AL-respuestas

# AL-respuestas - MAT240 Assignment 4 — Partial Solutions...

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Unformatted text preview: MAT240 Assignment 4 — Partial Solutions Arthur Fischer February 17, 2004 Exercise (#23, p.56). Let v 1 , . . . , v k , v be vectors in a finite-dimensional vector space V , and define W 1 = span( { v 1 , . . . , v k } ) , and W 2 = span( { v 1 , . . . , v k , v } ) . (a) Find necessary are sufficient conditions on v such that dim( W 1 ) = dim( W 2 ) . Solution. I claim that dim( W 1 ) = dim( W 2 ) iff v ∈ W 1 . If dim( W 1 ) = dim( W 2 ), then, as W 1 ⊆ W 2 , clearly, by Theorem 1.11 it follows that W 1 = W 2 , and since v ∈ W 2 it follows that v ∈ W 1 . If v ∈ W 1 , then there are scalars a 1 , . . . , a k such that v = a 1 v 1 + . . . + a k v k . For any w ∈ W 2 , by definition of span it follows that there are scalars b 1 , . . . , b k , b such that w = b 1 v 1 + . . . + b k v k + bv . Then w = b 1 v 1 + . . . + b k v k + bv = b 1 v 1 + . . . + b k v k + b ( a 1 v 1 + . . . + a k v k ) = ( ba 1 + b 1 ) v 1 + . . . + ( ba k + b k ) v k , and thus w ∈ W 1 , and so W 2 ⊆ W 1 . It is clear that W 1 ⊆ W 2 , and therefore W 1 = W 2 , and thus dim( W 1 ) = dim( W 2 ). (b) State and prove a relationship involving dim( W 1 ) and dim( W 2 ) in the case that dim( W 1 ) = dim( W 2 ) . Solution. I claim that if dim( W 1 ) = dim( W 2 ), then dim( W 2 ) = dim( W 1 ) + 1. By the Replacement Theorem it follows that there is a basis β ⊆ { v 1 , . . . , v k } of W 1 . Without loss of generality, we may assume that β = { v 1 , . . . , v l } for some l ≤ k . Then, for each i = l + 1 , . . . , k there are scalars a ( i ) 1 , . . . , a ( i ) l such that v i = a ( i ) 1 v 1 + . . . + a ( i ) l v l . I claim that β = β ∪ { v } spans W 2 . For any w ∈ W 2 , by definition there are scalars b 1 , . . . , b k , b such that w = b 1 v 1 + . . . b k v k + bv , and therefore w = b 1 v 1 + . . . + b l v l + b l +1 v l +1 + . . . + b k v k + bv = b 1 v 1 + . . . + b 1 v 1 + b l +1 a ( l +1) 1 v 1 + . . . + a ( l +1) k v k + . . . + b k a ( k ) 1 v 1 + . . . + a ( k ) k v k + bv = b 1 + b l +1 a ( l +1) 1 + . . . + b k a ( k ) 1 v 1 + . . . + b l + b l +1...
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AL-respuestas - MAT240 Assignment 4 — Partial Solutions...

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