Feb 16, 2007 - Vector Spaces. Reducing the augmented matrix of this system to row-echelon form, we obtain... 1 â2 4 x1. 0 1 â1. âx1 â x2. 0 0 ...

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23. Show that the set of all solutions to the nonhomogeneous differential equation

and let S1 + S2 = {v ∈ V : v = x + y for some x ∈ S1 and y ∈ S2 } .

y + a1 y + a2 y = F (x), where F (x) is nonzero on an interval I , is not a subspace of C 2 (I ). 24. Let S1 and S2 be subspaces of a vector space V . Let

(b) Show that S1 ∩ S2 is a subspace of V .

S1 ∪ S2 = {v ∈ V : v ∈ S1 or v ∈ S2 },

(c) Show that S1 + S2 is a subspace of V .

S1 ∩ S2 = {v ∈ V : v ∈ S1 and v ∈ S2 },

4.4

(a) Show that, in general, S1 ∪ S2 is not a subspace of V .

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 c1 v1 + c2 v2 + · · · + ck vk ,

(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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Solution: We must establish that for every v = (x1 , x2 ) in R2 , there exist constants c1 and c2 such that v = c1 v1 + c2 v2 . y

(4.4.2)

That is, in component form, (4/3, 4/3) (1, 1)

(x1 , x2 ) = c1 (1, 1) + c2 (2, −1).

(2, 1)

v1

Equating corresponding components in this equation yields the following linear system:

v x (2/3,1/3)

v2

c1 + 2c2 = x1 , c1 − c2 = x2 .

(2,1)

In this system, we view x1 and x2 as ﬁxed, while the variables we must solve for are c1 and c2 . The determinant of the matrix of coefﬁcients of this system is

Figure 4.4.1: The vector v = (2, 1) expressed as a linear combination of v1 = (1, 1) and v2 = (2, −1).

1 2 = −3. 1 −1 Since this is nonzero regardless of the values of x1 and x2 , the matrix of coefﬁcients is invertible, and hence for all (x1 , x2 ) ∈ R2 , 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 , so the given vectors do span R2 . Indeed, solving the linear system yields c1 = 13 (x1 + 2x2 ),

y c2v2 v2

v

c

v1 1

Hence,

v2 c2

v1

c2 = 13 (x1 − x2 ).

(x1 , x2 ) = 13 (x1 + 2x2 )v1 + 13 (x1 − x2 )v2 . For example, if v = (2, 1), then c1 = x illustrated in Figure 4.4.1.

c1v1

4 3

and c2 = 13 , so that v = 43 v1 + 13 v2 . This is

Figure 4.4.2: Any two noncollinear vectors in R2 span R2 .

More generally, any two nonzero and noncolinear vectors v1 and v2 in R2 span R2 , since, as illustrated geometrically in Figure 4.4.2, every vector in R2 can be written as a 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 .

Solution: Let v = (x1 , x2 , x3 ) be an arbitrary vector in R3 . We must determine whether there are real numbers c1 , c2 , c3 such that v = c1 v1 + c2 v2 + c3 v3

(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 −2 4 x1 0 1 −1 . −x1 − x2 0 0 0 7x1 + 11x2 + x3 It follows that the system is consistent if and only if x1 , x2 , x3 satisfy 7x1 + 11x2 + x3 = 0.

(4.4.4)

Consequently, Equation (4.4.3) holds only for those vectors v = (x1 , x2 , x3 ) in R3 whose components satisfy Equation (4.4.4). Hence, v1 , v2 , and v3 do not span R3 . 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 forms a subspace S of R3 , 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 they were coplanar. In general, any three noncoplanar vectors v1 , v2 , and v3 in R3 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

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, the question of whether or not the vectors v1 , v2 , and v3 span R3 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. Theorem 4.4.4

Let v1 , v2 , . . . , vk be vectors in Rn . Then {v1 , v2 , . . . , vk } spans Rn 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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Proof Rewriting the system Ac = v as the linear combination c1 v1 + c2 v2 + · · · + ck vk = v, we see that the existence of a solution (c1 , c2 , . . . , ck ) to this vector equation for each v in Rn is equivalent to the statement that {v1 , v2 , . . . , vk } spans Rn . Next, we consider a couple of examples involving vector spaces other than Rn . Example 4.4.5

Verify that A1 =

1 0 , 0 0

A2 =

1 1 , 0 0

A3 =

1 1 , 10

A4 =

11 11

span M2 (R).

Solution:

An arbitrary vector in M2 (R) is of the form a b A= . c d

If we write c1 A1 + c2 A2 + c3 A3 + c4 A4 = A, then equating the elements of the matrices on each side of the equation yields the system c1 + c2 + c3 + c4 c2 + c3 + c4 c3 + c4 c4

= a, = b, = c, = 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 M2 (R) is 1 0 0 1 0 0 0 0 , , , , 0 0 0 0 1 0 0 1

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 p(x) = a0 + a1 x + a2 x 2 .

