Lecture 20 - Feb 25

Lecture 20 - Feb 25 - Wednesday February 25 Lecture 20 :...

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Wednesday February 25 Lecture 20 : Reducing a spanning family to a basis. (Refers to section 4.3) Expectations: 1. State the Removal Lemma ( The spanning theorem). 2. Recognize every non-trivial vector space with a finite spanning family has a basis. We will consider methods to construct the basis of a vector space. 20.1 Recall If B = { v 1 , v 2 , ,..., v k } is both 1. a spanning family of V , and 2. linearly independent then we say that B is a basis of V . 20.2 Lemma The Removal Lemma . ( Sometimes called the spanning theorem ) Let U = { v 1 , v 2 ,..., v k } and V = span{ v 1 , v 2 ,..., v k } = span U . Suppose the vector v 1 in U = { v 1 , v 2 ,..., v k } is a linear combination of all the other vectors in that set then span{ v 1 , v 2 ,..., v k } = span{ v 2 ,..., v k }, i.e., v 1 can be removed from U without affecting the span of the set. Proof: We are given that v 1 is a linear combination of the other vectors in U . That is v 1 = β 2 v 2 + β 3 v 3 +..... + β k v k . . ± Let w V = span{ v 1 , v 2 ,..., v k }. Then, for some scalars α 1 , α 2 , , . .., α k . w = α 1 v 1 + α 2 v 2 +..... + α k v k = α 1 ( β 2 v 2 + β 3 v 3 +..... + β k v k )+ α 2 v 2 + α 3 v 3 + . ....+ α k v k . = ( α 1 β 2 + α 2 ) v 2 + ( α 1 β 3 + α 3 ) v 3 +..... +( α 1 β k + α k 2 ) v k , a vector in span{ v 2 ,..., v k }. So V is contained in the span{ v 2 ,..., v k }. Hence V = span{ v 2 ,..., v k }.
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Since V = span{ v 1 , v 2 ,..., v k } then span{ v 1 , v 2 ,..., v k } = span{ v 2 ,..., v k }. 20.2.2
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Lecture 20 - Feb 25 - Wednesday February 25 Lecture 20 :...

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