线性代数教学资料chapter3

1、li jie3 3 the vector space r the vector space rn n 3.2 vector space properties of rn 3.3 examples of subspaces 3.4 bases for subspaces 3.5 dimension 3.6 orthogonal bases for subspacescore sectionsli jiein mathematics and the physical sciences, the term vector is applied to a wide variety of objects.

2、 perhaps the most familiar application of the term is to quantities, such as force and velocity, that have both magnitude and direction. such vectors can be represented in two space or in three space as directed line segments or arrows. as we will see in chapter 5,the term vector may also be used to

3、 describe objects such as matrices , polynomials, and continuous real-valued functions. 3.1 introductionli jiein this section we demonstrate that rn, the set of n-dimensional vectors, provides a natural bridge between the intuitive and natural concept of a geometric vector and that of an abstract ve

4、ctor in a general vector space.li jie3.2 vector space properties of rn.numbers real ,:2121nnnxxxxxxxxrli jiethe definition of subspaces of rna subset w of rn is a subspace of rn if and only if the following conditions are met:(s1)* the zero vector, , is in w.(s2)x+y is in w whenever x and y are in w

5、.(s3)ax is in w whenever x is in w and a is any scalar.li jieexample 1: let w be the subset of r3 defined by.numbers realany and ,:32321321xxxxxxxxxxwverify that w is a subspace of r3 and give a geometric interpretation of w.solution:li jiestep 1. an algebraic specification for the subset w is given

6、, and this specification serves as a test for determining whether a vector in rn is or is not in w.step 2.test the zero vector, , of rn to see whether it satisfies the algebraic specification required to be in w. (this shows that w is nonempty.)verifying that w is a subspace of rnli jiestep 3.choose

7、 two arbitrary vectors x and y from w. thus x and y are in rn, and both vectors satisfy the algebraic specification of w.step 4. test the sum x+y to see whether it meets the specification of w.step 5. for an arbitrary scalar, a, test the scalar multiple ax to see whether it meets the specification o

8、f w. li jieexample 3: let w be the subset of r3 defined by.1:21numbers realany x and x ,21xxxxwshow that w is not a subspace of r3.example 2: let w be the subset of r3 defined by .,:321number realany x ,3xx ,2xx11312xxxxxwverify that w is a subspace of r3 and give a geometric interpretation of w.li

9、jieexample 4:let w be the subset of r2 defined by.,:21integersany x and x21xxxxwdemonstrate that w is not a subspace of r2.example 5:let w be the subset of r2 defined by.,:21 0 x or 0 x21eitherwherexxxxwdemonstrate that w is not a subspace of r2.exercise p175 18 32li jie3.3 examples of subspacesin t

10、his section we introduce several important and particularly useful examples of subspaces of rn.li jiethe span of a subsettheorem 3:if v1, vr are vectors in rn, then the set w consisting of all linear combinations of v1, ,vr is a subspace of rn.if s=v1, ,vr is a subset of rn, then the subspace w cons

11、isting of all linear combinations of v1, ,vr is called the subspace spanned by s and will be denoted by sp(s) or spv1, ,vr.li jiefor example:(1)for a single vector v in rn, spv is the subspace spv=av: a is any real number .(2)if u and v are noncollinear geometric vectors, then spu,v=au+bv: a,b any r

12、eal numbers(3) if u, v, w are vectors in r3,and are not on the same space,then spu,v,w=au+bv+cw : a,b,c any real numbersli jieexample 1: let u and v be the three-dimensional vectors210 vand 012udetermine w=spu,v and give a geometric interpretation of w.li jiethe null space of a matrixwe now introduc

13、e two subspaces that have particular relevance to the linear system of equations ax=b, where a is an (mn) matrix. the first of these subspaces is called the null space of a (or the kernel of a) and consists of all solutions of ax=.definition 1:let a be an (m n) matrix. the null space of a denoted n(

14、a) is the set of vectors in rn defined by n(a)=x:ax= , x in rn.theorem 4:if a is an (m n) matrix, then n(a) is a subspace of rn.li jieexample 2:describe n(a), where a is the (3 4) matrix.142145121311asolution: n(a) is determined by solving the homogeneous system ax= .this is accomplished by reducing

15、 the augmented matrix a| to echelon form. it is easy to verify that a| is row equivalent to.000000211003201li jiesolving the corresponding reduced system yields x1=-2x3-3x4 x2=-x3+2x4,1023011223243434343xxxxxxxxxwhere x3 and x4 are arbitrary; that is,.numbers realany and ,10230112:)(4343xxxxxxanli j

