xMath121PS3

# xMath121PS3 - Math 121 Homework Week 3 Michael Von Kor 1.6...

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Math 121 Homework, Week 3 Michael Von Korﬀ October 21, 2004 1.6: 31,33; 2.1: 9,14,15,19,35; 2.2: 3,5,13 Problem 1. 1.6 31. Let W 1 and W 2 be subspaces of a vector space V having dimensions m and n, respectively, where m n . (a) Prove that dim ( W 1 W 2 ) n. (b) Prove that dim ( W 1 + W 2 ) m + n . Proof. (a) v W 1 W 2 , by deﬁnition v W 2 . So W 1 W 2 W 2 . Thus, by theorem 1.11 (which states that the dimension of a subspace is less than the dimension of the larger space), dim ( W 1 W 2 ) n. (b) Let S 1 = { u 1 ,u 2 ,...,u m } be a basis for W 1 . Let S 2 = { v 1 ,v 2 ,...,v n } be a basis for W 2 . Then x W 1 + W 2 , there exist w 1 W 1 and w 2 W 2 such that w 1 + w 2 = x . Moreover, since S 1 spans W 1 and S 2 spans W 2 , we can write w 1 and w 2 as linear combinations of vectors in S 1 and S 2 , w 1 = a 1 u 1 + ... + a m u m and w 2 = b 1 v 1 + ...b n v n . So x = w 1 + w 2 = a 1 u 1 + ... + a m u m + b 1 v 1 + ...b n v n . Thus x span ( S 1 S 2 ). So S 1 S 2 is a set of size at most m+n (it could be smaller if the two sets have some vectors in common) that spans W 1 + W 2 . But any set spanning a ﬁnite-dimensional vector space must have size greater than or equal to the dimension of the space, so dim ( W 1 + W 2 ) m + n . Problem 2. 33. (a) Let W 1 and W 2 be subspaces of a vector space V such that V = W 1 W 2 . If β 1 andβ 2 are bases for W 1 and W 2 respectively, show that β 1 β 2 = and β 1 β 2 is a basis for V. (b) Conversely, let β 1 and β 2 be disjoint bases for subspaces W 1 and W 2 respec- tively. Prove that if β 1 β 2 is a basis for V, then V = W 1 W 2 . Proof. (a) V = W 1 W 2 , so by deﬁnition W 1 W 2 = { 0 } . But β 1 W 1 and β 2 W 2 , so β 1 β 2 W 1 W 2 = { 0 } . The zero vector is linearly dependent with itself and cannot be a basis vector; thus, β 1 β 2 = . Also, we showed in problem 31 that W 1 + W 2 is spanned by the union of its bases. In other words, β 1 β 2 spans V. To show that β 1 β 2 is a basis, all we need to do is show that it is linearly independent. Let β 1 = { u 1 ,u 2 ,...,u m } , β 2 = { v 1 ,v 2 ,...,v n } . Given a 1 ,...,a m ,b 1 ,...,b n such that a 1 u 1 + ... + a m u m + b 1 v 1 + ... + b n v n = 0, we can write a 1 u 1 + ... + a m u m = - b 1 v 1 - ... - b n v n = 0. The left-hand term is a vector in W 1 and 1

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the right-hand term is a vector in W 2 , but W 1 W 2 = { 0 } , so both terms must be zero. Thus a 1 = ... = a m = b 1 = ... = b n = 0. So β 1 β 2 is linearly independent and is thus a basis for V.
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xMath121PS3 - Math 121 Homework Week 3 Michael Von Kor 1.6...

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