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# Applied Linear Algebra MATH 2210

UCONN

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This 6 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 2210 at University of Connecticut taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/205836/math-2210-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15

43 Linearly Independent Sets Bases Definition A set of vectors V1v2 vp in a vector space Vis said to be linearly independent if the vector equation 01V102V2 CPVP 0 has only the trivial solution 01 0 cp 0 The set V1v2 vp is said to be linearly dependent ifthere exists weights c1 cp not all 0 such that 01V102V2 CPVP The following results from Section 17 are still true for more general vectors spaces IA set containing the zero vector is linearly dependentl A set of two vectors is linearly dependent if and only if one is a multiple of the other A set containing the zero vector is linearly independent EXAMPLE 1 2 0 0 3 2 isalinearly set 3 4 0 0 3 0 EXAMPLE 1 2 3 6 isalinearly set since 3 6 I 3 4 9 11 9 11 1 2 is not a multiple of 3 4 Theorem 4 An indexed set V1V2 vp of two or more vectors with v1 0 is linearly dependent if and only if some vector vj j gt 1 is a linear combination ofthe preceding vectors V1 Vj71 EXAMPLE Let p1 p2 p3 be a set of vectors in P2 where p1t 1 p20 t2 and p3t 4t 2 Is this a linearly dependent set Solution Since p3 p2 p1 p2 p3 is a linearly set p1 A Basis Set LetH be the plane illustrated below Which ofthe following are valid descriptions ofH aH Spanv1v2 bH Spanv1v3 cH Spanv2v3 dH Spanv1v2v3 X3 A basis set is an ef cient spanning set containing no unnecessary vectors In this case we would consider the linearly independent sets V1v2 and V1v3 to both be examples of basis sets or bases plural for basis for H DEFINITION LetH be a subspace ofa vector space V An indexed set of vectors 3 b1bp in Vis a basis forH if i 3 is a linearly independent set and ii H Spanb1bp 1 0 0 EXAMPLE Let e1 0 e2 1 e3 0 Show that e1e2e3 is a basis for 0 0 1 R3 The set e1e2e3 is called a standard basis for R3 Solutions Review the IMT page 129 Let Ae1e2 e3 Ooh i 0 0 1 0 Since A has 3 pivots the columns ofA are linearly 0 1 by the IMT and the columns ofA by IMT Therefore e1e2e3 is a basis for R3 EXAMPLE LetS 1 t 12 tquot Show thatS is a basis for Pquot Solution Any polynomial in Pquot is in span of S To show that S is linearly independent assume colcltcntquot0 Then c0 01 cquot 0 HenceS is a basis for Pquot 0 EXAMPLE Let v1 2 V2 1 V3 0 1 Is v1v2v3 a basis for R3 0 Solution Again letA v1 v2 v3 2 1 0 Using row reduction 0 3 1 0 1 1 2 M 1 72 M 72 U3 U1 and since there are 3 pivots the columns ofA are linearly independent and they span R3 by the IMT Therefore v1v2v3 is a basis for R3 EXAMPLE Explain why each of the following sets is not a basis for R3 1 4 0 1 1 4 a 2 5 1 73 b 2 a 5 3 7 0 7 3 6 Bases for Nul A EXAMPLE FindabasisforNulAwhereA 3 6 6 3 9 12 13 0 3 O Solution Row reduce A 0 x1 ZXZ 13X4 33xs 1 2 0 13 33 0 X36X415C5 0 0 1 76 715 0 x2 x4 and x5 are free x1 729m 713m 7 33xs 72 713 733 x2 x2 1 0 0 X3 6X4 15xs x2 0 x4 6 x5 15 X4 X4 0 1 0 x5 x5 0 0 1 T T T u V W Therefore uvw is a spanning set for NulA In the last section we observed that this set is linearly independent Therefore uvw is a basis for NulA The technique used here always provides a linearly independent set The Spanning Set Theorem A basis can be constructed from a spanning set of vectors by discarding vectors which are linear combinations of preceding vectors in the indexed set EXAMPLE Suppose v1 01 v2 01 and v3 T Solution If X is in Spanv1v2v3 then X C1V102V2C3V3 C1V102V2C3 V1 V2 V1 V2 Therefore Spanv1v2v3 Spanv1v2 THEOREM 5 The Spanning Set Theorem LetS V1 vp be a set in Vand letH Spanv1 vp a If one of the vectors in S say vk is a linear combination ofthe remaining vectors in S then the set formed from S by removing vk still spans H b lfH 0 some subset ofS is a basis for H Bases for Col A EXAMPLE Find a basis for ColA where 1 2 0 4 2 4 71 3 Aa1a2 a3 a4 3 6 2 22 4 8 0 16 Solution Row reduce 1 2 0 4 0 0 1 5 a1 a2 a3 a4 b1 b2 b3 b4 0 0 0 0 0 0 0 0 Notethat b2 b1 and a2 a1 b4 4b1 5b3 and a4 4a1 583 b1 and b3 are not multiples of each other a1 and a3 are not multiples of each other Elementary row operations on a matrix do not affect the linear dependence relations among the columns ofthe matrix Therefore Spana1 a2 a3 a4 Spana1 a3 and a1a3 is a basis for ColA THEOREM 6 The pivot columns of a matrixA form a basis for Col A 1 72 3 EXAMPLE Let v1 2 V2 74 v3 6 Find a basis for Spanv1v2v3 73 6 9 1 72 3 Solution LetA 2 74 6 and note that ColA Spanv1v2v3 73 6 9 1 72 0 By row reductionA 0 0 1 Therefore a basis 0 0 for SpanV1V2V3 is Review 1 To nd a basis for Nul A use elementary row operations to transform A 0 to an equivalent reduced row echelon form B 0 Use the reduced row echelon form to nd parametric form of the general solution to Ax 0 The vectors found in this parametric form of the general solution form a basis for Nul A 2 A basis for Col A is formed from the pivot columns ofA Warning Use the pivot columns of A not the pivot columns of B where B is in reduced echelon form and is row equivalent to A

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