If we let p0 (x) = 1,

p1 (x) = x,

p2 (x) = x 2 ,

then p(x) = a0 p0 (x) + a1 p1 (x) + a2 p2 (x). Thus, every vector in P2 is a linear combination of 1, x, and x 2 , and so a spanning set for P2 is {1, x, x 2 }. For practice, the reader might show that {x 2 , x + x 2 , 1 + x + x 2 } 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 }. We have span{v1 , v2 , . . . , vk } = {v ∈ V : v = c1 v1 + c2 v2 + · · · + ck vk , c1 , c2 , . . . , ck ∈ F }. (4.4.5) For example, suppose V = C 2 (I ), and let y1 (x) = sin x and y2 (x) = cos x. Then span{y1 , y2 } = {y ∈ C 2 (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 = a1 v1 + a2 v2 + · · · + ak vk

and

v = b1 v1 + b2 v2 + · · · + bk vk ,

for some scalars ai , bi . Thus, u + v = (a1 v1 + a2 v2 + · · · + ak vk ) + (b1 v1 + b2 v2 + · · · + bk vk ) = (a1 + b1 )v1 + (a2 + b2 )v2 + · · · + (ak + bk )vk = c1 v1 + c2 v2 + · · · + ck vk , 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(a1 v1 + a2 v2 + · · · + ak vk ) = (ra1 )v1 + (ra2 )v2 + · · · + (rak )vk = d1 v1 + d2 v2 + · · · + dk vk , 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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Example 4.4.8

Spanning Sets

263

If V = R2 and v1 = (−1, 1), determine span{v1 }.

Solution:

We have span{v1 } = {v ∈ R2 : v = c1 v1 , c1 ∈ R} = {v ∈ R2 : v = c1 (−1, 1), c1 ∈ R} = {v ∈ R2 : 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) c1v1 (1, 1)

The subspace of ⺢2 spanned by the vector v1 (1, 1) v1 x

Figure 4.4.4: The subspace of R2 spanned by v1 = (−1, 1).

Example 4.4.9

If V = R3 , v1 = (1, 0, 1), and v2 = (0, 1, 1), determine the subspace of R3 spanned by v1 and v2 . Does w = (1, 1, −1) lie in this subspace?

Solution:

We have

span{v1 , v2 } = {v ∈ R3 : v = c1 v1 + c2 v2 , c1 , c2 ∈ R} = {v ∈ R3 : v = c1 (1, 0, 1) + c2 (0, 1, 1), c1 , c2 ∈ R} = {v ∈ R3 : 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 ⺢3 spanned by v1 (1, 0, 1), v2 (0, 1, 1) v1

v2 y

x

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

Figure 4.4.5: The subspace of R3 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

A1 =

1 0 , 0 0

A2 =

0 1 , 1 0

A3 =

0 0 0 1

in M2 (R). Determine span{A1 , A2 , A3 }.

Solution:

By deﬁnition we have

span{A1 , A2 , A3 } = {A ∈ M2 (R) : A = c1 A1 + c2 A2 + c3 A3 , c1 , c2 , c3 ∈ R} 1 0 0 1 0 0 = A ∈ M2 (R) : A = c1 + c2 + c3 0 0 1 0 0 1 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 p1 (x) = 1 + 3x,

p2 (x) = x + x 2 ,

and decide whether {p1 , p2 } is a spanning set for P2 .

Solution:

We have

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

(4.4.6)

Since there is no x 2 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 + (3c1 + c2 )x + c2 x 2 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 Linear combination, Linear span, Spanning set.

8. If S is a spanning set for a vector space V , then any proper subset S of S is not a spanning set for V .

Skills

9. The vector space of 3 × 3 upper triangular matrices is spanned by the matrices Eij where 1 ≤ i ≤ j ≤ 3.

• Be able to determine whether a given set of vectors S spans a vector space V , and be able to prove your answer mathematically. • Be able to determine the linear span of a set of vectors. For vectors in Rn , be able to give a geometric description of the linear span. • If S is a spanning set for a vector space V , be able to write any vector in V as a linear combination of the elements of S. • Be able to construct a spanning set for a vector space V . As a special case, be able to determine a spanning set for the null space of an m × n matrix. • Be able to determine whether a particular vector v in a vector space V lies in the linear span of a set S of vectors in V .

True-False Review For Questions 1–12, decide if the given statement is true or false, and give a brief justiﬁcation for your answer. If true, you can quote a relevant deﬁnition or theorem from the text. If false, provide an example, illustration, or brief explanation of why the statement is false. 1. The linear span of a set of vectors in a vector space V forms a subspace of V . 2. If some vector v in a vector space V is a linear combination of vectors in a set S, then S spans V . 3. If S is a spanning set for a vector space V and W is a subspace of V , then S is a spanning set for W . 4. If S is a spanning set for a vector space V , then every vector v in V must be uniquely expressible as a linear combination of the vectors in S. 5. A set S of vectors in a vector space V spans V if and only if the linear span of S is V . 6. The linear span of two vectors in R3 is a plane through the origin. 7. Every vector space V has a ﬁnite spanning set.

