線性代數(shù) 英文講義

1、The operations of addition and scalar multiplication are used in many diverse contexts in mathematics. Regardless of the context, however, these operations usually obey the same set of algebraic rules. Thus a general theory of mathematical systems involving addition and scalar multiplication will ha

2、ve applications to many areas in mathematics. 1. Examples andDefinitionNew words and phrasesVector space 向量空間Polynomial 多項式Degree 次數(shù)Axiom 公理Additive inverse 加法逆1.1 ExamplesExamining the following sets:(1) V=: The set of all vectors (2) V=: The set of all mxn matrices(3) V=: The set of all continuous

3、 functions on the interval (4) V=: The set of all polynomials of degree less than n. Question 1: What do they have in common? We can see that each of the sets, there are two operations: addition and multiplication, i.e. with each pair of elements x and y in a set V, we can associate a unique element

4、 x+y that is also an element in V, and with each element x and each scalar , we can associate a unique element in V. And the operations satisfy some algebraic rules.More generally, we introduce the concept of vector space. .1.2 Vector Space AxiomsDefinitionLet V be a set on which the operations of a

5、ddition and scalar multiplication are defined. By this we mean that, with each pair of elements x and y in a set V, we can associate a unique element x+y that is also an element in V, and with each element x and each scalar , we can associate a unique element in V. The set V together with the operat

6、ions of addition and scalar multiplication is said to form a vector space if the following axioms are satisfied.A1. x+y=y+x for any x and y in V.A2. (x+y)+z=x+(y+z) for any x, y, z in V.A3. There exists an element 0 in V such that x+0=x for each x in V.A4. For each x in V, there exists an element x

7、in V such that x+(-x)=0.A5. (x+y)=x+y for each scalar and any x and y in V.A6. (+)x=x+x for any scalars and and any x in V.A7. ()x=(x) for any scalars and and any x in V.A8. 1x=x for all x in V.From this definition, we see that the examples in 1.1 are all vector spaces.In the definition, there is an

8、 important component, the closure properties of the two operations. These properties are summarized as follows:C1. If x is in V and is a scalar, then x is in VC2. If x, y are in V, then x+y is in V.An example that is not a vector space:Let , on this set, the addition and multiplication are defined i

9、n the usually way. The operation + and scalar multiplication are not defined on W. The sum of two vector is not necessarily in W, neither is the scalar multiplication. Hence, W together with the addition and multiplication is not a vector space. In the examples in 1.1, we see that the following stat

10、ements are true.Theorem If V is a vector space and x is any element of V, then (i) 0x=0 (ii) x+y=0 implies that y=-x (i.e. the additive inverse is unique).(iii) (-1)x=-x.But is this true for any vector space?Question: Are they obvious? Do we have to prove them? But if we look at the definition of ve

11、ctor space, we dont know what the elements are, how the addition and multiplication are defined. So theorem above is not very obvious.Proof (i) x=1x=(1+0)x=1x+0x=x+0x, (A6 and A8)Thus x+x=-x+(x+0x)=(-x+x)+0x (A2) 0=0+0x=0x (A1, A3, and A4)(ii) Suppose that x+y=0. then -x=-x+0=-x+(x+y)Therefore, -x=(

12、-x+x)+y=0+y=y(iii) 0=0x=(1+(-1)x=1x+(-1)x, thus x+(-1)x=0 It follows from part (ii) that (-1)x=-xAssignmentfor section 1, chapter 3Hand in: 9, 10, 12.2. SubspacesNew words and phrasesSubspace 子空間Trivial subspace 平凡子空間Proper subspace 真子空間Span 生成Spanning set生成集Nullspace 零空間2.1 DefinitionGiven a vector

13、 space V, it is often possible to form another vector space by taking a subset of V and using the operations of V. For a new subset S of V to be a vector space, the set S must be closed under the operations of addition and scalar multiplication. Examples (on page 124)The set together with the usual

14、addition and scalar multiplication is itself a vector space .The set S=together with the usual addition and scalar multiplication is itself a vector space. DefinitionIf S is a nonempty subset of a vector space V, and S satisfies the following conditions: (i)xS whenever xS for any scalar (ii) x+y S w

15、henever xS and ySthen S is said to be a subspace （子空間）of V.A subspace S of V together with the operations of addition and scalar multiplication satisfies all the conditions in the definition of a vector space. Hence, every subspace of a vector space is a vector space in its own right. Trivial Subspa

16、ces and Proper SubspacesThe set containing only the zero element forms a subspace, called zero subspace, and V is also a subspace of V. Those two subspaces are called trivial subspaces of V. All other subspaces are referred to as proper subspaces. Examples of Subspaces (1) the set of all differentia

17、ble functions on a,b is a subspace of (2) the set of all polynomials of degree less than n (1) with the property p(0) form a subspace of . (3) the set of matrices of the form forms a subspace of . (4) the set of all mxm symmetric matrices forms a subspace of (5) the set of all mxm skew-symmetric mat

