linear dependence independence

# Linear dependence - Linear Algebra Lesson 9 E.P Bautista December 6 2007 E.P Bautista Linear Algebra December 6 2007 1 26 Linear Combination

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Unformatted text preview: Linear Algebra Lesson 9 E.P. Bautista December 6, 2007 E.P. Bautista () Linear Algebra December 6, 2007 1 / 26 Linear Combination Definition Let V be a vector space and let S = { v 1 , v 2 , . . . v n } ⊂ V . A vector v is linear combination of the elements of S if v = a 1 v 1 + a 2 v 2 + ··· + a n v n for some real numbers a i where i = 1 , . . . , n . E.P. Bautista () Linear Algebra December 6, 2007 2 / 26 Span Theorem Let V be a vector space and let S = { v 1 , v 2 , . . . v n } ⊂ V . The span of S , denoted by [ S ] and defined by [ S ] = { a 1 v 1 + a 2 v 2 + ··· + a n v n | a i ∈ R } is a subspace of V . E.P. Bautista () Linear Algebra December 6, 2007 3 / 26 Span Proof: Let u , v ∈ [ S ], then there exists real numbers a i and b i where i = 1 , . . . n such that u = a 1 v 1 + a 2 v 2 + ··· + a n v n and v = b 1 v 1 + b 2 v 2 + ··· + b n v n . Then u + v = ( a 1 + b 1 ) v 1 + ( a 2 + b 2 ) v 2 + ··· + ( a n + b n ) v n ∈ [ S ] . Next we let c ∈ R and u ∈ V . Then, cu = ca 1 v 1 + ca 2 v 2 + ··· ca n v n ∈ [ S ] . We may then conclude that [ S ] ≤ V . E.P. Bautista () Linear Algebra December 6, 2007 4 / 26 Remarks If S = { v 1 , . . . v n } and W = [ S ], then we say that S spans W or that the vectors { v 1 , . . . , v n } span W . E.P. Bautista () Linear Algebra December 6, 2007 5 / 26 Remarks We are interested in spanning sets because is a set S spans a vector space V then we know that every vector in V is a linear combination of vectors in S . Since S is a much smaller set that V , we may then use S to represent or even generate the vectors in V . E.P. Bautista () Linear Algebra December 6, 2007 6 / 26 Example Let V = R 3 . Observe that if we consider S = 1 , 1 , 1 then it is easy to see that S spans V because every vector a b c ∈ V may be written as a b c = a 1 + b 1 + c 1 . E.P. Bautista () Linear Algebra December 6, 2007 7 / 26 Example Example Does the set of vectors S = { 1 2 ,- 1 1 } span R 2 ? Solution: If S spans R 2 then every vector [ a , b ] ∈ R 2 may be expressed as a linear combination of the vectors in S . That is, there must exist x , y ∈ R such that [ a , b ] = x 1 2 + y- 1 1 ....
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## This note was uploaded on 09/30/2011 for the course MATH 221 at The University of British Columbia.

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Linear dependence - Linear Algebra Lesson 9 E.P Bautista December 6 2007 E.P Bautista Linear Algebra December 6 2007 1 26 Linear Combination

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