16、ieexample 5:let s=v1,v2,v3,v4 be a subset of r3, where.1-52 v ,541 v, ,532 v,1214321andvshow that there exists a set t=w1,w2 consisting of two vectors in r3 such that sp(s)=sp(t).solution: let .155154322121ali jieset row operation to a and reduce a to the following matrix: 00001210450100001210212136

17、3012102121.155154322121aso, sp(s)=av1+bv2:a,b any real numberbecause sp(t)=sp(s), then sp(t)=av1+bv2:a,b any real numberfor example, we set li jie7015322121323211vvw31053212122212vvwli jie.155154322121athe solution on p184,152541532121taand the row vectors of at are precisely the vectors v1t,v2t,v3t

18、, and v4t. it is straightforward to see that at reduces to the matrix.000000310701tbso, by theorem 6, at and bt have the same row space. thus a and b have the same column space whereli jie.31-0 wand 70121w.003700100001bin particular, sp(s)=sp(t), where t=w1,w2,li jietwo of the most fundamental conce

19、pts of geometry are those of dimension and the use of coordinates to locate a point in space. in this section and the next, we extend these notions to an arbitrary subspace of rn by introducing the idea of a basis for a subspace.3.4 bases for subspacesli jiean example from r2 will serve to illustrat

20、e the transition from geometry to algebra. we have already seen that each vector v in r2,bavcan be interpreted geometrically as the point with coordinates a and b. recall that in r2 the vectors e1 and e2 are defined byli jie.10e and 0121eclearly the vector v in (1) can be expressed uniquely as a lin

21、ear combination of e1 and e2: v=ae1+be2 (2)li jieas we will see later, the set e1,e2 is an example of a basis for r2 (indeed, it is called the natural basis for r2). in eq.(2), the vector v is determined by the coefficients a and b (see fig.3.12). thus the geometric concept of characterizing a point

22、 by its coordinates can be interpreted algebraically as determining a vector by its coefficients when the vector is expressed as a linear combination of “basis” vectors.li jiespanning sets let w be a subspace of rn, and let s be a subset of w. the discussion above suggests that the first requirement

23、 for s to be a basis for w is that each vector in w be expressible as a linear combination of the vectors in s. this leads to the following definition.li jiedefinition 3:let w be a subspace of rn and let s=w1,wm be a subset of w. we say that s is a spanning set for w, or simply that s spans w, if ev

24、ery vector w in w can be expressed as a linear combination of vectors in s; w=a1w1+amwm.li jiea restatement of definition 3 in the notation of the previous section is that s is a spanning set of w provided that sp(s)=w. it is evident that the set s=e1,e2,e3, consisting of the unit vectors in r3, is

25、a spanning set for r3. specifically, if v is in r3,cbavthen v=ae1+be2+ce3. the next two examples consider other subset of r3.li jieexample 1: in r3, let s=u1,u2,u3, where.421u and ,132-u ,011321udetermine whether s is a spanning set for r3.,410231121|cbava.1003440107910001cbacbacbasolution: the augm

26、ented matrix this matrix is row equivalent toli jieexample 2: let s=v1,v2,v3 be the subset of r3 defined by.072 vand ,7-01- v,321321vdoes s span r3?,073702211|cbavasolution:and the matrix a|v is row equivalent toli jie.27000)2/1 (2/3102/2/701cbabab111 w vector theexample, .0c2b7a- where,:)(forcbavvs

27、spso,is in r3 but is not in sp(s); that is, w cannot be expressed as a linear combination of v1, v2,and v3.li jiethe next example illustrates a procedure for constructing a spanning set for the null space, n(a), of a matrix a.example 3:let a be the (34) matrix.142145121311aexhibit a spanning set for

28、 n(a), the null space of a.solution: the first step toward obtaining a spanning set for n(a) is to obtain an algebraic specification for n(a) by solving the homogeneous system ax=.li jie.numbers realany and ,232:)(43434343xxxxxxxxxxan)8.(102301120230223243444333434343xxxxxxxxxxxxxxxlet u1 and u2 be

29、the vectorsli jie.1023u and 011221utherefore, n(a)=spu1,u2li jieminimal spanning sets if w is a subspace of rn, w, then spanning sets for w abound. for example a vector v in a spanning set can always be replaced by av, where a is any nonzero scalar. it is easy to demonstrate, however, that not all s

30、panning sets are equally describe. for example, define u in r2 by.11uthe set s=e1,e2,u is a spanning set for r2. indeed, for an arbitrary vector v in r2,bavli jiev=(a-c)e1+(b-c)e2+cu, where c is any real number whatsoever. but the subset e1,e2 already spans r2, so the vector u is unnecessary .recall

