4.3 Linearly Independent Sets; Bases Math 2331 { Linear Algebra 4.3 Linearly Independent Sets; Bases Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/∼jiwenhe/math2331 Jiwen He, University of Houston Math 2331, Linear Algebra 1 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A 4.3 Linearly Independent Sets; Bases Linearly Independent Sets Definition Facts Examples A Basis Set: Definition A Basis Set: Examples Nul A: Examples and Theorem Col A: Examples and Theorem The Spanning Set Theorem Jiwen He, University of Houston Math 2331, Linear Algebra 2 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Linearly Independent Sets Linearly Independent Sets A set of vectors fv1; v2;:::; vpg in a vector space V is said to be linearly independent if the vector equation c1v1 + c2v2 + ··· + cpvp = 0 has only the trivial solution c1 = 0;:::; cp = 0. The set fv1; v2;:::; vpg is said to be linearly dependent if there exists weights c1;:::; cp;not all 0, such that c1v1 + c2v2 + ··· + cpvp = 0. Jiwen He, University of Houston Math 2331, Linear Algebra 3 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Linearly Independent Sets: Facts The following results from Section 1.7 are still true for more general vectors spaces. Fact 1 A set containing the zero vector is linearly dependent. Fact 2 A set of two vectors is linearly dependent if and only if one is a multiple of the other. Fact 3 A set containing the zero vector is linearly independent. Jiwen He, University of Houston Math 2331, Linear Algebra 4 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Linearly Independent Sets: Examples Example 1 2 0 0 3 2 ; ; is a linearly set. 3 4 0 0 3 0 Example 1 2 3 6 ; is a linearly set 3 4 9 11 3 6 1 2 since is not a multiple of . 9 11 3 4 Jiwen He, University of Houston Math 2331, Linear Algebra 5 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Linearly Independent Sets: Examples Theorem (4) An indexed set fv1; v2;:::; vpg of two or more vectors, with v1 6= 0, is linearly dependent if and only if some vector vj (j > 1) is a linear combination of the preceding vectors v1;:::; vj−1. Example Let fp1, p2, p3g be a set of vectors in P2 where p1(t) = t, 2 2 p2(t) = t , and p3(t) = 4t + 2t . Is this a linearly dependent set? Solution: Since p3 = p1 + p2, fp1, p2, p3g is a linearly set. Jiwen He, University of Houston Math 2331, Linear Algebra 6 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A A Basis Set Let H be the plane illustrated below. Which of the following are valid descriptions of H? (a) H =Spanfv1; v2g (b) H =Spanfv1; v3g (c) H =Spanfv2; v3g (d) H =Spanfv1; v2; v3g A basis set is an “efficient” spanning set containing no unnecessary vectors. In this case, we would consider the linearly independent sets fv1; v2g and fv1; v3g to both be examples of basis sets or bases (plural for basis) for H. Jiwen He, University of Houston Math 2331, Linear Algebra 7 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A A Basis Set: Definition and Examples A Basis Set Let H be a subspace of a vector space V . An indexed set of vectors β = fb1;:::; bpg in V is a basis for H if i. β is a linearly independent set, and ii. H = Spanfb1;:::; bpg. Example 2 1 3 2 0 3 2 0 3 Let e1 = 4 0 5 ; e2 = 4 1 5 ; e3 = 4 0 5. Show that 0 0 1 3 fe1; e2; e3g is a basis for R . The set fe1; e2; e3g is called a standard basis for R3. Solutions: (Review the IMT, page 112) 2 1 0 0 3 Let A = e1 e2 e3 = 4 0 1 0 5. 