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Example 179. Let V be the set of all real-valued continuous functions defined on R1 . If f and g are in V, we define f ⊕ g by (f ⊕ g)(t) = f (t) + g (t). If f in V and c is a scalar, we define c ⊙ f by (c ⊙ f )(t) = cf (t). Then V is a vector space, which is denoted by C(−∞, ∞). Example 180. Let V be the set of all real numbers with the operations u⊕v = u−v(⊕ is ordinary subtraction) and c⊙u = cu(⊙ ordinary multiplication). Is V a vector space? If it is not, which properties in Definition 172 fail to hold? Solution: If u and v are in V, and c is a scalar, then u ⊕ v and c ⊙ u are in V, so that (a) and (b) in Definition 172 hold. However, property (1) fails to hold, since u ⊕ v = u − v and v ⊕ u = v − u and these are not the same, in general. (Find u and v such that u − v 6= v − u). Also, we shall let the reader verify that properties (2), (3), and (4) fail to hold. Properties (5), (7), and (8) hold, but property (6) does not hold, because (c + d) ⊙ u = (c + d)u = cu + du, whereas c ⊙ u ⊕ d ⊙ u = cu ⊕ du = cu − du and these are not equal, in general. Thus V is not a vector space. Example 181. Let V be the set of all integers; define ⊕ as ordinary addition and ⊙ as ordinary multiplication. Here V is not a vector, because if u is any nonzero vector √ in V and c = 3, then c ⊙ u is not in V . Thus (b) fails to hold. Remark 182. To verify that a given set V with two operations ⊕ and ⊙ is a real vector space, we must show that it satisfies all the properties of Definition 172. • If both (a) and (b) hold, it is recommended that (3), the existence of a zero element, be verified next. Naturally, if (3) fails to hold, we do not have a vector space and do not have to check the remaining properties.

6.1

Subspaces

In this section we begin to analyze the structure of a vector space. First, it is convenient to have a name for a subset of a given vector space that is itself a vector space with respect to the same operations as those in V. Thus we have a definition. Definition 183. Let V be a vector space and W a nonempty subset of V . If W is a vector space with respect to the operations in V , then W is called a subspace of V .

69

Theorem 184. Let V be a vector space with operations ⊕ and ⊙ and let W be a nonempty subset of V . Then W is a subspace of V if and only if the following conditions hold: (a) If u and v are any vectors in W, then u ⊕ v is in W.

(b) If c is any real number and u is any vector in W, then c ⊙ u is in W.

Example 185. Every vector space has at least two subspaces, itself and the subspace {0} consisting only of the zero vector. (Recall that 0 ⊕ 0 = 0 and c ⊙ 0 = 0 in any vector space.) Thus {0} is closed for both operations and hence is a subspace of V . The subspace {0} is called the zero subspace of V . Example 186. Let P2 be the set consisting of all polynomials of degree ≤ 2 and the zero polynomial; P2 is a subset of P , the vector space of all polynomials. To verify that P2 is a subspace of P , show it is closed for ⊕ and ⊙. In general, the set Pn consisting of all polynomials of degree ≤ n and the zero polynomial is a subspace of P . Also, Pn is a subspace of Pn+1. Example 187. Let V be the set of all polynomials of degree exactly = 2; V is a subset of P , the vector space of all polynomials; but V is not a subspace of P , because the sum of the polynomials 2t2 + 3t + 1 and −2t2 + t + 2 is not in V , since it is a polynomial of degree 1.   a Example 188. Let W be the set of all vectors in R3 of the form  b  where a a+b and b are any real numbers. To verify Theorem 184 a) and (b), we let     a1 a2   and v =  b2 u =  b1 a 1 + b1 a 2 + b2

be two vectors in W . Then 

   a1 + a2 a1 + a2   = b1 + b2 b1 + a 2 u⊕v = (a1 + b1 ) + (a2 + b2 ) (a1 + a2 ) + (b1 + b2 )

is in W, for W consists of all those vectors whose third entry is the sum of the first two entries. Similarly,       a1 ca1 ca1   = = cb1 cb1 c ⊙  b1 a 1 + b1 c (a1 + b1 ) ca1 + cb1

is in W . Hence W is a subspace of R3 .