10. A spanning set for the vector space P2 must contain a polynomial of each degree 0, 1, and 2. 11. If m < n, then any spanning set for Rn must contain more vectors than any spanning set for Rm . 12. The vector space P of all polynomials with real coefﬁcients cannot be spanned by a ﬁnite set S.

Problems For Problems 1–3, determine whether the given set of vectors spans R2 . 1. {(1, −1), (2, −2), (2, 3)}. 2. {(2, 5), (0, 0)}. 3. {(6, −2), (−2, 2/3), (3, −1)}. Recall that three vectors v1 , v2 , v3 in R3 are coplanar if and only if det([v1 , v2 , v3 ]) = 0. For Problems 4–6, use this result to determine whether the given set of vectors spans R3 . 4. {(1, −1, 1), (2, 5, 3), (4, −2, 1)}. 5. {(1, −2, 1), (2, 3, 1), (0, 0, 0), (4, −1, 2)}. 6. {(2, −1, 4), (3, −3, 5), (1, 1, 3)}. 7. Show that the set of vectors {(1, 2, 3), (3, 4, 5), (4, 5, 6)} does not span R3 , but that it does span the subspace of R3 consisting of all vectors lying in the plane with equation x − 2y + z = 0. 8. Show that v1 = (2, −1), v2 = (3, 2) span R2 , and express the vector v = (5, −7) as a linear combination of v1 , v2 . 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 combination of v1 , v2 , v3 .

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10. Show that v1 = (1, 1), v2 = (−1, 2), v3 = (1, 4) span R2 . Do v1 , v2 alone span R2 also?

For Problems 22–24, determine whether the given vector v lies in span{v1 , v2 }.

11. Let S be the subspace of R3 consisting of all vectors of the form v = (c1 , c2 , c2 − 2c1 ). Show that S is spanned by v1 = (1, 0, −2), v2 = (0, 1, 1).

22. v = (3, 3, 4), v1 = (1, −1, 2), v2 = (2, 1, 3) 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 a set of vectors that spans S. 13. Let S be the subspace of R3 consisting of all solutions to the linear system x − 2y − z = 0. Determine a set of vectors that spans S. For Problems 14–15, determine a spanning set for the null space of the given matrix A. 1 2 3 14. A = 3 4 5 . 5 6 7 1 2 3 5 15. A = 1 3 4 2 . 2 4 6 −1 16. Let S be the subspace of M2 (R) consisting of all symmetric 2 × 2 matrices with real elements. Show that S is spanned by the matrices 1 0 0 0 0 1 A1 = , A2 = , A3 = . 0 0 0 1 1 0 17. Let S be the subspace of M2 (R) consisting of all skewsymmetric 2 × 2 matrices with real elements. Determine a matrix that spans S. 18. Let S be the subset of M2 (R) consisting of all upper triangular 2 × 2 matrices.

23. v = (5, 3, −6), v1 = (−1, 1, 2), v2 = (3, 1, −4) in R3 . 24. v = (1, 1, −2), v1 = (3, 1, 2), v2 = (−2, −1, 1) in R3 . 25. If p1 (x) = x − 4 and p2 (x) = x 2 − x + 3, determine whether p(x) = 2x 2 − x + 2 lies in span{p1 , p2 }. 26. Consider the vectors 1 −1 0 1 , A2 = , A1 = 2 0 −2 1

27. Consider the vectors 12 A1 = , −1 3

20. v1 = (1, 2, −1), v2 = (−2, −4, 2). R3

spanned by the vectors 21. Let S be the subspace of v1 = (1, 1, −1), v2 = (2, 1, 3), v3 = (−2, −2, 2). Show that S also is spanned by v1 and v2 only.

A2 =

−2 1 1 −1

in M2 (R). Find span{A1 , A2 }, and determine whether or not 3 1 B= −2 4 lies in this subspace. 28. Let V = C ∞ (I ) and let S be the subspace of V spanned by the functions f (x) = cosh x,

g(x) = sinh x.

(a) Give an expression for a general vector in S. (b) Show that S is also spanned by the functions h(x) = ex ,

(b) Determine a set of 2 × 2 matrices that spans S.

19. v1 = (1, −1, 2), v2 = (2, −1, 3).

A3 =

30 1 2

in M2 (R). Determine span{A1 , A2 , A3 }.

(a) Verify that S is a subspace of M2 (R). For Problems 19–20, determine span{v1 , v2 } for the given vectors in R3 , and describe it geometrically.

j (x) = e−x .

For Problems 29–32, give a geometric description of the subspace of R3 spanned by the given set of vectors. 29. {0}. 30. {v1 }, where v1 is any nonzero vector in R3 . 31. {v1 , v2 }, where v1 , v2 are nonzero and noncollinear vectors in R3 .

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32. {v1 , v2 }, where v1 , v2 are collinear vectors in R3 . 33. Prove that if S and S are subsets of a vector space V such that S is a subset of S , then span(S) is a subset of span(S ).

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267

34. Prove that span{v1 , v2 , v3 } = span{v1 , v2 } if and only if v3 can be written as a linear combination of v1 and v2 .

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 Example 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 Example 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

3 Since 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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