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Spanning Sets the Only Algebraic Operations That Are Defined in a Vector Space V Are Those of Addition and Scalar Multiplication

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Spanning Sets the Only Algebraic Operations That Are Defined in a Vector Space V Are Those of Addition and Scalar Multiplication i i “main” 2007/2/16 page 258 i i 258 CHAPTER 4 Vector Spaces 23. Show that the set of all solutions to the nonhomoge- and let neous differential equation S1 + S2 = {v ∈ V : y + a1y + a2y = F(x), v = x + y for some x ∈ S1 and y ∈ S2} . where F(x) is nonzero on an interval I, is not a sub- 2 space of C (I). (a) Show that, in general, S1 ∪ S2 is not a subspace of V . 24. Let S1 and S2 be subspaces of a vector space V . Let (b) Show that S1 ∩ S2 is a subspace of V . S ∪ S ={v ∈ V : v ∈ S or v ∈ S }, 1 2 1 2 (c) Show that S1 + S2 is a subspace of V . S1 ∩ S2 ={v ∈ V : v ∈ S1 and v ∈ S2}, 4.4 Spanning Sets The only algebraic operations that are defined in a vector space V are those of addition and scalar multiplication. Consequently, the most general way in which we can combine the vectors v1, v2,...,vk in V is c1v1 + c2v2 +···+ckvk, (4.4.1) where c1,c2,...,ck are scalars. An expression of the form (4.4.1) is called a linear combination of v1, v2,...,vk. Since V is closed under addition and scalar multiplica- tion, it follows that the foregoing linear combination is itself a vector in V . One of the questions we wish to answer is whether every vector in a vector space can be obtained by taking linear combinations of a finite set of vectors. The following terminology is used in the case when the answer to this question is affirmative: DEFINITION 4.4.1 If every vector in a vector space V can be written as a linear combination of v1, v2, ..., vk, we say that V is spanned or generated by v1, v2, ..., vk and call the set of vectors {v1, v2,...,vk} a spanning set for V . In this case, we also say that {v1, v2,...,vk} spans V . This spanning idea was introduced in the preceding section within the framework of differential equations. In addition, we are all used to representing geometric vectors in R3 in terms of their components as (see Section 4.1) v = ai + bj + ck, where i, j, and k denote the unit vectors pointing along the positive x-, y-, and z-axes, respectively, of a rectangular Cartesian coordinate system. Using the above terminology, we say that v has been expressed as a linear combination of the vectors i, j, and k, and that the vector space of all geometric vectors is spanned by i, j, and k. We now consider several examples to illustrate the spanning concept in different vector spaces. Example 4.4.2 Show that R2 is spanned by the vectors v1 = (1, 1) and v2 = (2, −1). i i i i i i “main” 2007/2/16 page 259 i i 4.4 Spanning Sets 259 2 Solution: We must establish that for every v = (x1,x2) in R , there exist constants c1 and c2 such that v = c1v1 + c2v2. (4.4.2) y That is, in component form, (4/3, 4/3) = + − (1, 1) (x1,x2) c1(1, 1) c2(2, 1). (2, 1) v 1 Equating corresponding components in this equation yields the following linear system: v x c1 + 2c2 = x1, (2/3,Ϫ1/3) c1 − c2 = x2. v2 (2,Ϫ1) In this system, we view x1 and x2 as fixed, while the variables we must solve for are c1 Figure 4.4.1: The vector and c2. The determinant of the matrix of coefficients of this system is v = (2, 1) expressed as a linear = combination of v1 (1, 1) and 12 = − =−3. v2 (2, 1). 1 −1 Since this is nonzero regardless of the values of x1 and x2, the matrix of coefficients is 2 invertible, and hence for all (x1,x2) ∈ R , the system has a (unique) solution according to Theorem 2.6.4. Thus, Equation (4.4.2) can be satisfied for every vector v ∈ R2,sothe given vectors do span R2. Indeed, solving the linear system yields = 1 + = 1 − y c1 3 (x1 2x2), c2 3 (x1 x2). c v Hence, 2 2 v 2 c 2 = 1 + + 1 − ϩ (x1,x2) 3 (x1 2x2)v1 3 (x1 x2)v2. v 1 v c 1 2 ϭ v = = 4 = 1 = 4 + 1 v 1 c1v1 For example, if v (2, 1), then c1 3 and c2 3 , so that v 3 v1 3 v2. This is x illustrated in Figure 4.4.1. 