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Note 189. Henceforth, we shall usually denote u ⊕ v and c ⊙ u in a vector space V as u + v and cu, respectively. We can also show that a nonempty subset W of a vector space V is a subspace of V if and only if cu + dv is in W for any vectors u and v in W and any scalars c and d. Definition 190. Let v1 , v2 , . . . , vk be vectors in a vector space V. A vector v in V is called a linear combination of v1 , v2 , . . . , vk if v = a1 v1 + a2 v2 + · · · + ak vk =

k X

aj vj

j=1

for some real numbers a1 , a2 , . . . , ak . Example 191. In R3 let   1  v1 = 2  , 1 The vector



 1 v2 =  0  , 2

and



 1 v3 =  1  . 0



 2 v= 1 5

is a linear combination of v1 , v2 , and v3 if we can find real numbers a1 , a2 , and a3 so that a1 v1 + a2 v2 + a3 v3 = v. Substituting for v, v1 , v2 , and v3 , we have         1 1 1 2 a1  2  + a2  0  + a3  1  =  1  . 1 2 0 5

Equating corresponding entries leads to the linear system (verity) a1 + a2 + a3 = 2 2a1 + a3 = 1 a1 + 2a2 = 5.

Solving this linear system by the methods of previous Chapters gives (verify) a1 = 1, a2 = 2, and a3 = −1, which means that v is a linear combination of v1 , v2 , and v3 . Thus v = v1 + 2v2 − v3 . 71

6.2

Span

Linear combinations play an important role in describing vector spaces. The set of all possible linear combinations of a pair of vectors in a vector space V gives us a subspace. We have the following definition to help with such constructions: Definition 192. If S = {v1 , v2 , . . . , vk } is a set of vectors in a vector space V , then the set of all vectors in V that are linear combinations of the vectors in S is denoted by span S = hSi = {a1 v1 + a2 v2 + . . . + ak vk | a1 , a2 , . . . , ak ∈ R}. Example 193. Consider the set S of 2 × 3 matrices given by         1 0 0 0 1 0 0 0 0 0 0 0 S= , , , . 0 0 0 0 0 0 0 1 0 0 0 1 Then span S is the set in M23 consisting of all vectors of the form           a b 0 0 1 0 0 0 0 0 0 0 1 0 0 +b +c +d = , where a, b, c, d ∈ R. a 0 0 1 0 c d 0 0 0 0 0 0 0 1 0 That is, span S is the subset of M23 consisting of all matrices of the form   a b 0 0 c d where a, b, c, and d are real numbers. Theorem 194. Let S = {v1 , v2 , . . . , vk } be a set of vectors in a vector space V . Then span S is a subspace of V . Proof. Let u=

k X

and

aj vj

w=

k X

bj vj

j=1

j=1

for some real numbers a1 , a2 , . . . , ak and b1 , b2 , . . . , bk . We have u+w =

k X

aj vj +

j=1

k X

bj vj =

j=1

k X

(aj + bj ) vj .

j=1

Moreover, for any real number c, cu = c

k X j=1

aj vj

!

=

k X

(caj ) vj .

j=1

Since the sum u + w and the scalar multiple cu are linear combinations of the vectors in S, then span S is a subspace of V . 72

Definition 195. Let S be a set of vectors in a vector space V . If every vector in V is a linear combination of the vectors in S , then the set S is said to span V , or V is spanned (generated) by the set S; that is, span S = V . Remark 196. If span S = V , S is called a spanning set of V . A vector space can have many spanning sets. Example 197. In R3 , let 

   2 1 v1 =  1  and v2 =  −1  . 1 3

Determine whether the vector

belongs to span{v1 , v2 }.



 1 v = 5  −7

Solution: If we can find scalars a1 , a2 such that v = a1 v1 + a2 v2 , then v belongs to span{v1 , v2 }. Substituting for v1 , v2 and v, we have       2 1 1 a1  1  + a2  −1  =  5  . 1 3 −7 This expression corresponds to the linear system whose augmented matrix is (verify)   2 1 1  1 −1 5 . 1 3 −7 The reduced row echelon form of this system is (verify)   1 0 2  0 1 −3  , 0 0 0

which indicates that the linear system is consistent, a1 = 2, and a2 = 3. Hence v belongs to span{v1 , v2 }. Example 198. In P2 , let v1 = 2t2 + t + 2,

v2 = t2 − 2t,

v3 = 5t2 − 5t + 2,

Determine whether the vector v = t2 + t + 2 73

v4 = −t2 − 3t − 2.

belongs to span {v1 , v2 , v3 , v4 } .

