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Bases and Dimension 281 i i “main” 2007/2/16 page 281 i i 4.6 Bases and Dimension 281 2 3 40. (a) Show that {1,x,x ,x } is linearly independent (a) If n>m, prove that {u1, u2,...,un} is linearly on every interval. dependent on V . k (b) If fk(x) = x for k = 0, 1,...,n, show that (b) If n = m, prove that {u1, u2,...,un} is linearly {f0,f1,...,fn} is linearly independent on every independent in V if and only if det[aij ] = 0. interval for all fixed n. (c) If n<m, prove that {u1, u2,...,un} is linearly 41. (a) Show that the functions independent in V if and only if rank(A) = n, r1x r2x r3x =[ ] f1(x) = e ,f2(x) = e ,f3(x) = e where A aij . have Wronskian (d) Which result from this section do these results generalize? 111 (r1+r2+r3)x W[f1,f2,f3](x) = e r1 r2 r3 r2 r2 r2 45. Prove from the definition of “linearly independent” 1 2 3 { } + + that if v1, v2,...,vn is linearly independent and = (r1 r2 r3)x − − − e (r3 r1)(r3 r2)(r2 r1), if A is an invertible n × n matrix, then the set {Av ,Av ,...,Av } is linearly independent. and hence determine the conditions on r1,r2,r3 1 2 n such that {f1,f2,f3} is linearly independent on { } every interval. 46. Prove that if v1, v2 is linearly independent and v3 { } { } (b) More generally, show that the set of functions is not in span v1, v2 , then v1, v2, v3 is linearly independent. {er1x,er2x,...,ernx} is linearly independent on every interval if and 47. Generalizing the previous exercise, prove that if { } only if all of the r are distinct. [Hint: Show that v1, v2,...,vk is linearly independent and vk+1 is i { } { } the Wronskian of the given functions is a multiple not in span v1, v2,...,vk , then v1, v2,...,vk+1 is of the n × n Vandermonde determinant, and then linearly independent. use Problem 21 in Section 3.3.] 48. Prove Theorem 4.5.2. 42. Let {v1, v2} be a linearly independent set in a vector = + = + space V , and let v αv1 v2, w v1 αv2, where 49. Prove Proposition 4.5.7. α is a constant. Use Definition 4.5.4 to determine all values of α for which {v, w} is linearly independent. 50. Prove that if {v1, v2,...,vk} spans a vector space V , { } 43. If v1 and v2 are vectors in a vector space V , and then for every vector v in V , v, v1, v2,...,vk is lin- u1, u2, u3 are each linear combinations of them, prove early dependent. that {u1, u2, u3} is linearly dependent. 51. Prove that if V = Pn and S ={p1,p2,...,pk} is a 44. Let v1, v2,...,vm be a set of linearly independent vec- set of vectors in V each of a different degree, then S is tors in a vector space V and suppose that the vectors linearly independent. [Hint: Assume without loss of u1, u2,...,un are each linear combinations of them. generality that the polynomials are ordered in descend- It follows that we can write ing degree: deg(p1)>deg(p2)>··· > deg(pk). m Assuming that c1p1 + c2p2 +···+ckpk = 0, first u = a v ,k= 1, 2,...,n, k ik i show that c1 is zero by examining the highest degree. i=1 Then repeat for lower degrees to show successively for appropriate constants aik. that c2 = 0, c3 = 0, and so on.] 4.6 Bases and Dimension The results of the previous section show that if a minimal spanning set exists in a (nontrivial) vector space V , it cannot be linearly dependent. Therefore if we are looking for minimal spanning sets for V , we should focus our attention on spanning sets that are linearly independent. One of the results of this section establishes that every spanning set for V that is linearly independent is indeed a minimal spanning set. Such a set will be i i i i i i “main” 2007/2/16 page 282 i i 282 CHAPTER 4 Vector Spaces called a basis. This is one of the most important concepts in this text and a cornerstone of linear algebra. DEFINITION 4.6.1 4 A set of vectors {v1, v2,...,vk} in a vector space V is called a basis for V if (a) The vectors are linearly independent. (b) The vectors span V . Notice that if we have a finite spanning set for a vector space, then we can always, in principle, determine a basis for V by using the technique of Corollary 4.5.12. Fur- thermore, the computational aspects of determining a basis have been covered in the previous two sections, since all we are really doing is combining the two concepts of linear independence