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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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 multiplication, 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).

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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, (2/3,1/3) c1 − c2 = x2. v2 (2,1) 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 − (x1,x2) 3 (x1 2x2)v1 3 (x1 x2)v2. v 1 v c 1 2 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.

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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 v c1v1 c2v2 c3v3

v1 c3v3 c1v1

v3 c2v2 v2 y

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.

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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.

Example 4.4.6 Determine a spanning set for P2, the vector space of all polynomials of degree 2 or less.

Solution: The general polynomial in P2 is 2 p(x) = a0 + a1x + a2x . If we let 2 p0(x) = 1,p1(x) = x, p2(x) = x , then p(x) = a0p0(x) + a1p1(x) + a2p2(x). 2 Thus, every vector in P2 is a linear combination of 1, x, and x , and so a spanning set 2 2 2 2 for P2 is {1,x,x }. For practice, the reader might show that {x ,x+ x , 1 + x + x } is another spanning set for P2, by making the appropriate modiﬁcations to the calculations in this example.

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The Linear Span of a Set of Vectors

Now let v1, v2, ..., vk be vectors in a vector space V . Forming all possible linear combinations of v1, v2, ..., vk generates a subset of V called the linear span of {v1, v2,...,vk}, denoted span{v1, v2,...,vk}.Wehave

span{v1, v2,...,vk}={v ∈ V : v = c1v1 + c2v2 +···+ckvk,c1,c2,...,ck ∈ F }. (4.4.5)

2 For example, suppose V = C (I), and let y1(x) = sin x and y2(x) = cos x. Then

2 span{y1,y2}={y ∈ C (I) : y(x) = c1 cos x + c2 sin x,c1,c2 ∈ R}.

From Example 1.2.16, we recognize y1 and y2 as being nonproportional solutions to the differential equation y + y = 0. Consequently, in this example, the linear span of the given functions coincides with the set of all solutions to the differential equation y + y = 0 and therefore is a subspace of V . Our next theorem generalizes this to show that any linear span of vectors in any vector space forms a subspace.

Theorem 4.4.7 Let v1, v2, ..., vk be vectors in a vector space V . Then span{v1, v2,...,vk} is a subspace of V .

Proof Let S = span{v1, v2,...,vk}. Then 0 ∈ S (corresponding to c1 = c2 = ··· = ck = 0 in (4.4.5)), so S is nonempty. We now verify closure of S under addition and scalar multiplication. If u and v are in S, then, from Equation (4.4.5),

u = a1v1 + a2v2 +···+akvk and v = b1v1 + b2v2 +···+bkvk,

for some scalars ai,bi. Thus,

u + v = (a1v1 + a2v2 +···+akvk) + (b1v1 + b2v2 +···+bkvk) = (a1 + b1)v1 + (a2 + b2)v2 +···+(ak + bk)vk = c1v1 + c2v2 +···+ckvk,

where ci = ai + bi for each i = 1, 2,...,k. Consequently, u + v has the proper form for membership in S according to (4.4.5), so S is closed under addition. Further, if r is any scalar, then

ru = r(a1v1 + a2v2 +···+akvk) = (ra1)v1 + (ra2)v2 +···+(rak)vk = d1v1 + d2v2 +···+dkvk,

where di = rai for each i = 1, 2,...,k. Consequently, ru ∈ S, and so S is also closed under scalar multiplication. Hence, S = span{v1, v2,...,vk} is a subspace of V .

Remarks

1. We will also refer to span{v1, v2,...,vk} as the subspace of V spanned by v1, v2,...,vk. 2. As a special case, we will declare that span(∅) ={0}.

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4.4 Spanning Sets 263

2 Example 4.4.8 If V = R and v1 = (−1, 1), determine span{v1}. Solution: We have 2 span{v1}={v ∈ R : v = c1v1,c1 ∈ R} 2 ={v ∈ R : v = c1(−1, 1), c1 ∈ R} 2 ={v ∈ R : v = (−c1,c1), c1 ∈ R}.