Solution: If we can find scalars a1 , a2 , a3 , and a4 so that a1 v1 + a2 v2 + a3 v3 + a4 v4 = v then v belongs to span {v1 , v2 , v3 , v4 } . Substituting for v1 , v2 , v3 , and v4 , we have         a1 2t2 + t + 2 + a2 t2 − 2t + a3 5t2 − 5t + 2 + a4 −t2 − 3t − 2 = t2 + t + 2,

or

(2a1 + a2 + 5a3 − a4 ) t2 + (a1 − 2a2 − 5a3 − 3a4 ) t + (2a1 + 2a3 − 2a4 ) = t2 + t + 2. Now two polynomials agree for all values of t only if the coefficients of respective powers of t agree. Thus we get the linear system 2a1 + a2 + 5a3 − a4 = 1

a1 − 2a2 − 5a3 − 3a4 = 1 2a1 + 2a3 − 2a4 = 2. To determine whether this system of linear equations is consistent, we form the augmented matrix and transform it to reduced row echelon form, obtaining (verify)   1 0 1 −1 0  0 1 3 1 0 , 0 0 0 0 1

which indicates that the system is inconsistent; that is, it has no solution. Hence v does not belong to span{v1 , v2 , v3 , v4 }. Remark 199. In general, to determine whether a specific vector v belongs to span S , we investigate the consistency of an appropriate linear system. Example 200. Let V be the vector space R3 . Let     1 1 v1 =  2  , v2 =  0  , and 1 2

 1 v3 =  1  . 0 



 a To find out whether v1 , v2 , v3 span V , we pick any vector v =  b  in V (a, b and c c are arbitrary real numbers) and determine whether there are constants a1 , a2 , and a3 such that a1 v1 + a2 v2 + a3 v3 = v. 74

This leads to the linear system (verify) a1 + a2 + a3 = a 2a1 + a3 = b a1 + 2a2 = c. A solution is (verify) −2a + 2b + c a−b+c 4a − b − 2c . , a2 = , a3 = 3 3 3 Thus v1 , v2 , v3 span V . This is equivalent to saying that span {v1 , v2 , v3 } = R3 . a1 =

Example 201. Let V be the vector space P2 . Let v1 = t2 + 2t + 1 and v2 = t2 + 2. Does {v1 , v2 } span V ? Solution: Let v = at2 + bt + c be any vector in V, where a, b, and c are any real numbers. We must find out whether there are constants a1 and a2 such that a1 v1 + a2 v2 = v, or Thus

    a1 t2 + 2t + 1 + a2 t2 + 2 = at2 + bt + c.

(a1 + a2 ) t2 + (2a1 ) t + (a1 + 2a2 ) = at2 + bt + c. Equating the coefficients of respective powers of t, we get the linear system a1 + a2 = a 2a1 = b a1 + 2a2 = c. Transforming the augmented matrix  1  0 0

of this linear system, we obtain (verify)  0 2a − c . 1 c−a 0 b − 4a + 2c

If b − 4a + 2c 6= 0, then the system is inconsistent and there is no solution. Hence {v1 , v2 } does not span V.

6.3

Linear Independence

Definition 202. The vectors v1 , v2 , . . . , vk in a vector space V are said to be linearly dependent if there exist constants a1 , a2 , . . . , ak , not all zero, such that k X j=1

aj vj = a1 v1 + a2 v2 + · · · + ak vk = 0. 75

(15)

Otherwise, v1 , v2 , . . . , vk are called linearly independent. That is, v1 , vk , . . . , vk are linearly independent if, whenever a1 v1 + a2 v2 + · · · + ak vk = 0, a1 = a2 = · · · = ak = 0. If S = {v1 , v2 , . . . , vk } , then we also say that the set S is linearly dependent or linearly independent if the vectors have the corresponding property. It should be emphasized that for any vectors v1 , v2 , . . . , vk , Equation (15) always holds if we choose all the scalars a1 , a2 , · · · , ak equal to zero. The important point in this definition is whether it is possible to satisfy (15) with at least one or the scalars different from zero. Example 203. Determine whether the vectors     3 1 v1 =  2  , v2 =  2  , 1 0

are linearly independent.

