线性代数 英文讲义

Chapter 3 V ector 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

V ector 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=2R : The set of all vectors

12x x ?? ???

(2) V=m n R ?: The set of all mxn matrices (3) V=[,]

a b C

: The set of all continuous functions on the interval [,]a b

(4) V=n P : 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 x+y that is also an element in V, and with each element x and each scalar α, we can associate a unique element αin V. And the operations satisfy some algebraic rules.

x

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 α, we can associate a unique element xαin 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. α(x+y)= αx+αy for each scalar αand any x and y in V.

A6. (α+β)x=αx+βx for any scalars αandβand any x in V.

A7. (αβ)x=α(βx) for any scalars αandβand 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 αis a scalar, then αx 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 {}

=, on this set, the addition and W a a

(,1)| is a real number

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?

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.

§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

1212|2x S x x x ??????==??

???????

together with the usual addition and

scalar multiplication is itself a vector space .

The set S=

| and are real num bers a a a b 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)

α

x ∈S whenever x ∈S for any scalar

α

(ii) x+y ∈S whenever x ∈S and y ∈S

then S is said to be a subspace （子空间）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 Subspaces and Proper Subspaces

The 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 differentiable functions on [a,b] is a subspace of

[,]a b C

(2) the set of all polynomials of degree less than n (>1) with the property p(0) form a subspace of

n P .

(3) the set of matrices of the form a b b

c ?? ?-??

forms a subspace of

22

R

?.

(4) the set of all mxm symmetric matrices forms a subspace of

m m

R

?

(5) the set of all mxm skew-symmetric matrices form a subspace of

m m

R

?

2.2 The Nullspace of a Matrix

Let A be an mxn matrix, and

{}()|,0n

N A X X R AX =∈=.

Then N(A) form a subspace of n

R

. 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 Vectors

In this part, we give a method for forming a subspace of V with finite number of vectors in V . Given n vectors

12n

v ,v ,,v in a vector space of V , we can form a new

subset of V as the following.

{}12n 1122n n Span(v ,v ,,v )v v v |' are scalars i s αααα=+++

It is easy to show that this set forms a subset of V . We call this subspace the span of

12n

v ,v ,,v , or the subspace of V spanned by

12n v ,v ,,v .

Theorem 3.2.1 If

12n

v ,v ,,v are elements of a vector space of V , then

{}

12n 1122n n Span(v ,v ,,v )v v v |' are scalars i s αααα=+++ is a subspace

of V .

For example, the subspace spanned by two vectors

100?? ? ? ???

and

010?? ? ? ???

is

the subspace consisting of the elements

120x x ?? ? ? ???

. 2.4 Spanning Set for a Vector Space

★Definition If

12n

v ,v ,,v are vectors of V and

V=12

n Span(v ,v

,,v )

, then the set {}12n v ,v ,,v is called a spanning set

（生成集）for V .

In other words, the set {}12n v ,v ,,v is a spanning set for V if and only if every element can be written as a linear combination of

12n v ,v ,,v .

The spanning sets for a vector space are not unique. Examples (Determining if a set spans for 3

R

)

(a)

()

{}1

231,2,3,T

e

e e

(b) ()()()

{}1,1,1,1,1,0,1,

0,

0T

T

T

(c) ()()

{}1,0,1,

0,

1,

0T T

(d) ()()()

{}1,

2,

4,

2,

1,

3,

4,

1,

1T

T

T

-

To do this, we have to show that every vector in 3

R

can be written

as a linear combination of the given vectors.

Assignment for section 2, chapter 3 Hand in: 6, 8, 13, 16, 17, 18, 20

Not required: 21

§3. Linear Independence

New words and phrases

Linear independence 线性无关性 Linearly independent 线性无关的 Linear dependence 线性相关性 Linearly dependent 线性相关的

3.1 Motivation

In this section, we look more closely at the structure of vector spaces. 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=12

n Span(v ,v

,,v )

The set {}12n v ,v ,,v 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 with 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 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

3

R

.

11x 12?? ?=- ? ??? 22x 31-?? ?

= ?

?

??

31x 3

8-??

?= ? ???

These three vectors satisfy

(1)

312x =3x +2x

Any linear combination of 123

x ,x ,x can be reduced to a linear

combination of 12

x ,x . Thus S= Span(123x ,x ,x )=Span(12x ,x ).