18、rices form a subspace of 2.2 The Nullspace of a Matrix Let A be an mxn matrix, and.Then N(A) form a subspace of . The subspace N(A) is called the nullspace of A. The proof is a straightforward verification of the definition.2.3 The Span of a Set of VectorsIn this part, we give a method for forming a

19、 subspace of V with finite number of vectors in V.Given n vectors in a vector space of V, we can form a new subset of V as the following.It is easy to show that this set forms a subset of V. We call this subspace the span of , or the subspace of V spanned by .Theorem If are elements of a vector spac

20、e of V, then is a subspace of V.For example, the subspace spanned by two vectors and is the subspace consisting of the elements .2.4 Spanning Set for a Vector SpaceDefinitionIf are vectors of V and V=, then the set is called a spanning set （生成集）for V. In other words, the set is a spanning set for V

21、if and only if every element can be written as a linear combination of .The spanning sets for a vector space are not unique. Examples (Determining if a set spans for )(a) (b) (c) (d) To do this, we have to show that every vector in can be written as a linear combination of the given vectors.Assignme

22、nt for section 2, chapter 3Hand in: 6, 8, 13, 16, 17, 18, 20Not required: 213. Linear IndependenceNew words and phrasesLinear independence 線性無關(guān)性Linearly independent 線性無關(guān)的Linear dependence 線性相關(guān)性Linearly dependent 線性相關(guān)的3.1 MotivationIn this section, we look more closely at the structure of vector spac

23、es. We restrict ourselves to vector spaces that can be generated from a finite set of elements, or vector spaces that are spans of finite number of vectors. V=The set is called a generating set or spanning set(生成集). It is desirable to find a minimal spanning set. By minimal, we mean a spanning set w

24、ith no unnecessary element. To see how to find a minimal spanning set, it is necessary to consider how the vectors in the collection depend on each other. Consequently we introduce the concepts of linear dependence and linear independence. These simple concepts provide the keys to understanding the

25、structure of vector spaces. Give an example in which we can reduce the number of vectors in a spanning set. Consider the following three vectors in .These three vectors satisfy (1) Any linear combination of can be reduced to a linear combination of . Thus S= Span()=Span(). (2) (a dependency relation

26、)Since the three coefficients are nonzero, we could solve for any vector in terms of the other two. It follows that Span()=Span()=Span()=Span()On the other hand, no such dependency relationship exists between . In deed, if there were scalars and , not both 0, such that (3) then we could solve for on

27、e of the two vectors in terms of the other. However, neither of the two vectors in question is a multiple of the other. Therefore, Span() and Span() are both proper subspaces of Span(), and the only way that (3) can hold is if .Observations:(I) If span a vector space V and one of these vectors can b

28、e written as a linear combination of the other n-1 vectors, then those n-1 vectors span V.(II) Given n vectors , it is possible to write one of the vectors as a linear combination of the other n-1 vectors if and only if there exist scalars not all zero such that Proof of I: Suppose that can be writt

29、en as a linear combination of the vectors . Proof of II: The key point here is that there at least one nonzero coefficient. 3.2 DefinitionsDefinition The vectors in a vector space V are said to be linearly independent（線性獨立的） if implies that all the scalars must equal zero. Example: are linearly inde

30、pendent. Definition The vectors in a vector space V are said to be linearly dependent （線性相關(guān)的）if there exist scalars not all zero such that.Let be vector in . Then are linearly dependent. If there are nontrivial choices of scalars for which the linear combination equals the zero vector, then are line

31、arly dependent. If the only way the linear combination can equal the zero vector is for all scalars to be 0, then are linearly independent.3.3 Geometric InterpretationThe linear dependence and independence in and . Each vector in or represents a directed line segment originated at the origin. Two ve

32、ctor are linearly dependent in or if and only if two vectors are collinear. Three or more vector in must be linearly dependent. Three vectors in are linearly dependent if and only if three vectors are coplanar. Four or more vectors in must be linearly dependent. 3.4 Theorems and ExamplesIn this part

33、, we learn some theorems that tell whether a set of vectors is linearly independent. Example: (Example 3 on page 138) Which of the following collections of vectors are linearly independent?(a) (b) (c) (d) The problem of determining the linear dependency of a collection of vectors in can be reduced t

34、o a problem of solving a linear homogeneous system. If the system has only the trivial solution, then the vectors are linearly independent, otherwise, they are linearly dependent, We summarize the this method in the following theorem: Theorem n vectors in are linearly dependent if the linear system

35、Xc=0 has a nontrivial solution, where .Proof: Xc=0.Theorem Let be n vectors in and let . The vectors will be linearly dependent if and only if X is singular. (the determinant of X is zero)Proof: Xc=0 has a nontrivial solution if and only X is singular. Theorem Let be vectors in a vector space V. A v

36、ector v in Span() can be written uniquely as a linear combination of if and only if are linearly independent.(A vector v in Span() can be written as two different linear combinations of if and only if are linearly dependent.)（Note: If-sufficient condition ; Only if- necessary condition）Proof: Let vS

37、pan(), then Necessity: (contrapositive law for propositions) Suppose that vector v in Span() can be written as two different linear combination of , then prove that are linearly dependent. The difference of two different linear combinations gives a dependency relation of Suppose that are linearly de

