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Chapter 3---Section 1 Examples and Definition

Chapter 3 Vector Spaces

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 have applications to many areas in mathematics.

§1. Examples and Definition

New words and phrases

Vector space 向量空间 Polynomial 多项式 Degree 次数 Axiom 公理

Additive inverse 加法逆

1.1 Examples

Examining the following sets: (1) V=R2: The set of all vectors

?x1??? ?x2? (2) V=Rm?n: The set of all mxn matrices (3) V=C[a,b]: The set of all continuous functions on the interval [a,b]

(4) V=Pn: The set of all polynomials of degree less than n. Question 1: What do they have in common?

Chapter 3---Section 1 Examples and Definition

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 x+y that is also an element in V, and with each element x and each scalar

?x

?, 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 Axioms

★Definition Let V be a set on which the operations of addition 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 associate a unique element

?x?, we can

in V. The set V together with the

operations 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 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.

Chapter 3---Section 1 Examples and Definition

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 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 V

C2. If x, y are in V, then x+y is in V. An example that is not a vector space: Let

W??(a,1)|a is a real number?, on this set, the addition and

multiplication are defined in 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 statements are true.

Theorem 3.1.1 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?

Chapter 3---Section 1 Examples and Definition

Question: Are they obvious? Do we have to prove them?

But if we look at the definition of vector space, we don’t 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=(-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=-x

Assignment for section 1, chapter 3

Hand in: 9, 10, 12.

Chapter 3---Section 2 Subspaces

§2. Subspaces

New words and phrases

Subspace 子空间

Trivial subspace 平凡子空间 Proper subspace 真子空间 Span 生成

Spanning set生成集 Nullspace 零空间

2.1 Definition

Given a vector 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

??x1????S???|x?2x?21?x????2?? together with the usual addition and

scalar multiplication is itself a vector space .

The set S=

??a??????a|a and b are real numbers??????b??????together with the usual

addition and scalar multiplication is itself a vector space.

★Definition If S is a nonempty subset of a vector space V, and S satisfies the following conditions: (i)

?x?S whenever x?S for any scalar

?

Chapter 3---Section 3 Linear Independence

different representations. The sum of the original relation plus the dependency relation gives a new representation.

Assignment for section 3, chapter 3

Hand in : 5, 11, 13, 14, 15, ; Not required: 6, 7, 8, 9, 10,

Chapter 3---Section 4 Basis and Dimension

§4. Basis and Dimension

New words and phrases

Basis 基

Dimension 维数

Minimal spanning set 最小生成集 Standard Basis 标准基

4.1 Definitions and Theorems

A minimal 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 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 whole vector space and, consequently, we say that they form a basis for the vector space（向量空间的基）.

★Definition The vectors if and only if (i) (ii)

A basis of V actually is a minimal spanning set （最小张成集）for V.

v1,v2,?,vnv1,v2,?,vn form a basis for a vector space V

are linearly independent span V.

v1,v2,?,vnChapter 3---Section 4 Basis and Dimension

We know that spanning sets for 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 elements.

Theorem 3.4.1 If ?v1,v2,?,vn? is a spanning set for a vector space V, then any collection of m vectors in V, where m>n, is linearly dependent.

Proof Let ?u1,u2,?,um? be a collection of m vectors in V. Then each can be written as a linear combination of

v1,v2,?,vnui

.

ui=ai1v1+ai2v2+?+ainvn

A linear combination

nnc1u1+ c2u2???cmumncan be written in the form

c1?a1jvj+ c2?a2jvj???cm?anjvj

j=1j=1j=1Rearranging the terms, we see that

nmijc1u1+ c2u2???cmum??(?aj?1i?1ci)vj

Then we consider the equation c1u1+ c2u2???cmum?0 to see if we can

find a nontrivial solution (c1,c2,?,cn). The left-hand side of the equation can be written as a linear combination of

v1,v2,?,vn. We show that there

Chapter 3---Section 4 Basis and Dimension

are scalars

c1,c2,?,cn, not all zero, such that c1u1+ c2u2???cmum?0.

Here, we have to use a theorem: A homogeneous linear system must have a nontrivial solution if it has more unknowns than equations. Corollary 3.4.2 If ?v1,v2,?,vn?and ?u1,u2,?,um? are both bases for a vector space V, then n=m. (all the bases must have the same number of vectors.)

Proof Since

v1,v2,?,vn span V, if m>n, then ?u1,u2,?,um? must be

linearly dependent. This contradicts the hypothesis that ?u1,u2,?,um? is linearly independent. Hence 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.

★Definition Let V be a vector space. If V has a basis consisting of n vectors, we say that V has dimension 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.

m?n. By the same reasoning,

n?m. So

Chapter 3---Section 4 Basis and Dimension

Recall that a set of n vector 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 3.4.3 If V is a vector space of dimension n>0: I.

Any set of n linearly independent vectors spans V (so this set forms a basis for the vector 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

v1,v2,?,vn are linearly independent and v is

any vector in V. Since V has dimension n, the collection of vectors

v1,v2,?,vn,v must be linearly dependent. Then we show that v can be

v1,v2,?,vnexpressed in terms of Proof of II: If

.

v1,v2,?,vn are linearly dependent, then one of v’s 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 k

Theorem 3.4.4 If V is a vector space of dimension n>0:

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