1.6 Bases and Dimension Math 4377/6308 Advanced Linear Algebra 1.6 Bases and Dimension Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/∼jiwenhe/math4377 Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 1 / 24 1.6 Bases and Dimension A Basis Set 1.6 Bases and Dimension A Basis Set: Definition A Basis Set: Examples The Spanning Set Theorem Properties of Bases The Replacement Theorem The Dimension of a Vector Space: Definition The Dimension of a Vector Space: Examples Dimensions of Subspaces of R3 Dimensions of Subspaces: Theorem Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 2 / 24 1.6 Bases and Dimension A Basis Set 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 4377/6308, Advanced Linear Algebra Spring, 2015 3 / 24 1.6 Bases and Dimension A Basis Set A Basis Set: Definition Definition A basis β for a vector space V is a linearly independent subset of V that generates V . The vectors of β form a basis for V . A Basis Set of Subspace 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 Since span(;) = f0g and ; is linearly independent, ; is a basis for the zero vector space. Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 4 / 24 1.6 Bases and Dimension A Basis Set A Basis Set: Examples 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 . Solutions: 2 1 0 0 3 Let A = e1 e2 e3 = 4 0 1 0 5. 0 0 1 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 . n The basis fe1; ··· ; eng is called a standard basis for F : e1 = (1; 0; ··· ; 0), e2 = (0; 1; 0; ··· ; 0), ··· , en = (0; ··· ; 0; 1). Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 5 / 24 1.6 Bases and Dimension A Basis Set A Basis Set: Examples Example 2 n Let S = 1; x; x ;:::; x . 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 · x + ··· + cn · x = 0: Then c0 = c1 = ··· = cn = 0. Hence S is a basis for Pn. The basis f1; x; x2; ··· ; xng is called the standard basis for Pn(F ). Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 6 / 24 1.6 Bases and Dimension A Basis Set 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 3 is a basis for R . Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 7 / 24 1.6 Bases and Dimension A Basis Set A Basis Set: Example Example 3 Explain why each of the following sets is not a basis for R : 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 4377/6308, Advanced Linear Algebra Spring, 2015 8 / 24 1.6 Bases and Dimension A Basis 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 −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 4377/6308, Advanced Linear Algebra Spring, 2015 9 / 24 1.6 Bases and Dimension A Basis Set The Spanning Set Theorem Theorem (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 4377/6308, Advanced Linear Algebra Spring, 2015 10 / 24 1.6 Bases and Dimension A Basis Set Bases for Spanning Set: Theorem and Examples Example Find a basis for H = spanfa1; a2; a3; a4g, where 2 1 2 0 4 3 6 2 4 −1 3 7 [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 4377/6308, Advanced Linear Algebra Spring, 2015 11 / 24 1.6 Bases and Dimension A Basis Set Bases for Spanning Set: Theorem and 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. Therefore Span fa1, a2, a3, a4g = Span fa1, a3g and fa1, a3g is a basis for H: Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 12 / 24 1.6 Bases and Dimension A Basis Set Bases for Spanning Set: Theorem and Example Theorem The pivot columns of a matrix A = [a1; ··· ; a2] form a basis for span(a1; ··· ; an). Example 2 1 3 2 −2 3 2 3 3 Let v1 = 4 2 5 ; v2 = 4 −4 5 ; v3 = 4 6 5. Find a basis for −3 6 9 Spanfv1; v2; v3g . Solution: Let 2 1 −2 3 3 A = [v1; v2; v3] = 4 2 −4 6 5 −3 6 9 Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 13 / 24 1.6 Bases and Dimension A Basis Set Bases for Spanning Set: Theorem and Example (cont.) 2 1 −2 0 3 By row reduction, [v1; v2; v3] s 4 0 0 1 5. Therefore a basis 0 0 0 82 3 2 39 < = for Spanfv1; v2; v3g is 4 5 ; 4 5 . : ; Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 14 / 24 1.6 Bases and Dimension A Basis Set Properties of Bases Theorem (1.8) Let V be a vector space and β = fu1; ··· ; ung be a subset of V . Then β is a basis for V if and only if each v 2 V can be uniquely expressed as a linear combination of vectors of β: v = a1u1 + a2u2 + ··· + anun for unique scalars a1, ··· , an. Theorem (1.9) If a vector space V is generated by a finite set S, then some subset of S is a basis for V . Hence V has a finite basis. Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 15 / 24 1.6 Bases and Dimension A Basis Set The Replacement Theorem Theorem (1.10 The Replacement Theorem) Let V be a vector space that is generated by a set G containing exactly n vectors, and let L be a linearly independent subset of V containing exactly m vectors. Then m ≤ n and there exists a subset H of G containing exactly n − m vectors such that L [ H generates V . Corollary 0 If a vector space V has a basis β = fb1;:::; bng, then any set in V containing more than n vectors must be linearly dependent. Proof: Suppose S = fu1;:::; upg is a set of p vectors in V where p > n. If S is a linearly independent subset of V , the Replacement Theorem implies that p ≤ n, a contradiction.