38、pendent, then there exist two different representations. The sum of the original relation plus thedependency relation gives a new representation. Assignment for section 3, chapter 3Hand in : 5, 11, 13, 14, 15, ; Not required: 6, 7, 8, 9, 10, 4. Basis and DimensionNew words and phrasesBasis 基Dimensio

39、n 維數(shù)Minimal spanning set 最小生成集Standard Basis 標(biāo)準(zhǔn)基4.1 Definitions and TheoremsAminimal spanning set for a vector space V is a spanning set with no unnecessary elements (i.e., all the elements in the set are needed in order to span the vector space). If a spanning set is minimal, then its elements are

40、linearly independent. This is because if they were linearly dependent, then we could eliminate a vector from the spanning set, the remaining elements still span the vector space, this would contradicts the assumption of minimality. The minimal spanning set forms the basic building blocks for the who

41、le vector space and, consequently, we say that they form a basis for the vector space（向量空間的基）.Definition The vectors form a basis for a vector space V if and only if (i) are linearly independent (ii) span V.A basis of V actually is a minimal spanning set （最小張成集）for V. We know that spanning sets for

42、a vector space are not unique. Minimal spanning sets for a vector space are also not unique. Even though, minimal spanning sets have something in common. That is, the number of elements in minimal spanning sets. We will see that all minimal spanning sets for a vector space have the same number of el

43、ements.Theorem If is a spanning set for a vector space V, then any collection of m vectors in V, where mn, is linearly dependent.Proof Let be a collection of m vectors in V. Then each can be written as a linear combination of . A linear combinationcan be written in the formRearranging the terms, we

44、see that Then we consider the equation to see if we can find a nontrivial solution (). The left-hand side of the equation can be written as a linear combination of . We show that there are scalars , not all zero, such that .Here, we have to use a theorem: A homogeneous linear system must have a nont

45、rivial solution if it has more unknowns than equations. Corollary If and are both bases for a vector space V, then n=m. (all the bases must have the same number of vectors.)Proof Since span V, if mn, then must be linearly dependent. This contradicts the hypothesis that is linearly independent. Hence

46、 . By the same reasoning, . So m=n.From the corollary above, all the bases for a vector space have the same number of elements (if it is finite). This number is called the dimension of the vector space. DefinitionLet V be a vector space. If V has a basis consisting of n vectors, we say that V has di

47、mension n (the dimension of a vector space of V is the number of elements in a basis.) The subspace 0 of V is said to have dimension 0. V is said to be finite-dimensional if there is a finite set of vectors that spans V; otherwise, we say that V is infinite-dimensional. Recall that a set of n vector

48、 is a basis for a vector space if two conditions are satisfied. If we know that the dimension of the vector space is n, then we just need to verify one condition. Theorem If V is a vector space of dimension n0:I. Any set of n linearly independent vectors spans V (so this set forms a basis for the ve

49、ctor space).II. Any n vectors that span V are linearly independent (so this set forms a basis for the vector space).Proof Proof of I: Suppose that are linearly independent and v is any vector in V. Since V has dimension n, the collection of vectors must be linearly dependent. Then we show that v can

50、 be expressed in terms of .Proof of II: If are linearly dependent, then one of vs can be written as a linear combination of the other n-1 vectors. It follows that those n-1 vectors still span V. Thus, we will obtain a spanning set with k0:(i) No set of less than n vectors can span V.(ii) Any subset

51、of less than n linearly independent vectors can be extended to form a basis for V.(iii) Any spanning set containing more than n vectors can be pared down (to reduce or remove by or as by cutting) to form a basis for V.Proof (i): If there are m (n) vectors that can span V, then we can argue that dimV

52、n. this contradicts the assumption.(ii) We assume that are linearly independent ( kn, we can continue to eliminate vectors in this manner until we arrive at a spanning set containing n vectors.4.2 Standard BasesThe standard bases（標(biāo)準(zhǔn)基）for , .Although the standard bases appear to be the simplest and m

53、ost natural to use, they are not the most appropriate bases for many applied problems. Once the application is solved in terms of the new basis, it is a simple matter to switch back and represent the solution in terms of the standard basis. Assignment for section 4, chapter 3Hand in : 4, 7, 9，10，12，

54、16，17,18 Not required: 11，13，14, 15， 5. Change of BasisNew words and phrasesTransition matrix 過渡矩陣5.1 MotivationMany applied problems can be simplified by changing from one coordinate system to another. Changing coordinate systems in a vector space is essentially the same as changing from one basis

55、to another. For example, in describing the motion of a particle in the plane at a particular time, it is often convenient to use a basis for consisting of a unit tangent vector t and a unit normal vector n instead of the standard basis.In this section we discuss the problem of switching from one coo

56、rdinate system to another. We will show that this can be accomplished by multiplying a given coordinate vector x by a nonsingular matrix S.5.2 Changing Coordinates in The standard basis for is . Any vector in can be written as a linear combination.The scalars can be thought of as the coordinates （坐標(biāo)） of x w