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Lab 6 Solutions 4.1 a. b. c. d. e. f. Additive inverse Transitive property of equality Add v to both sides Associative property of addition Commutative property and Additive inverse Additive identity 1 of 4 4.2 ( ) ( ) 2 of 4 4.2 #47. Prove that 0v = 0 for any element v of a vector space π. 0v 0v 0v 0v + (ο0v) 0v + (ο0v) 0 0 = = = = = = = 0v (0 + 0)v 0v + 0v (0v + 0v) + (ο0v) 0v + (0v + (ο0v)) 0v + 0 0v Reflexive property of equality Additive identity in β Distributive property Add ο0v to both sides Associative property Additive inverses Additive identity in π 4.2 #48 Prove that (β1)v = βv. Distributive property Additive inverses in R Theorem 4.4(1), i.e. 0v = 0 Scalar identity Add βv to both sides Associative property Additive inverse Commutative property Additive identity (β1)v + (1)v = (β1 + 1)v (β1)v + (1)v = 0v (β1)v + (1)v = 0 (β1)v + v = 0 ((β1)v + v) + (βv) = 0 + (βv) (β1)v + (v + (βv)) = 0 + (βv) (β1)v + 0 = 0 + (βv) (β1)v + 0 = (βv) + 0 (β1)v = βv 4.3 #48 1 To show that ο = { f β C[0, 1] such that β« f (x)dx = 0 } is a subspace of C[0, 1], we need to 0 show that ο is not empty and ο is closed under vector addition and scalar multiplication. 1 ο is not empty because the zero function z(x) = 0 is in ο: β« z(x)dx = 0 0 ο is closed under addition: if f β ο and g β ο then the new function f + g β ο because 1 β« [ f + g](x)dx = 0 1 β« 1 f (x) + g(x)dx = 0 β« 0 1 f (x)dx + β« g(x)dx = 0 + 0 = 0 0 Finally, ο is closed under scalar multiplication: if c is a scalar and f β ο then the new function cf β ο because 1 1 1 0 0 0 β« [cf ](x)dx = β« c f (x)dx = c β« f (x)dx = c0 = 0 3 of 4 4.3 #52 Prove that W is a subspace. 4.3 #58 a) To show that ο + ο a subspace of ο when ο and ο are subspaces of ο, we need to show that ο + ο is not empty and ο + ο is closed under vector addition and scalar multiplication. ο + ο is not empty because 0 β ο and 0 β ο so 0 = 0 + 0 β ο + ο. ο + ο is closed under addition: if u1 β ο + ο and u2 β ο + ο then u1 = v1 + w1 and u2 = v2 + w2 for some v1, v2 β ο and w1, w2 β ο. So u1 + u2 = (v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) β ο + ο because ο and ο are closed under addition. Finally, ο + ο is closed under scalar multiplication: if c is a scalar and u β ο + ο then u1 = v1 + w1 for some v β ο and w β ο. So cu = c(v + w) = cv1 + cv2 β ο + ο because ο and ο are closed under scalar multiplication. b) If ο = {(x, 0): x is a real number} and ο = {(0, y): y is a real number} then ο + ο = {(x, y): x and y are a real numbers} so ο + ο = ο2. 4 of 4