Geometrically, this is the line through the origin with parametric equations x =−c1, y = c1, so that the Cartesian equation of the line is y =−x. (See Figure 4.4.4.)

y (—c1, c1) c v 1 1 The subspace of ޒ2 spanned by the vector v (1, 1) (1, 1) 1 v1 x

2 Figure 4.4.4: The subspace of R spanned by v1 = (−1, 1).

3 3 Example 4.4.9 If V = R , v1 = (1, 0, 1), and v2 = (0, 1, 1), determine the subspace of R spanned by v1 and v2. Does w = (1, 1, −1) lie in this subspace? Solution: We have 3 span{v1, v2}={v ∈ R : v = c1v1 + c2v2,c1,c2 ∈ R} 3 ={v ∈ R : v = c1(1, 0, 1) + c2(0, 1, 1), c1,c2 ∈ R} 3 ={v ∈ R : v = (c1,c2,c1 + c2), c1,c2 ∈ R}.

Since the vector w = (1, 1, −1) is not of the form (c1,c2,c1 + c2), it does not lie in span{v1, v2}. Geometrically, span{v1, v2} is the plane through the origin determined by the two given vectors v1 and v2. It has parametric equations x = c1, y = c2, z = c1 +c2, which implies that its Cartesian equation is z = x + y. Thus, the fact that w is not in span{v1, v2} means that w does not lie in this plane. The subspace is depicted in Figure 4.4.5.

z The subspace of ޒ3 spanned by v1 (1, 0, 1), v2 (0, 1, 1)

v2 v1

x w (1, 1, 1) does not lie in span{v1, v2}

3 Figure 4.4.5: The subspace of R spanned by v1 = (1, 0, 1) and v2 = (0, 1, 1) is the plane with Cartesian equation z = x + y.

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Example 4.4.10 Let 10 01 00 A = ,A= ,A= 1 00 2 10 3 01

in M2(R). Determine span{A1,A2,A3}. Solution: By deﬁnition we have

{ }={ ∈ R : = + + ∈ R} span A1,A2,A3 A M2( ) A c1A1 c2A2 c3A3,c 1,c2,c3 10 01 00 = A ∈ M2(R) : A = c1 + c2 + c3 00 10 01 c1 c2 = A ∈ M2(R) : A = ,c1,c2,c3 ∈ R . c2 c3

This is the set of all real 2 × 2 symmetric matrices.

Example 4.4.11 Determine the subspace of P2 spanned by

2 p1(x) = 1 + 3x, p2(x) = x + x ,

and decide whether {p1,p2} is a spanning set for P2. Solution: We have

span{p1,p2}={p ∈ P2 : p(x) = c1p1(x) + c2p2(x), c1,c2 ∈ R} 2 ={p ∈ P2 : p(x) = c1(1 + 3x) + c2(x + x ), c1,c2 ∈ R} 2 ={p ∈ P2 : p(x) = c1 + (3c1 + c2)x + c2x ,c1,c2 ∈ R}.

Next, we will show that {p1,p2} is not a spanning set for P2. To establish this, we need give only one example of a polynomial in P2 that is not in span{p1,p2}. There are many such choices here, but suppose we consider p(x) = 1 + x. If this polynomial were in span{p1,p2}, then we would have to be able to ﬁnd values of c1 and c2 such that

2 1 + x = c1 + (3c1 + c2)x + c2x . (4.4.6)

2 Since there is no x term on the left-hand side of this expression, we must set c2 = 0. But then (4.4.6) would reduce to

1 + x = c1(1 + 3x).

Equating the constant terms on each side of this forces c1 = 1, but then the coefﬁcients of x do not match. Hence, such an equality is impossible. Consequently, there are no values of c1 and c2 such that the Equation (4.4.6) holds, and therefore, span{p1,p2} = P2.

Remark In the previous example, the reader may well wonder why we knew from the beginning to select p(x) = 1 + x as a vector that would be outside of span{p1,p2}. In truth, we only need to ﬁnd a polynomial that does not have the form p(x) = c1 + 2 (3c1 + c2)x + c2x and in fact, “most” of the polynomials in P2 would have achieved the desired result here.