and linear span. Consequently, this section is somewhat more the- oretically oriented than the preceding ones. The reader is encouraged not to gloss over the theoretical aspects, as these really are fundamental results in linear algebra. There do exist vector spaces V for which it is impossible to find a finite set of linearly independent vectors that span V . The vector space Cn(I), n ≥ 1, is such an example (Example 4.6.19). Such vector spaces are called infinite-dimensional vector spaces. Our primary interest in this text, however, will be vector spaces that contain a finite span- ning set of linearly independent vectors. These are known as finite-dimensional vector spaces, and we will encounter numerous examples of them throughout the remainder of this section. n 2 We begin with the vector space R .InR , the most natural basis, denoted {e1, e2}, consists of the two vectors e1 = (1, 0), e2 = (0, 1), (4.6.1) 3 and in R , the most natural basis, denoted {e1, e2, e3}, consists of the three vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). (4.6.2) The verification that the sets (4.6.1) and (4.6.2) are indeed bases of R2 and R3, respec- tively, is straightforward and left as an exercise.5 These bases are referred to as the standard basis on R2 and R3, respectively. In the case of the standard basis for R3 given in (4.6.2), we recognize the vectors e1, e2, e3 as the familiar unit vectors i, j, k pointing along the positive x-, y-, and z-axes of the rectangular Cartesian coordinate system. n More generally, consider the set of vectors {e1, e2,...,en} in R defined by e1 = (1, 0,...,0), e2 = (0, 1,...,0), ..., en = (0, 0,...,1). These vectors are linearly independent by Corollary 4.5.15, since det([e1, e2,...,en]) = det(In) = 1 = 0. n n Furthermore, the vectors span R , since an arbitrary vector v = (x1,x2,...,xn) in R can be written as v = x1(1, 0,...,0) + x2(0, 1,...,0) +···+xn(0, 0,...,1) = x1e1 + x2e2 +···+xnen. 4 The plural of basis is bases. 5Alternatively, the verification is a special case of that given shortly for the general case of Rn. i i i i i i “main” 2007/2/16 page 283 i i 4.6 Bases and Dimension 283 n Consequently, {e1, e2,...,en} is a basis for R . We refer to this basis as the standard basis for Rn. The general vector in Rn has n components, and the standard basis vectors arise as the n vectors that are obtained by sequentially setting one component to the value 1 and the other components to 0. In general, this is how we obtain standard bases in vector spaces whose vectors are determined by the specification of n independent constants. We illustrate with some examples. Example 4.6.2 Determine the standard basis for M2(R). Solution: The general matrix in M2(R) is ab . cd Consequently, there are four independent parameters that give rise to four special vectors in M2(R). Sequentially setting one of these parameters to the value 1 and the others to 0 generates the following four matrices: 10 01 00 00 A = ,A= ,A= ,A= . 1 00 2 00 3 10 4 01 We see that {A1,A2,A3,A4} is a spanning set for M2(R). Furthermore, c1A1 + c2A2 + c3A3 + c4A4 = 02 holds if and only if 10 01 00 00 00 c + c + c + c = 1 00 2 00 3 10 4 01 00 —that is, if and only if c1 = c2 = c3 = c4 = 0. Consequently, {A1,A2,A3,A4} is a linearly independent spanning set for M2(R), hence it is a basis. This is the standard basis for M2(R). Remark More generally, consider the vector space of all m × n matrices with real entries, Mm×n(R).IfweletEij denote the m×n matrix with value 1 in the (i, j)-position and zeros elsewhere, then we can show routinely that {Eij : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is a basis for Mm×n(R), and it is the standard basis. Example 4.6.3 Determine the standard basis for P2. Solution: We have 2 P2 ={a0 + a1x + a2x : a0,a1,a2 ∈ R}, so that the vectors in P2 are determined by specifying values for the three parameters a0,a1, and a2. Sequentially setting one of these parameters to the value 1 and the other two to the value 0 yields the following vectors in P2: 2 p0(x) = 1,p1(x) = x, p2(x) = x . i i i i i i “main” 2007/2/16 page 284 i i 284 CHAPTER 4 Vector Spaces We have shown in Example 4.4.6 that {p0,p1,p2} is a spanning set for P2.
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