44.3© 2012 Pearson Education, Inc.Vector SpacesLINEARLY INDEPENDENT SETS; BASESSlide 4.3- 2© 2012 Pearson Education, Inc.LINEAR INDEPENDENT SETS; BASES An indexed set of vectors {v1, …, vp} in V is said to be linearly independent if the vector equation----(1)has only the trivial solution, . The set {v1, …, vp} is said to be linearly dependent if (1) has a nontrivial solution, i.e., if there are some weights, c1, …, cp, not all zero, such that (1) holds. In such a case, (1) is called a linear dependence relation among v1, …, vp.11 2 2v v ... v 0ppcc c 10,..., 0pccSlide 4.3- 3© 2012 Pearson Education, Inc.LINEAR INDEPENDENT SETS; BASES Theorem 4: An indexed set {v1, …, vp} of two or more vectors, with , is linearly dependent if and only if some vj(with ) is a linear combination of the preceding vectors, . Definition: Let H be a subspace of a vector space V. An indexed set of vectors  in V is a basis for H if (i)  is a linearly independent set, and (ii) The subspace spanned by  coincides with H; that is, 1v01j 1{b ,...,b}p1Span{b ,...,b}pH11v ,...,vjSlide 4.3- 4© 2012 Pearson Education, Inc.LINEAR INDEPENDENT SETS; BASES The definition of a basis applies to the case when , because any vector space is a subspace of itself. Thus a basis of V is a linearly independent set that spans V. When , condition (ii) includes the requirement that each of the vectors b1, …, bpmust belong to H, because Span {b1, …, bp} contains b1, …, bp.HVHVSlide 4.3- 5© 2012 Pearson Education, Inc.STANDARD BASIS Let e1, …, enbe the columns of the matrix, In.  That is, The set {e1, …, en} is called the standard basis for . See the following figure.nn1210 001e ,e ,...,e000 1n       nSlide 4.3- 6© 2012 Pearson Education, Inc.THE SPANNING SET THEOREM Theorem 5: Let be a set in V, and let . 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 vkstill spans H.b. If , some subset of S is a basis for H. Proof:a. By rearranging the list of vectors in S, if necessary, we may suppose that vpis a linear combination of —say, 1{v ,...,v}pS1Span{v ,...,v}pH{0}H11v ,...,vpSlide 4.3- 7© 2012 Pearson Education, Inc.THE SPANNING SET THEOREM----(2) Given any x in H, we may write----(3)for suitable scalars c1, …, cp. Substituting the expression for vpfrom (2) into (3), it is easy to see that x is a linear combination of .  Thus spans H, because x was an arbitrary element of H.11 1 1vv...vpppaa11 1 1xv... v vpp ppccc 11v,...vp11{v ,...,v}pSlide 4.3- 8© 2012 Pearson Education, Inc.THE SPANNING SET THEOREMb. If the original spanning set S is linearly independent, then it is already a basis for H. Otherwise, one of the vectors in S depends on the others and can be deleted, by part (a). So long as there are two or more vectors in the spanning set, we can repeat this process until the spanning set is linearly independent and hence is a basis for H. If the spanning set is eventually reduced to one vector, that vector will be nonzero (and hence linearly independent) because .{0}HSlide 4.3- 9© 2012 Pearson Education, Inc.THE SPANNING SET THEOREM Example 1: Let , , and .Note that , and show that . Then find a basis for the subspace H. Solution: Every vector in Span {v1, v2} belongs to Hbecause 10v2122v2036v165123Span{v,v,v}H312v5v3v123 12Span{v,v,v}Span{v,v}11 2 2 11 2 2 3vvvv0vcc cc Slide 4.3- 10© 2012 Pearson Education, Inc.THE SPANNING SET THEOREM Now let x be any vector in H—say,.  Since , we may substitute Thus x is in Span {v1, v2}, so every vector in H already belongs to Span {v1, v2}. We conclude that H and Span {v1, v2} are actually the set of vectors. It follows that {v1, v2} is a basis of H since {v1, v2} is linearly independent.11 2 2 33xv v vcc c312v5v3v11 2 2 3 1 2131232xv v (5v3v)( 5 )v ( 3 )vcc ccc cc   Slide 4.3- 11© 2012 Pearson Education, Inc.BASIS FOR COL B Example 2: Find a basis for Col B, where Solution: Each nonpivot column of B is a linear combination of the pivot columns. In fact, and .  By the Spanning Set Theorem, we may discard b2and b4, and {b1, b3, b5} will still span Col B.125140 20001 10bb b000 01000 00B21b4b413b2b bSlide 4.3- 12© 2012 Pearson Education, Inc.BASIS FOR COL B Let Since and no vector in S is a linear combination of the vectors that precede it, S is linearly independent. (Theorem 4). Thus S is a basis for Col B. 135100010{b ,b ,b } , ,001000S1b0Slide 4.3- 13© 2012 Pearson Education, Inc.BASES FOR NUL A AND COL A Theorem 6: The pivot columns of a matrix A form a basis for Col A. Proof: Let B be the reduced echelon form of A. The set of pivot columns of B is linearly independent, for no vector in the set is a linear combination of the vectors that precede it. Since A is row equivalent to B, the pivot columns of A are linearly independent as well, because any linear dependence relation among the columns of Acorresponds to a linear dependence relation among the columns of B.Slide 4.3- 14© 2012 Pearson Education, Inc.BASES FOR NUL A AND COL A For this reason, every nonpivot column of A is a linear combination of the pivot columns of A. Thus the nonpivot columns of a may be discarded from the spanning set for Col A, by the Spanning Set Theorem. This leaves the pivot columns of A as a basis for Col A. Warning: The pivot columns of a