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Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Homework 5 solutions Section 2.5 5.2. There are equivalences { v 1 ,...,v n } is linearly independent dim Span { v 1 ,...,v n } = n Span { v 1 ,...,v n } = V. The last equivalence follows from dim V = n ; the only ndimensional subspace of V is V itself. 5.4. (2 pts) No, it is not possible. We have Span { w 1 ,w 2 ,w 3 } = Span { v 1 + v 2 ,v 2 + v 3 ,v 3 + v 1 } Span { v 1 ,v 2 ,v 3 } and we know { v 1 ,v 2 ,v 3 } is linearly dependent , so it spans at most a 2dimensional space. We conclude dim Span { w 1 ,w 2 ,w 3 } dim Span { v 1 ,v 2 ,v 3 } < 3 so that { w 1 ,w 2 ,w 3 } is also linearly dependent. 5.5. (1 pt check) Via the isomorphism V R 3 sending the basis { u,v,w } to the standard basis of R 3 , the problem consists of showing that the vectors 1 1 1 , 1 1 , 1 form a basis of R 3 . They do, because 0 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 0 1 has a pivot in every row and column. 5.6. (1 pt check) We take advantage of the trailing zeros by writing the vectors in the order v 2 ,v 3 ,v 1 . Consider the matrix v 2 v 3 v 1 = 3 1 2 2 1 1 50 1 921 5 3 ....
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This note was uploaded on 11/07/2011 for the course MATH 416 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Math, Linear Algebra, Algebra

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