2 2 Figure 4.4.2: Any two More generally, any two nonzero and noncolinear vectors v1 and v2 in R span R , noncollinear vectors in R2 span since, as illustrated geometrically in Figure 4.4.2, every vector in R2 can be written as a R2 . linear combination of v1 and v2. Example 4.4.3 Determine whether the vectors v1 = (1, −1, 4), v2 = (−2, 1, 3), and v3 = (4, −3, 5) span R3. 3 Solution: Let v = (x1,x2,x3) be an arbitrary vector in R . We must determine whether there are real numbers c1, c2, c3 such that v = c1v1 + c2v2 + c3v3 (4.4.3) or, in component form, (x1,x2,x3) = c1(1, −1, 4) + c2(−2, 1, 3) + c3(4, −3, 5). Equating corresponding components on either side of this vector equation yields c1 − 2c2 + 4c3 = x1, −c1 + c2 − 3c3 = x2, 4c1 + 3c2 + 5c3 = x3. i i i i i i “main” 2007/2/16 page 260 i i 260 CHAPTER 4 Vector Spaces Reducing the augmented matrix of this system to row-echelon form, we obtain 1 −24 x1 01−1 −x1 − x2 . 0007x1 + 11x2 + x3 It follows that the system is consistent if and only if x1, x2, x3 satisfy 7x1 + 11x2 + x3 = 0. (4.4.4) 3 Consequently, Equation (4.4.3) holds only for those vectors v = (x1,x2,x3) in R 3 whose components satisfy Equation (4.4.4). Hence, v1, v2, and v3 do not span R . Geometrically, Equation (4.4.4) is the equation of a plane through the origin in space, and so by taking linear combinations of the given vectors, we can obtain only those vectors which lie on this plane. We leave it as an exercise to verify that indeed the three given vectors lie in the plane with Equation (4.4.4). It is worth noting that this plane 3 forms a subspace S of R , and that while V is not spanned by the vectors v1, v2, and v3, S is. The reason that the vectors in the previous example did not span R3 was because 3 they were coplanar. In general, any three noncoplanar vectors v1, v2, and v3 in R span R3, since, as illustrated in Figure 4.4.3, every vector in R3 can be written as a linear combination of v1, v2, and v3. In subsequent sections we will make this same observation from a more algebraic point of view. z ϭ ϩ ϩ v c1v1 c2v2 c3v3 v1 c3v3 c1v1 v3 c2v2 v2 y x Figure 4.4.3: Any three noncoplanar vectors in R3 span R3. Notice in the previous example that the linear combination (4.4.3) can be written as the matrix equation Ac = v, where the columns of A are the given vectors v1, v2, and v3: A =[v1, v2, v3]. Thus, 3 the question of whether or not the vectors v1, v2, and v3 span R can be formulated as follows: Does the system Ac = v have a solution c for every v in R3? If so, then the column vectors of A span R3, and if not, then the column vectors of A do not span R3. This reformulation applies more generally to vectors in Rn, and we state it here for the record. n n Theorem 4.4.4 Let v1, v2,...,vk be vectors in R . Then {v1, v2,...,vk} spans R if and only if, for the matrix A =[v1, v2,...,vk], the linear system Ac = v is consistent for every v in Rn. i i i i i i “main” 2007/2/16 page 261 i i 4.4 Spanning Sets 261 Proof Rewriting the system Ac = v as the linear combination c1v1 + c2v2 +···+ckvk = v, we see that the existence of a solution (c1,c2,...,ck) to this vector equation for each v n n in R is equivalent to the statement that {v1, v2,...,vk} spans R . Next, we consider a couple of examples involving vector spaces other than Rn. Example 4.4.5 Verify that 10 11 11 11 A = ,A= ,A= ,A= 1 00 2 00 3 10 4 11 span M2(R). Solution: An arbitrary vector in M (R) is of the form 2 ab A = . cd If we write c1A1 + c2A2 + c3A3 + c4A4 = A, then equating the elements of the matrices on each side of the equation yields the system c1 + c2 + c3 + c4 = a, c2 + c3 + c4 = b, c3 + c4 = c, c4 = d. Solving this by back substitution gives c1 = a − b, c2 = b − c, c3 = c − d, c4 = d. Hence, we have A = (a − b)A1 + (b − c)A2 + (c − d)A3 + dA4. Consequently every vector in M2(R) can be written as a linear combination of A1, A2, A3, and A4, and therefore these matrices do indeed span M2(R). Remark The most natural spanning set for M (R) is 2 10 01 00 00 , , , , 00 00 10 01 a fact that we leave to the reader as an exercise.
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