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Exercises for 4.4 Key Terms 8. If S is a spanning set for a vector space V , then any S S V Linear combination, Linear span, Spanning set. proper subset of is not a spanning set for . 9. The vector space of 3 × 3 upper triangular matrices is Skills spanned by the matrices Eij where 1 ≤ i ≤ j ≤ 3.

• Be able to determine whether a given set of vectors 10. A spanning set for the vector space P2 must contain a S spans a vector space V , and be able to prove your polynomial of each degree 0, 1, and 2. answer mathematically. 11. If m

1. The linear span of a set of vectors in a vector space V 5. {(1, −2, 1), (2, 3, 1), (0, 0, 0), (4, −1, 2)}. forms a subspace of V . 6. {(2, −1, 4), (3, −3, 5), (1, 1, 3)}. 2. If some vector v in a vector space V is a linear combination of vectors in a set S, then S spans V . 7. Show that the set of vectors

3. If S is a spanning set for a vector space V and W is a {(1, 2, 3), (3, 4, 5), (4, 5, 6)} subspace of V , then S is a spanning set for W. does not span R3, but that it does span the subspace 4. If S is a spanning set for a vector space V , then every of R3 consisting of all vectors lying in the plane with vector v in V must be uniquely expressible as a linear equation x − 2y + z = 0. combination of the vectors in S. 8. Show that v = (2, −1), v = (3, 2) span R2, and ex- 5. A set S of vectors in a vector space V spans V if and 1 2 press the vector v = (5, −7) as a linear combination only if the linear span of S is V . of v1, v2. 6. The linear span of two vectors in R3 is a plane through = − = − = the origin. 9. Show that v1 ( 1, 3, 2), v2 (1, 2, 1), v3 (2, 1, 1) span R3, and express v = (x,y,z)as a linear 7. Every vector space V has a ﬁnite spanning set. combination of v1, v2, v3.

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10. Show that v1 = (1, 1), v2 = (−1, 2), v3 = (1, 4) span For Problems 22–24, determine whether the given vector v 2 2 R .Dov1, v2 alone span R also? lies in span{v1, v2}.

11. Let S be the subspace of R3 consisting of all vectors 3 22. v = (3, 3, 4), v1 = (1, −1, 2), v2 = (2, 1, 3) in R . of the form v = (c1,c2,c2 − 2c1). Show that S is = − = spanned by v1 (1, 0, 2), v2 (0, 1, 1). 23. v = (5, 3, −6), v1 = (−1, 1, 2), v2 = (3, 1, −4) in R3. 12. Let S be the subspace of R4 consisting of all vectors of the form v = (c1,c2,c2 − c1,c1 − 2c2). Determine 24. v = (1, 1, −2), v1 = (3, 1, 2), v2 = (−2, −1, 1) in a set of vectors that spans S. R3. 13. Let S be the subspace of R3 consisting of all solutions = − = 2 − + to the linear system 25. If p1(x) x 4 and p2(x) x x 3, determine 2 whether p(x) = 2x − x + 2 lies in span{p1,p2}. x − 2y − z = 0. 26. Consider the vectors Determine a set of vectors that spans S. 1 −1 01 30 A = ,A= ,A= For Problems 14–15, determine a spanning set for the null 1 20 2 −21 3 12 space of the given matrix A.   R { } 123 in M2( ). Determine span A1,A2,A3 . 14. A =  345. 27. Consider the vectors 567   12 −21 123 5 A = ,A= 1 −13 2 1 −1 15. A =  134 2. 246−1 in M2(R). Find span{A1,A2}, and determine whether 16. Let S be the subspace of M2(R) consisting of all sym- or not × 31 metric 2 2 matrices with real elements. Show that S B = is spanned by the matrices −24 10 00 01 lies in this subspace. A = ,A= ,A= . 1 00 2 01 3 10 28. Let V = C∞(I) and let S be the subspace of V

17. Let S be the subspace of M2(R) consisting of all skew- spanned by the functions symmetric 2 × 2 matrices with real elements. Deter- = = mine a matrix that spans S. f(x) cosh x, g(x) sinh x.

