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1、Li Jie3 The Vector Space Rn 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. Perhaps the most famil
2、iar 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 describe objects such
3、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 vector in a general vecto
4、r space.Li Jie3.2 VECTOR SPACE PROPERTIES OF RnLi 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.(s3)aX is in W whenever X is in W and a is any scalar.
5、Li JieExample 1: Let W be the subset of R3 defined byVerify 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, and this specification serves as a test for determining whether a vector in Rn is or is not in
6、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 two arbitrary vectors X and Y from W. Thus X and Y are in Rn, and both vectors satisfy the algeb
7、raic 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 of W. Li JieExample 3: Let W be the subset of R3 defined byShow that W is not a subspace of R3.Exa
8、mple 2: Let W be the subset of R3 defined by Verify that W is a subspace of R3 and give a geometric interpretation of W.Li JieExample 4:Let W be the subset of R2 defined byDemonstrate that W is not a subspace of R2.Example 5:Let W be the subset of R2 defined byDemonstrate that W is not a subspace of
9、 R2.Exercise P175 18 32Li Jie3.3 EXAMPLES OF SUBSPACESIn this 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 R
10、n.If S=v1, ,vr is a subset of Rn, then the subspace W consisting 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 n
11、oncollinear geometric vectors, then Spu,v=au+bv: a,b any real 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 vectorsDetermine W=Spu,v and give a geometric interpretation of W.Li
12、 JieThe null space of a matrixWe now introduce 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
13、 (m n) matrix. The null space of A denoted N(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) matrixSolution: N(A) is determined by solving the homogeneous system Ax= .T
14、his is accomplished by reducing the augmented matrix A| to echelon form. It is easy to verify that A| is row equivalent toLi JieSolving the corresponding reduced system yields x1=-2x3-3x4 x2=-x3+2x4Where x3 and x4 are arbitrary; that is,Li JieExample 5:Let S=v1,v2,v3,v4 be a subset of R3, whereShow
15、that there exists a set T=w1,w2 consisting of two vectors in R3 such that Sp(S)=Sp(T).Solution: let Li JieSet row operation to A and reduce A to the following matrix: So, 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 JieLi JieThe s
16、olution on P184And the row vectors of AT are precisely the vectors v1T,v2T,v3T, and v4T. It is straightforward to see that AT reduces to the matrixSo, by Theorem 6, AT and BT have the same row space. Thus A and B have the same column space whereLi JieIn particular, Sp(S)=Sp(T), where T=w1,w2,Li JieT
17、wo of the most fundamental concepts 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
18、 from R2 will serve to illustrate the transition from geometry to algebra. We have already seen that each vector v in R2,can be interpreted geometrically as the point with coordinates a and b. Recall that in R2 the vectors e1 and e2 are defined byLi JieClearly the vector v in (1) can be expressed un
19、iquely as a linear 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 charact
20、erizing a point 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 fi
21、rst requirement 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
22、 spans W, if every 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 vec
23、tors in R3, is a spanning set for R3. Specifically, if v is in R3,Then v=ae1+be2+ce3. The next two examples consider other subset of R3.Li JieExample 1: In R3, let S=u1,u2,u3, whereDetermine whether S is a spanning set for R3.Solution: The augmented matrix this matrix is row equivalent toLi JieExamp
24、le 2: Let S=v1,v2,v3 be the subset of R3 defined byDoes S span R3?Solution:and the matrix A|v is row equivalent toLi JieSo,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
25、set for the null space, N(A), of a matrix A.Example 3:Let A be the (34) matrixExhibit a spanning set for 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 JieLet u1 and
26、 u2 be the vectorsLi JieTherefore, 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 spanning set
27、s are equally describe. For example, define u in R2 byThe set S=e1,e2,u is a spanning set for R2. indeed, for an arbitrary vector v in R2,Li 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 that a set v1,vm
28、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 that a linearly de
29、pendent 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, whereShow that S is a linearly dependent set, and exhibit a subset T of S such that T contains only two vectors but Sp(T)=Sp(S).Li JieSolution: The vector equa
30、tion x1v1+x2v2+x3v3= (10)is equivalent to the (3 3) homogeneous system of equations with augmented matrixLi JieMatrix is row equivalent toSo v3=-1v1+2v2Li Jie On the other hand, if B=v1,vm is a linearly independent spanning set for W, then no vector in B is a linear combination of the other m-1 vect
31、ors 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 at least one vector from S is a linear combination of the other r-1 vectors and can be discarded fr
32、om 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 expressed as a linear combination of the vectors retained). In this sense a linearly independent spanning
33、 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 basis for W is a linearly independent spanning set for W.Li JieUniqueness of representationRemark L