 −1 v3 =  2  −1 

Solution: Forming Equation (15),         3 1 −1 0        2 = 0 , a1 2 + a2 2 + a3 1 0 −1 0

we obtain the homogeneous system (verify)

3a1 + a2 − a3 = 0 2a1 + 2a2 + 2a3 = 0 a1 − a3 = 0 The corresponding augmented matrix is   3 1 −1 0  2 2 2 0 , 1 0 −1 0 whose reduced row echelon form is (verify)   1 0 −1 0  0 1 2 0 . 0 0 0 0 76

Thus there is a nontrivial solution 

 k  −2k  , k 6= 0 (verify), k

so the vectors are linearly dependent.     Example 204.   Are4 the vectors v1 = 1 0 1 2 , v2 = 0 1 1 2 , and v3 = 1 1 1 3 in R linearly dependent or linearly independent? Solution: We form Equation (15),

a1 v1 + a2 v2 + a3 v3 = 0 and solve for a1 , a2 , and a3 . The resulting homogeneous system is (verify) and solve for a1 , a2 , and a3 . The resulting homogeneous system is (verify) a1 + a3 = 0 a2 + a3 = 0 a1 + a2 + a3 = 0 2a1 + 2a2 + 3a3 = 0. The corresponding augmented matrix is  1  0   1 2

(verify) 0 1 1 2

1 1 1 3

 0 0   0  0

and its reduced row echelon form is (verify)  1 0 0  0 1 0   0 0 1 0 0 0

 0 0  . 0  0

Example 205. Are the vectors   2 1 , v1 = 0 1



Thus the only solution is the trivial solution a1 = a2 = a3 = 0, so the vectors are linearly independent.

v2 =



1 2 1 0

in M22 linearly independent? 77

,

v3 =



0 −3 −2 1



Solution: We form Equation (15),         0 −3 1 2 0 0 2 1 = + a3 , + a2 a1 −2 1 1 0 0 0 0 1

and solve for a1 , a2 , and a3 . Performing the scalar multiplications and adding the resulting matrices gives     0 0 2a1 + a2 a1 + 2a2 − 3a3 = . a2 − 2a3 a1 + a3 0 0

Using the definition for equal matrices, we have the linear system 2a1 + a2 = 0 a1 + 2a2 − 3a3 = 0 . a2 − 2a3 = 0 a1 + a3 = 0

The corresponding augmented matrix is  2 1 0 0  1 2 −3 0   0 1 −2 0 1 0 1 0

and its reduced row echelon form is (verify)  1 0 1  0 1 −2   0 0 0 0 0 0

Thus there is a nontrivial solution   −k  2k  , k



 , 

 0 0  . 0  0

k 6= 0 (verify)

so the vectors are linearly dependent.

Example 206. Are the vectors v1 = t2 + t + 2, v2 = 2t2 + t, and v3 = 3t2 + 2t + 2 in P2 linearly dependent or linearly independent? Solution: Forming Equation (15), we have (verify) a1 + 2a2 + 3a3 = 0 a1 + a2 + 2a3 = 0 2a1 + 2a3 = 0, which has infinitely many solutions (verify). A particular solution is a1 = 1, a2 = 1, a3 = −1, so, v1 + v2 − v3 = 0. Hence the given vectors are linearly dependent. 78

Theorem 207. Let S = {v1 , v2 , . . . , vn } be a set of n vectors in Rn (Rn ) . Let A be the matrix whose columns (rows) are the elements of S. Then S is linearly independent if and only if det(A) 6= 0.       Example 208. Is S = 1 2 3 , 0 1 2 , 3 0 −1 a linearly independent 3 set of vectors in R ? Solution: We form the matrix A whose rows are the vectors in S :   1 2 3 A = 0 1 2 . 3 0 −1

Since det(A) = 2 (verify), we conclude that S is linearly independent.