31、 that a set v1,vm of vectors in rn is linearly independent if the vector equation x1v1+xmvm= (9)has only the trivial solution x1=xm=0; if eq.(9) has a nontrivial solution, then the set is linearly dependent. the set s=e1,e2,u is linearly dependent because e1+e2-u=.li jieour next example illustrates

32、that a linearly dependent set is not an efficient spanning set; that is, fewer vectors will span the same space.example 4: let s=v1,v2,v3 be the subset of r3, where.153 vand ,132,111321vvshow that s is a linearly dependent set, and exhibit a subset t of s such that t contains only two vectors but sp

33、(t)=sp(s).li jiesolution: the vector equation x1v1+x2v2+x3v3= (10)is equivalent to the (3 3) homogeneous system of equations with augmented matrix011105310321ali jiematrix is row equivalent to000002100101bso v3=-1v1+2v2)(,:)(21sspnumberrealanybabvavtspli jie on the other hand, if b=v1,vm is a linear

34、ly independent spanning set for w, then no vector in b is a linear combination of the other m-1 vectors in b.the lesson to be drawn from example 4 is that a linearly dependent spanning set contains redundant information. that is, if s=w1,wr is a linearly dependent spanning set for a subspace w, then

35、 at least one vector from s is a linear combination of the other r-1 vectors and can be discarded from s to produce a smaller spanning set.li jie hence if a vector is removed from b, this smaller set cannot be a spanning set for w (in particular, the vector removed from b is in w but cannot be expre

36、ssed as a linear combination of the vectors retained). in this sense a linearly independent spanning set is a minimal spanning set and hence represents the most efficient way of characterizing the subspace. this idea leads to the following definition.definition 4:let w be a nonzero subspace of rn. a

37、 basis for w is a linearly independent spanning set for w.li jieuniqueness of representationremark let b=v1,v2, ,vp be a basis for w, where w is a subspace of rn. if x is in w, then x can be represented uniquely in terms of the basis b. that is, there are unique scalars a1,a2, ,ap such that x=a1v1+a

38、2v2+apvp.as we see later, these scalars are called the coordinates x with respect to the basis.example of basesit is easy to show that the unit vectors ,100 ,010 ,001321eeeis a basis for r3li jiein general, the n-dimensional vectors e1,e2,en form a basis for rn, frequently called the natural basis.

39、,111 ,011 ,001321vvvprovide another basis for r3.and the vectorsli jieexample 6:let w be the subspace of r4 spanned by the set s=v1,v2,v3,v4,v5, where ,2052 ,1401 ,5141 ,1121 ,121154321vvvvvfind a subset of s that is a basis for w.solution: 000001100020310102-01 21511041125042121111aso v1,v2,v4is a

40、basis for w. li jiethe procedure demonstrated in the preceding example can be outlined as follows:1.a spanning set sv1,vm for a subspace w is given.2.solve the vector equation x1v1+xmvm= (20)3.if eq.(20) has only the trivial solution x1=xm=0, then s is a linearly independent set and hence is a basis

41、 for w.4.if eq.(20) has nontrivial solutions, then there are unconstrained variables. for each xj that is designated as an unconstrained variable, delete the vector vj from the set s. the remaining vectors constitute a basis for w.li jietheorem 7:if the nonzero matrix a is row equivalent to the matr

42、ix b in echelon form, then the nonzero rows of b form a basis for the row space of a.li jieli jietheorem 8: let w be a subspace of rn, and let b=w1,w2,wp be a spanning set for w containing p vectors. then an set of p+1 or more vectors in w is linearly dependent.as an immediate corollary of theorem 8

43、, we can show that all bases for a subspace contain the same number of vectors.corollary: let w be a subspace of rn, and let b=w1,w2,wp be a basis for w containing p vectors. then every basis for w contains p vectors.3.5 dimension li jiedefinition 5 : let w be a subspace of rn. if w has a basis b=w1

44、,w2,wp of p vectors, then we say that w is a subspace of dimension p, and we write dim(w)=p. since r3 has a basis e1,e2,e3 containing three vectors, we see that dim(r3)=3. in general, rn has a basis e1,e2,en that contains n vectors; so dim(rn)=n. thus the definition of dimension the number of vector

45、s in a basis agrees with the usual terminology; r3 is threedimensional, and in general, rn is n-dimensional.li jieexample 1: let w be the subspace of r3 defined by.arbitrary ,2,:33231321xxxxxxxxxxwexhibit a basis for w and determine dim(w).li jiesolution: a vector x in w can be written in the form.1

