4.3 Linearly Independent Sets; Bases Definition A set of vectors v - PDF document

4.3 Linearly Independent Sets; Bases Definition A set of vectors  v 1 , v 2 , … , v p  in a vector space V is said to be linearly independent if the vector equation c 1 v 1 + c 2 v 2 + ⋯ + c p v p = 0 has only the trivial solution c 1 = 0, … , c p = 0 . The set  v 1 , v 2 , … , v p  is said to be linearly dependent if there exists weights c 1 , … , c p , not all 0 , such that c 1 v 1 + c 2 v 2 + ⋯ + c p v p = 0 . The following results from Section 1.7 are still true for more general vectors spaces. A set containing the zero vector is linearly dependent. 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. 1

1 2 0 0 3 2 EXAMPLE: is a , , 3 4 0 0 3 0 linearly __________________ set. 1 2 3 6 EXAMPLE: is a linearly , 3 4 9 11 3 6 _________________ set since is not a 9 11 1 2 multiple of . 3 4 Theorem 4 An indexed set  v 1 , v 2 , … , v p  of two or more vectors, with v 1 ≠ 0 , is linearly dependent if and only if some vector v j ( j > 1 ) is a linear combination of the preceding vectors v 1 , … , v j − 1 . EXAMPLE: Let p 1 , p 2 , p 3 be a set of vectors in P 2 where p 1  t  = t , p 2  t  = t 2 , and p 3  t  = 4 t + 2 t 2 . Is this a linearly dependent set? Solution: Since p 3 = _____ p 1 + _____ p 2 , p 1 , p 2 , p 3 is a linearly _______________________ set. 2

A Basis Set Let H be the plane illustrated below. Which of the following are valid descriptions of H ? (a) H = Span  v 1 , v 2  (b) H = Span  v 1 , v 3  (c) H = Span  v 2 , v 3  (d) H = Span  v 1 , v 2 , v 3  x 3 v 3 v 2 v x 1 2 x 1 A basis set is an “efficient” spanning set containing no unnecessary vectors. In this case, we would consider the linearly independent sets  v 1 , v 2  and  v 1 , v 3  to both be examples of basis sets or bases (plural for basis) for H. DEFINITION Let H be a subspace of a vector space V . An indexed set of vectors β =  b 1 , … , b p  in V is a basis for H if (i) β is a linearly independent set, and (ii) H = Span  b 1 , … , b p  . 3

1 0 0 EXAMPLE: Let e 1 = , e 2 = , e 3 = . 0 1 0 0 0 1 Show that  e 1 , e 2 , e 3  is a basis for R 3 . The set  e 1 , e 2 , e 3  is called a standard basis for R 3 . Solutions: (Review the IMT, page 129) Let 1 0 0 . Since A has 3 pivots, A = e 1 e 2 e 3 0 1 0 = 0 0 1 the columns of A are linearly ______________________ by the IMT and the columns of A _____________ _________ by IMT. Therefore,  e 1 , e 2 , e 3  is a basis for R 3 . 1, t , t 2 , … , t n EXAMPLE: Let S = . Show that S is a basis for P n . Solution: Any polynomial in P n is in span of S . To show that S is linearly independent, assume c 0 ⋅ 1 + c 1 ⋅ t + ⋯ + c n ⋅ t n = 0 Then c 0 = c 1 = ⋯ = c n = 0 . Hence S is a basis for P n . 4

1 0 1 EXAMPLE: Let v 1 = , v 2 = , v 3 = . 2 1 0 0 1 3 Is  v 1 , v 2 , v 3  a basis for R 3 ? 1 0 1 Solution: Again, let A = v 1 v 2 v 3 . Using row 2 1 0 = 0 1 3 reduction, 1 0 1 1 0 1 1 0 1 2 1 0 0 1 − 2 0 1 − 2 ∽ ∽ 0 1 3 0 1 3 0 0 5 and since there are 3 pivots, the columns of A are linearly independent and they span R 3 by the IMT. Therefore  v 1 , v 2 , v 3  is a basis for R 3 . 5

EXAMPLE: Explain why each of the following sets is not a basis for R 3 . 1 4 0 1 (a) , , , 2 5 1 − 3 3 7 0 7 1 4 (b) , 2 5 3 6 6

Bases for Nul A EXAMPLE: Find a basis for Nul A where 3 6 6 3 9 A = . 6 12 13 0 3 Solution: Row reduce : A 0 x 1 = − 2 x 2 − 13 x 4 − 33 x 5 1 2 0 13 33 0 x 3 = 6 x 4 + 15 x 5 0 0 1 − 6 − 15 0 x 2 , x 4 and x 5 are free x 1 − 2 x 2 − 13 x 4 − 33 x 5 x 2 x 2 x 3 6 x 4 + 15 x 5 = = x 4 x 4 x 5 x 5 − 2 − 13 − 33 1 0 0 x 2 x 4 x 5 0 6 15 + + 0 1 0 0 0 1 ↑ ↑ ↑ u v w 7

Therefore  u , v , w  is a spanning set for Nul A . In the last section we observed that this set is linearly independent. Therefore  u , v , w  is a basis for Nul A . 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. − 1 0 EXAMPLE: Suppose v 1 = , v 2 = and 0 − 1 − 2 v 3 = . − 3 Solution: If x is in Span  v 1 , v 2 , v 3  , then x = c 1 v 1 + c 2 v 2 + c 3 v 3 = c 1 v 1 + c 2 v 2 + c 3  _____ v 1 + _____ v 2  = _____________ v 1 + _____________ v 2 Therefore, Span  v 1 , v 2 , v 3  = Span  v 1 , v 2  . 8

THEOREM 5 The Spanning Set Theorem Let S =  v 1 , … , v p  be a set in V and let H = Span  v 1 , … , v p  . a. If one of the vectors in S - say v k - is a linear combination of the remaining vectors in S , then the set formed from S by removing v k still spans H . b. If H ≠  0  , some subset of S is a basis for H . 9

Bases for Col A EXAMPLE: Find a basis for Col A , where 1 2 0 4 2 4 − 1 3 a 1 a 2 a 3 a 4 . A = = 3 6 2 22 4 8 0 16 Solution: Row reduce: 1 2 0 4 0 0 1 5 a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 ∼ ⋯  = 0 0 0 0 0 0 0 0 Note that b 2 = ____ b 1 and a 2 = ____ a 1 b 4 = 4 b 1 + 5 b 3 and a 4 = 4 a 1 + 5 a 3 b 1 and b 3 are not multiples of each other a 1 and a 3 are not multiples of each other Elementary row operations on a matrix do not affect the linear dependence relations among the columns of the matrix. Therefore Span a 1 , a 2 , a 3 , a 4 = Span a 1 , a 3 and a 1 , a 3 is a basis for Col A . 10

THEOREM 6 The pivot columns of a matrix A form a basis for Col A . 1 − 2 3 EXAMPLE: Let v 1 = , v 2 = , v 3 = . 2 − 4 6 − 3 6 9 Find a basis for Span  v 1 , v 2 , v 3  . 1 − 2 3 Solution: Let A = and note that 2 − 4 6 − 3 6 9 Col A = Span  v 1 , v 2 , v 3  . 1 − 2 0 By row reduction, A  . Therefore a basis 0 0 1 0 0 0 for Span  v 1 , v 2 , v 3  is . , 11

Review: 1. To find 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 find parametric form of the general solution to A x = 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 of A . 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 . 12