18. Let S be the subset of M2(R) consisting of all upper (a) Give an expression for a general vector in S. triangular 2 × 2 matrices. (b) Show that S is also spanned by the functions (a) Verify that S is a subspace of M2(R). = x = −x (b) Determine a set of 2 × 2 matrices that spans S. h(x) e ,j(x)e . { } For Problems 19–20, determine span v1, v2 for the given For Problems 29–32, give a geometric description of the sub- 3 vectors in R , and describe it geometrically. space of R3 spanned by the given set of vectors. 19. v = (1, −1, 2), v = (2, −1, 3). 1 2 29. {0}. 20. v = (1, 2, −1), v = (−2, −4, 2). 1 2 3 30. {v1}, where v1 is any nonzero vector in R . 21. Let S be the subspace of R3 spanned by the vectors v1 = (1, 1, −1), v2 = (2, 1, 3), v3 = (−2, −2, 2). 31. {v1, v2}, where v1, v2 are nonzero and noncollinear 3 Show that S also is spanned by v1 and v2 only. vectors in R .

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3 32. {v1, v2}, where v1, v2 are collinear vectors in R . 34. Prove that

33. Prove that if S and S are subsets of a vector space V span{v1, v2, v3}=span{v1, v2} such that S is a subset of S, then span(S) is a subset of span(S). if and only if v3 can be written as a linear combination of v1 and v2.

4.5 Linear Dependence and Linear Independence As indicated in the previous section, in analyzing a vector space we will be interested in determining a spanning set. The reader has perhaps already noticed that a vector space V can have many such spanning sets.

Example 4.5.1 Observe that {(1, 0), (0, 1)}, {(1, 0), (1, 1)}, and {(1, 0), (0, 1), (1, 2)} are all spanning sets for R2.

As another illustration, two different spanning sets for V = M2(R) were given in Exam- ple 4.4.5 and the remark that followed. Given the abundance of spanning sets available for a given vector space V , we are faced with a natural question: Is there a “best class of” spanning sets to use? The answer, to a large degree, is “yes”. For instance, in Exam- ple 4.5.1, the spanning set {(1, 0), (0, 1), (1, 2)} contains an “extra” vector, (1, 2), which seems to be unnecessary for spanning R2, since {(1, 0), (0, 1)} is already a spanning set. In some sense, {(1, 0), (0, 1)} is a more efﬁcient spanning set. It is what we call a minimal spanning set, since it contains the minimum number of vectors needed to span the vector space.3 But how will we know if we have found a minimal spanning set (assuming one exists)? Returning to the example above, we have seen that

span{(1, 0), (0, 1)}=span{(1, 0), (0, 1), (1, 2)}=R2.

Observe that the vector (1, 2) is already a linear combination of (1, 0) and (0, 1), and therefore it does not add any new vectors to the linear span of {(1, 0), (0, 1)}. As a second example, consider the vectors v1 = (1, 1, 1), v2 = (3, −2, 1), and v3 = 4v1 + v2 = (7, 2, 5). It is easily veriﬁed that det([v1, v2, v3]) = 0. Consequently, the three vectors lie in a plane (see Figure 4.5.1) and therefore, since they are not collinear, the linear span of these three vectors is the whole of this plane. Furthermore, the same plane is generated if we consider the linear span of v1 and v2 alone. As in the previous example, the reason that v3 does not add any new vectors to the linear span of {v1, v2} is that it is already a linear combination of v1 and v2. It is not possible, however, to generate all vectors in the plane by taking linear combinations of just one vector, as we could generate only a line lying in the plane in that case. Consequently, {v1, v2} is a minimal spanning set for the subspace of R3 consisting of all points lying on the plane. As a ﬁnal example, recall from Example 1.2.16 that the solution space to the differential equation y + y = 0

3Since a single (nonzero) vector in R2 spans only the line through the origin along which it points, it cannot span all of R2; hence, the minimum number of vectors required to span R2 is 2.

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