0 0 1 Jiwen He, University of Houston Math 2331, Linear Algebra 8 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A A Basis Set: Definition and Examples Since A has 3 pivots, the columns of A are linearly by the IMT, and the columns of A by IMT; 3 therefore, fe1; e2; e3g is a basis for R . Example 2 n Let S = 1; t; t ;:::; t . Show that S is a basis for Pn. Solution: Any polynomial in Pn is in span of S:To show that S is linearly independent, assume n c0 · 1 + c1 · t + ··· + cn · t = 0: Then c0 = c1 = ··· = cn = 0. Hence S is a basis for Pn. Jiwen He, University of Houston Math 2331, Linear Algebra 9 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A A Basis Set: Example Example 2 1 3 2 0 3 2 1 3 Let v1 = 4 2 5 ; v2 = 4 1 5 ; v3 = 4 0 5. 0 1 3 3 Is fv1; v2; v3g a basis for R ? 2 1 0 1 3 Solution: Let A = [v1 v2 v3] = 4 2 1 0 5. By row reduction, 0 1 3 2 1 0 1 3 2 1 0 1 3 2 1 0 1 3 4 2 1 0 5 v 4 0 1 −2 5 v 4 0 1 −2 5 0 1 3 0 1 3 0 0 5 and since there are 3 pivots, the columns of A are linearly 3 independent and they span R by the IMT. Therefore fv1; v2; v3g is a basis for R3 . Jiwen He, University of Houston Math 2331, Linear Algebra 10 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A A Basis Set: Example Example Explain why each of the following sets is not a basis for R3: 82 3 2 3 2 3 2 39 < 1 4 0 1 = (a) 4 2 5 ; 4 5 5 ; 4 1 5 ; 4 −3 5 : 3 7 0 7 ; Example 82 3 2 39 < 1 4 = (b) 4 2 5 ; 4 5 5 : 3 6 ; Jiwen He, University of Houston Math 2331, Linear Algebra 11 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Bases for Nul A: Example Example Find a basis for Nul A where 3 6 6 3 9 A = : 6 12 13 0 3 Solution: Row reduce A 0 : x1 = −2x2 − 13x4 − 33x5 1 2 0 13 33 0 =) x = 6x + 15x 0 0 1 −6 −15 0 3 4 5 x2, x4 and x5 are free Jiwen He, University of Houston Math 2331, Linear Algebra 12 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Bases for Nul A: Example (cont.) 2 3 2 3 x1 −2x2 − 13x4 − 33x5 6 x2 7 6 x2 7 6 7 6 7 6 x3 7 = 6 6x4 + 15x5 7 = 6 7 6 7 4 x4 5 4 x4 5 x5 x5 2 −2 3 2 −13 3 2 −33 3 6 1 7 6 0 7 6 0 7 6 7 6 7 6 7 x2 6 0 7 + x4 6 6 7 + x5 6 15 7 6 7 6 7 6 7 4 0 5 4 1 5 4 0 5 0 0 1 " " " u v w Therefore fu; v; wg is a spanning set for Nul A. In the last section we observed that this set is linearly independent. Therefore fu; v; wg is a basis for Nul A. The technique used here always provides a linearly independent set. Jiwen He, University of Houston Math 2331, Linear Algebra 13 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A 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 −1 0 −2 Suppose v = ; v = and v = . 1 0 2 −1 3 −3 Solution: If x is in Spanfv1; v2; v3g, then x =c1v1 + c2v2 + c3v3 = c1v1 + c2v2 + c3 ( v1 + v2) = v1 + v2 Therefore, Spanfv1; v2; v3g =Spanfv1; v2g : Jiwen He, University of Houston Math 2331, Linear Algebra 14 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A The Spanning Set Theorem Theorem (5 The Spanning Set Theorem) Let S = fv1;:::; vpg be a set in V and let H = Span fv1;:::; vpg : a. If one of the vectors in S - say vk - is a linear combination of the remaining vectors in S, then the set formed from S by removing vk still spans H. b. If H 6= f0g, some subset of S is a basis for H. Jiwen He, University of Houston Math 2331, Linear Algebra 15 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Bases for Col A: Examples Example Find a basis for Col A, where 2 1 2 0 4 3 6 2 4 −1 3 7 A = [a1 a2 a3 a4] = 6 7. 4 3 6 2 22 5 4 8 0 16 Solution: Row reduce: 2 1 2 0 4 3 6 0 0 1 5 7 [a1 a2 a3 a4] ∼ · · · s 6 7 = [b1 b2 b3 b4] 4 0 0 0 0 5 0 0 0 0 Jiwen He, University of Houston Math 2331, Linear Algebra 16 / 20 4.3 Linearly Independent Sets; Bases Linearly Independent Sets A Basis Set Nul A Col A Bases for Col A: Examples (cont.) Note that b2 = b1 and a2 = a1 b4 = 4b1 + 5b3 and a4 = 4a1 + 5a3 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 of the matrix.