0 0 1 is a basis of the subspace; the dimension is 2. 92
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SSM: Linear Algebra Section 4.1 23. Proceeding as in Exercise 21, we find the basis 1 0 0 0 , 0 0 1 0 , 0 0 0 1 ; the dimension is 3.
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