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Math 2270 Spring 2004 Homework 10 Solutions Section 4.1) 1, 2, 6, 10, 13, 14, 20, 25, 30, 35 (1) General elements of P2 have the form a + bt + ct2 where a, b and c are arbitrary real numbers. The subset V consists of elements of P2 that also satisfy p(0) = a = 2, or have the general form 2 + bt + ct2 . We first check that the zero element (0 + 0 · t + 0 · t2 , or a = b = c = 0) is in the subset V . But, if a = 2, then the zero element is obviously not in V , so V is not a subspace of P2 . (2) General elements of P2 have the form a + bt + ct2 where a, b and c are arbitrary real numbers. The subset V consists of elements of P2 that also satisfy p(2) = a + 2b + 4c = 0, or a = −2b − 4c. This gives the general form (−2b − 4c) + bt + ct2 = b(−2 + t) + c(−4 + t2 ). We first check that the zero element (0 + 0 · t + 0 · t2) is in the subset V . By setting b = c = 0, we see that the zero element is in this subset. Next we must check that the subset is closed under linear combinations. Take p(t) = b1 (−2 + t) + c1 (−4 + t2 ) and q(t) = b2 (−2 + t) + c2 (−4 + t2 ) to be elements of V , and k1 and k2 to be scalars. Then, k1 p(t) + k2 q(t) = k1 b1 (−2 + t) + c1 (−4 + t2 ) + k2 b2 (−2 + t) + c2 (−4 + t2 ) = (k1 b1 + k2 b2 )(−2 + t) + (k1 c1 + k2 c2 )(−4 + t2 ) which is in the set V . Since V contains the zero polynomial and is closed under linear combinations, it must be a subspace of P2 . One possible basis for the subspace V is {−2 + t, −4 + t2 } (6) The 3 x 3 zero matrix is not invertible, and is therefore not in the subset of ℜ3x3 consisting of all invertible matrices. Therefore, this subset is not a subspace of ℜ3x3 . 1 (10) Let V be the set of all 3 x 3 matrices A such that A 2 = ~0. We see that the zero 3 ~ matrix must be in this set since 0~x = 0 for any vector ~x. Let A and B be elements in V , and let k1 and k2 be scalars. Then, 1 1 1 1 1 (k1 A + k2 B) 2 = k1 A 2 + k2 B 2 = k1 A 2 + k2 B 2 = k1~0 + k2~0 = ~0 3 3 3 3 3 so the linear combination of matrices A and B is also in the subset V . Therefore, V is a subspace of ℜ3x3 . 1 Math 2270 Spring 2004 Homework 10 Solutions (13) Consider the following geometric sequences: (1, 1, 1, 1, . . .) where a = 1 and r = 1, and (1, 2, 4, 8, . . .) where a = 1 and r = 2. Then, the sum of these two sequences is (2, 3, 5, 9, . . .). From the general form of these sequences we know that a is the first element in the list, so a = 2. The second element is ar, so ar = 2r = 3 implying that r = 3/2. If we use these values to calculate the third element in the list we expect to see ar 2 = 2(9/4) = 9/2. Instead, we see a value of 5. Therefore, the sum of the given geometric sequences does not produce another geometric sequence. The given subset is not closed under addition and is therefore not a subspace of all infinite sequences of real numbers. (14) Let V be the set of all infinite sequences of real numbers that converge to zero. The zero sequence, (0, 0, 0, 0, . . .), converges to zero so is in the set V . Next take two sequences (x0 , x1 , x2 . . .) and (y0 , y1 , y2 , . . .) from V and two scalars k1 and k2 . Then, k1 (x0 , x1 , x2 , . . .) + k2 (y0, y1 , y2 , . . .) = (k1 x0 , k1 x1 , k1 x2 , . . .) + (k2 y0 , k2 y1 , k2 y2 , . . .) = (k1 x0 + k2 y0 , k1 x1 + k2 y1 , k1 x2 + k2 y2 , . . .) We must look at the limit of this new sequence. lim (k1 xn + k2 yn ) = k1 n→∞ lim xn + k2 n→∞ lim yn = k1 · 0 + k2 · 0 = 0 n→∞ The new sequence also converges to zero so is in the subset V . Therefore, V is closed under linear combinations and is a subspace of the space of all infinite sequences of real numbers. (20) The criteria that a + d = 0 is the same as d = −a, so general matrices are of the form " a b c −a # =a " 1 0 0 −1 # +b " # +c 0 0 1 0 #) 0 1 0 0 " 0 0 1 0 # The space has dimension 3 and one possible basis is: (" 1 0 0 −1 # " , 0 1 0 0 # " , (25) General polynomials in P2 have the form a + bt + ct2 where a, b and c are arbitrary real numbers. If f (1) = 0, then a + b + c = 0, or a = −b − c. This gives the general form (−b − c) + bt + ct2 = b(−1 + t) + c(−1 + t2 ). The dimension of the space is 2 and one possible basis is {−1 + t, −1 + t2 }. 2 Math 2270 Spring 2004 Homework 10 Solutions (30) We want the set of all matrices A that satisfy " 1 2 3 6 #" a b c d # = " a + 2c b + 2d 3a + 6c 3b + 6d # = " 0 0 0 0 # This reduces to a system of four equations and four unknowns: a + 2c = 0 3a + 6c = 3(a + 2c) = 0 b + 2d = 0 3b + 6d = 3(b + 2d) = 0 which is the same as the two equations a = −2c and b = −2d. Therefore, matrices in the set have the general form " −2c −2d c d # =c " −2 0 1 0 # +d " 0 −2 0 1 # The dimension of the space is 2 and a possible basis is (" −2 0 1 0 # " , 0 −2 0 1 #) (35) Yes, the set of all linear transformations from ℜn to ℜm is a subspace of the set of all functions from ℜn to ℜm . Let ~x and ~y be two vectors in ℜn and let k be a scalar. First, we look at the zero function f (~x) = ~0. f (~x + ~y ) = ~0 = ~0 + ~0 = f (~x) + f (~y ) f (k~x) = ~0 = k · ~0 = kf (~x) so the zero function is a linear transformation. Take any two linear transformations from ℜn to ℜm , say T (~x) and G(~x). If T and G are linear, then by definition we know that: T (~x + ~y ) = T (~x) + T (~y ) T (k~x) = kT (~x) G(~x + ~y ) = G(~x) + G(~y ) G(k~x) = kG(~x) We must show that the linear transformations (T + G) and kT are also linear. First, we look at (T + G): (T + G)(~x + ~y ) = T (~x + ~y ) + G(~x + ~y ) = T (~x) + T (~y ) + G(~x) + G(~y ) = (T + G)(~x) + (T + G)(~y ) (T + G)(k~x) = T (k~x) + G(k~x) = kT (~x) + kG(~x) = k ((T + G)(~x)) 3 Math 2270 Spring 2004 Homework 10 Solutions Let c be another scalar and we look at the function kT : (kT )(~x + ~y ) = k (T (~x + ~y )) = k (T (~x) + T (~y )) = kT (~x) + kT (~y ) = (kT )(~x) + (kT )(~y ) (kT )(c~x) = k (T (c~x)) = k (cT (~x)) = c(kT )(~x) Therefore, the set of linear transformations is closed under linear combinations and is therefore a subspace of all functions. Since we are only looking at linear transformations from ℜn to ℜm this proof could also be done by defining matrices A and B corresponding to the transformations T and G. 4