34、et 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+a2v2+apvp.As we see later, these scalars are called the coordinates x with respect to the basis.Examp
35、le of basesIt is easy to show that the unit vectorsis a basis for R3Li JieIn general, the n-dimensional vectors e1,e2,en form a basis for Rn, frequently called the natural basis.Provide 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,v
36、5, whereFind a subset of S that is a basis for W.Solution:So v1,v2,v4is a 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 s
37、olution x1=xm=0, then S is a linearly independent set and hence is a basis 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 f
38、or W.Li JieTheorem 7:If the nonzero matrix A is row equivalent to the matrix 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
39、 vectors in W is linearly dependent.As an immediate corollary of Theorem 8, 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 DIMEN
40、SION Li JieDefinition 5 : Let W be a subspace of Rn. If W has a basis B=w1,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 ve
41、ctors; so dim(Rn)=n. Thus the definition of dimension the number of vectors 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 byExhibit a basis for W and determine dim(W).Li JieSolution: A ve
42、ctor x in W can be written in the formTherefore, the set u is a basis for W, whereLi JieExample 2 Let W be the subspace of R3, W=spanu1,u2,u3,u4,whereFind 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 vectors
43、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 Example
44、 2, and let v1,v2,v3 be the subset of W defined byDetermine 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 when a system of linear e
45、quation 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, whereLi JieSolution: To find the dimension of
46、the row space of A, observe that A is row equivalent to the matrixand 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.Li JieIt now follows that the nullity of A is 1 because the vectorA is row equivalent to matrix C, whereform
47、 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 subspace of R3. Thus they are entirely different subspaces; even so, the dimensions are the same, and the nex
48、t 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 JieTheorem 11: An (m n) system of linear equations , Ax=b, is consistent if and only if rank(A)=rank(A|b).Theo
49、rem 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 system of equations, Ax=b, to be consistent.Li Jie3.6 ORTHOGONAL BASES FOR SUBSPACES Orthogonal Bases The i
50、dea 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 example, consider the vectors u and v given by Li Jie Clearly uTv=0 , and these two vectors are perpendicula
51、r 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 Rn, we say that u and v are orthogonal if uTv=0 We will also find the concept of an orthogonal set of vectors to b
52、e 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 isExample 1 verify that S is an orthogonal set of vectors , whereLi JieTheorem 13 : let S =u1 u2 , , up, be a set of nonz
53、ero 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, 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, ifThen
54、B is said to be an orthonormal basis for WThe word orthonormal suggests both orthogonal and normalized. Thus an orthonormal basis is an orthogonal basis 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
55、 orthonormal basis for Rn.Li JieExample 2 Verify that the set B=v1 v2 v3, is an orthogonal basis for R3,whereCorollary: Let W be a subspace 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
56、=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 only nonzero vectors and if we define the scalars a i byThen C is an orthonormal set. That is , we can convert an orthogonal set of nonzero vectors into an orthonorma
57、l set by dividing each vector by its length.Li JieExample3: Recall that the set B in Example 2 is an orthogonal basis for R3. Modify B so that it is an orthonormal basis.Solution:Li JieDetermining CoordinatesSuppose that W is a p-dimensional subspace of Rn, and B=w1 w2 , ,.wp is a basis for W. if v
58、is any vector in W, then v can be written uniquely in the form v=a1w1+a2w2+ apwp (3)The scalars a1,a2,ap, in Eq.(3) are called the coordinates of v with respect to the basis B As we will see, it is fairly easy to determine the coordinates of a vector with respect to an orthogonal basis. To appreciat
59、e the savings in computation, consider how coordinates are found when the basis is not orthogonal.Li JieExample4: Express the vector v in terms of the orthogonal basis B=w1 , w2 ,w3,whereLi JieIn general , let W be a subspace of Rn, and let B=w1 w2 , , wp be an orthogonal basis for W. If v is any ve
60、ctor in W, then v can be expressed uniquely in the form v=a1w1+a2w2+apwp; (5a)WhereConstructing an Orthogonal BasisThe next theorem gives a procedure that can be used to generate an orthogonal basis from any given basis. This procedure, called the Gram-Schmidt process, is quite practical from a comp
61、utational standpointLi JieTheorem 14 Gram-SchmidtLet W be a p-dimensional subspace of Rn, and let w1 w2 , , wp be any basis for W. Then the set of vectors u1 u2 , , up is an orthogonal basis for W, whereAnd where, in generalLi JieExample 3 : let W be the subspace of R3 defined by W=Spw1,w2,whereUse the Gram-Schmidt process to construct an orthogonal basis for WLi JieExample 6 : Use the Gram-Schmidt orthogonalization process to generate an orthogonal basis for W=Spw1w2 w3,whereExercise P224 15, 19