46、1223333xxxxxtherefore, the set u is a basis for w, where.112uli jieexample 2 let w be the subspace of r3, w=spanu1,u2,u3,u4,where252,253,042,2114321uuuufind three different bases for w and give the dimension of w.li jietheorem 9:let w be a subspace of rn with dim(w)=p.1.any set of p+1 or more vector

47、s in w is linearly dependent.2.any set of fewer than p vectors in w does not span w.3.any set of p linearly independent vectors in w is a basis for w.4.any set of p vectors that spans w is a basis for w.properties of a p-dimensional subspaceli jieexample 3: let w be the subspace of r3 given in examp

48、le 2, and let v1,v2,v3 be the subset of w defined by.612 v,021 v,611321vdetermine which of the subsets v1 v2 v1,v2 v1,v3 v2,v3,and v1,v2,v3 is a basis for w.li jiethe rank of matrix in this subsection we use the concept of dimension to characterize nonsingular matrices and to determine precisely whe

49、n a system of linear equation ax=b is consistent. for an (mn) matrix a, the dimension of the null space is called the nullity of a, and the dimension of the range of a is called the rank of a.example 4: find the rank, nullity, and dimension of the row space for the matrix a, where.584232012111ali ji

50、esolution: to find the dimension of the row space of a, observe that a is row equivalent to the matrix,100003100201band b is in echelon form. since the nonzero rows of b form a basis for the row space of a, the row space of a has dimension 3.number realany ,032:)(3333xxxxxxanli jieit now follows tha

51、t the nullity of a is 1 because the vector0132va is row equivalent to matrix c, where,010000100001cform a basis for r(a). thus the rank of a is 3forms a basis for n(a).li jienote in the previous example that the row space of a is a subspace of r4, whereas the column space (or range) of a is a subspa

52、ce of r3. thus they are entirely different subspaces; even so, the dimensions are the same, and the next theorem states that this is always the case.theorem 10: if a is an (mn) matrix, then the rank of a is equal to the rank of at.remark: if a is an (m n) matrix, then n=rank(a)+nullity(a).li jietheo

53、rem 11: an (m n) system of linear equations , ax=b, is consistent if and only if rank(a)=rank(a|b).theorem 12: an (n n) matrix a is nonsingular if and only if the rank of a is n.the following theorem uses the concept of the rank of a matrix to establish necessary and sufficient conditions for a syst

54、em of equations, ax=b, to be consistent.li jie3.6 orthogonal bases for subspaces orthogonal bases the idea of orthogonality is a generalization of the vector geometry concept of perpendicularity. if u and v are two vectors in r2 or r3, then we know that u and v are perpendicular if utv=0 . for examp

55、le, consider the vectors u and v given by 36,21vanduli jie clearly utv=0 , and these two vectors are perpendicular when viewed as directed line segments in the plane. in general , for vectors in rn, we use the term orthogonal rather than the term perpendicular. specially, if u and v are vectors in r

56、n, we say that u and v are orthogonal if utv=0 we will also find the concept of an orthogonal set of vectors to be useful.li jiedefinition 6: let s =u1 u2 , , up, be a set of vectors in rn, the set s is said to be an orthogonal set if each pair of distinct vectors form s is orthogonal; that isjiwhen

57、uujti0example 1 verify that s is an orthogonal set of vectors , where0121,0111,2101sli jietheorem 13 : let s =u1 u2 , , up, be a set of nonzero vectors in rn,. if s is an orthogonal set of vectors , then s is a linearly independent set of vectors.proof:li jiedefinition 7: let w be a subspace of rn,

58、and let b=u1,u2 ,up be a basis for w. if b is an orthogonal set of vectors, then b is called an orthogonal basis for w. furthermore, ifpiforui11then b is said to be an orthonormal basis for wthe word orthonormal suggests both orthogonal and normalized. thus an orthonormal basis is an orthogonal basi

59、s consisting of vectors having length 1, where a vector of length 1 is a unit vector or a normalized vector. observe that the unit vectors ei form an orthonormal basis for rn.li jieexample 2 verify that the set b=v1 v2 v3, is an orthogonal basis for r3,where741113,121321vvvcorollary: let w be a subs

60、pace of rn, where dim(w)=p, if s is an orthogonal set of p nonzero vectors and is also a subset of w, then s is an orthogonal basis for w.li jieorthonormal basesif b=u1, u2 , ,up is an orthogonal set, then c=a1u1, a2u2 , , apup is also an orthogonal set for any scalars a1 a2, , , ap. if b contains o