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Math416_HW5_sol

# Math416_HW5_sol - Math 416 Abstract Linear Algebra Fall...

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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 n -dimensional 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 2-dimensional 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 , 0 1 1 , 0 0 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 0 50 1 0 - 921 5 0 0 - 3 . 1

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Let us insert columns to make the new columns into a basis of R 5 : e 1 v 2 e 3 v 3 v 1 = 1 3 0 1 2 0 - 2 0 1 - 1 0 0 1 50 1 0 0 0 - 921 5 0 0 0 0 - 3 has a pivot in every row and column, so that { v 1 , v 2 , v 3 , e 1 , e 3 }
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Math416_HW5_sol - Math 416 Abstract Linear Algebra Fall...

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