Theorem 209. Let S1 and S2 be finite subsets of a vector space and let S1 be a subset of S2 . Then the following statements are true: a) If S1 is linearly dependent , so is S2 . b) If S2 is linearly independent , so is S1 . Summary 210. At this point, we have established the following results: • The set S = {0} consisting only of 0 is linearly dependent, since, for example, 50 = 0, and 5 6= 0. • From this it follows that if S is any set of vectors that contains 0 , then S must be linearly dependent. • A set of vectors consisting of a single nonzero vector is linearly independent (verify). • If v1 , v2 , . . . , vk are vectors in a vector space V and any two of them are equal, then v1 , v2 , . . . , vk are linearly dependent (verify). Theorem 211. The nonzero vectors v1 , v2 , . . . , vn in a vector space V are linearly dependent if and only if one of the vectors vj (j ≥ 2) is a linear combination of the preceding vectors v1 , v2 , . . . , vj−1 .     Example 212.   Let V = R3 and also  v1 = 1 2 −1 , v2 = 1 −2 1 , v3 = −3 2 −1 and v4 = 2 0 0 . We find (verify) that v1 + v2 + 0v3 − v4 = 0

so v1 , v2 , v3 , and v4 are linearly dependent. We then have v4 = v1 + v2 + 0v3 . 79

Remark 213. 1) We observe that Theorem 211 does not say that every vector v is a linear combination of the preceding vectors. Thus, in Example 212, we also have v1 + 2v2 + v3 + 0v4 = 0. We cannot solve, in this equation, for v4 as a linear combination of v1 , v2 , and v3 , since its coefficient is zero. 2) We can also prove that if S = {v1 , v2 , . . . , vk } is a set of vectors in a vector space V, then S is linearly dependent if and only if one of the vectors in S is a linear combination of all the other vectors in S. For instance, in Example 212 v1 = −v2 − 0v3 + v4 ;

1 1 v2 = − v1 − v3 − 0v4 . 2 2

3) Observe that if v1 , v2 , . . . , vk are linearly independent vectors in a vector space, then they must be distinct and nonzero. Suppose that S = {v1 , v2 , . . . , vn } spans a vector space V, and vj is a linear combination of the preceding vectors in S. Then the set S1 = {v1 , v2 , . . . , vj−1 , vj+1, . . . , vn } consisting of S with vj deleted, also spans V. To show this result, observe that if v is any vector in V, then, since S spans V, we can find scalars a1 , a2 , . . . , an such that v = a1 v1 + a2 v2 + · · · + aj−1 vj−1 + aj vj + aj+1vj+1 + · · · + an vn . Now if vj = b1 v1 + b2 v2 + · · · + bj−1 vj−1 , then v = a1 v1 + a2 v2 + · · · + aj−1 vj−1 + aj (b1 v1 + b2 v2 + · · · + bj−1 vj−1 )

+ aj+1vj+1 + · · · + an vn =c1 v1 + c2 v2 + · · · + cj−1 vj−1 + cj+1vj+1 + · · · + cn vn

which means that span S1 = V.

Example 214. Consider the set of vectors S = {v1 , v2 , v3 , v4 } in        2 0 1 1 1  1   0  1        v1 =   0  , v2 =  1  , v3 =  1  , v4 =  1 0 0 0 0 and let W = span S. Since

v4 = v1 + v2 , we conclude that W = span S1 , where S1 = {v1 , v2 , v3 } . 80

R4 , where    

7

Basis and Dimension

In this section we continue our study of the structure of a vector space V by determining a set of vectors in V that completely describes V .

7.1

Basis

Definition 215. The vectors v1 , v2 , . . . , vk in a vector space V are said to form a basis for V if a) v1 , v2 , . . . , vk span V and b) v1 , v2 , . . . , vk are linearly independent. Remark 216. If v1 , v2 , . . . , vk forms a basis for a vector space V , then they must be distinct and nonzero.       1 0 0 Example 217. Let V = R3 . The vectors  0  ,  1  ,  0  form a basis for R3 , 0 0 1 3 called the natural basis or standard basis, for R . We can readily   see how to gener-  alize this to obtain the natural basis for Rn . Similarly, 1 0 0 , 0 1 0 , 0 0 1 is the natural basis for R3 . The natural basis for Rn is denoted by {e1 , e2 , . . . , en } , where  0  ...     0      1  ← ith row;  0    .   ..  0 

that is, ei is an n × 1 matrix with a 1 in the ith row and zeros elsewhere. The natural basis for R3 is also often denoted by       1 0 0      i = 0 , j = 1 , and k = 0  . 0 0 1

Example 218. Show that the set S = {t2 + 1, t − 1, 2t + 2} is a basis for the vector space P2 . Solution: To do this, we must show that S spans V and is linearly independent. To show that it spans V, we take any vector in V, that is, a polynomial at2 + bt + c, and 81

find constants a1 , a2 and a3 such that   at2 + bt + c = a1 t2 + 1 + a2 (...