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Unformatted text preview: 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 (1) = f (2) = 0. • If f and g are in V (so that f (1) = f (2) and g (1) = g (2)), then ( f + g ) (1) = ( f + g )(1) = f (1) + g (1) = f (2) + g (2) = ( f + g )(2), so that f + g is in V . • If f is in V (so that f (1) = f (2)) and k is any constant, then ( kf ) (1) = ( kf )(1) = kf (1) = kf (2) = ( kf )(2), so that kf is in V . If f ( t ) = a + bt + ct 2 then f ( t ) = b + 2 ct , and f is in V if f (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 is in V , b. If A = a b c and B = p q r are in V , then so is their sum A + B = a + p b + q c + r . 91 Chapter 4 SSM: Linear Algebra c. If A = a b c is in V , then so is kA = ka kb 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 squaresummable sequences is a subspace of V : • The sequence (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 ....
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This homework help was uploaded on 01/23/2008 for the course MATH 253B taught by Professor Bretsch during the Spring '08 term at Colby.
 Spring '08
 BRETSCH
 Linear Algebra, Algebra

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