(2)

1233x +2x +(1)x 0

-= (a dependency relation)

Since the three coefficients are nonzero, we could solve for any vector in terms of the other two. It follows that

Span(123x ,x ,x )=Span(12x ,x )=Span(13x ,x )=Span(23x ,x )

On the other hand, no such dependency relationship exists between

12x and x . In deed, if there were scalars 1

c and

2c , not both 0, such that

(3)

1122c x +c x 0

=

then we could solve for one 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(1x ) and Span(

2

x ) are both proper subspaces of

Span(12x ,x ), and the only way that (3) can hold is if 12c =c =0

.

Observations: (I)

If

12n

v ,v ,,v span a vector space V and one of these vectors can

be written as a linear combination of the other n-1 vectors, then those n-1 vectors span V .

(II) Given n vectors

12n

v ,v ,,v , 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

12n c ,c ,,c

not all zero such that

1122n n v v v 0

c c c +++=

Proof of I: Suppose that n v

can be written as a linear combination of

the vectors

12n-1v ,v ,,v .

Proof of II: The key point here is that there at least one nonzero coefficient.

3.2 Definitions

★Definition The vectors

12n

v ,v ,,v in a vector space V are said to

be linearly independent （线性独立的） if

1122n n v v v 0

c c c +++=

implies that all the scalars 12n c ,c ,,c

must equal zero.

Example: 12n e ,e ,,e

are linearly independent.

Definition The vectors

12n

v ,v ,,v in a vector space V are said to be

linearly dependent （线性相关的）if there exist scalars 12n c ,c ,,c

not all

zero such that

1122n n v v v 0

c c c +++= .

Let

12n e ,e ,,e ,x

be vector in n

R

. Then

12n e ,e ,,e ,x

are linearly

dependent.

If there are nontrivial choices of scalars for which the linear combination

1122

n v v v c c c +++

equals the zero vector, then

12n

v ,v ,,v

are linearly dependent. If the only way the linear combination

1122n n

v v v c c c +++ can equal the zero vector is for all scalars

12n c ,c ,,c

to be 0, then

12n

v ,v ,,v are linearly independent.

3.3 Geometric Interpretation

The linear dependence and independence in 2

R

and

3

R

.

Each vector in

2

R

or

3

R

represents a directed line segment

originated at the origin.

Two vector are linearly dependent in

2

R

or

3

R

if and only if two

vectors are collinear. Three or more vector in 2

R

must be linearly

dependent. Three vectors in

3

R

are linearly dependent if and only if three

vectors are coplanar. Four or more vectors in 3

R

must be linearly

dependent.

3.4 Theorems and Examples

In this part, 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)

()

{

}1231,2,3,T

e e e

(b) ()()()

{}1,1,1,1,1,0,1,

0,

0T

T

T

(c) ()()

{}1,0,1,

0,

1,

0T T

(d) ()()()

{}1,2,

4,

2,

1,

3,

4,

1,

1T

T

T

-

The problem of determining the linear dependency of a collection of vectors in

m

R

can be reduced to 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

12n

x ,x ,,x in

m

R

are linearly dependent if the

linear system Xc=0 has a nontrivial solution, where 12n X =(x ,x ,,x )

.

Proof: 1122n n c x +c x +c x 0

+=

?

Xc=0.

Theorem 3.3.1 Let

12n

x ,x ,,x be n vectors in

n

R

and let

12n X =(x ,x ,,x )

. The vectors

12n x ,x ,,x

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 3.3.2 Let

12n

v ,v ,,v be vectors in a vector space V . A vector v

in Span(12n v ,v ,,v ) can be written uniquely as a linear combination of

12n v ,v ,,v

if and only if

12n

v ,v ,,v are linearly independent.

(A vector v in Span(12n v ,v ,,v ) can be written as two different linear combinations of 12n

v ,v ,,v if and only if

12n

v ,v ,,v are linearly

dependent.)

（Note: If---sufficient condition ; Only if--- necessary condition ） Proof: Let v ∈ Span(12n v ,v ,,v ), then

1122n n

v v v v ααα=+++

Necessity: (contrapositive law for propositions)

Suppose that vector v in Span(12n v ,v ,,v ) can be written as two different linear combination of

12n

v ,v ,,v , then prove that

12n

v ,v ,,v are

linearly dependent. The difference of two different linear combinations gives a dependency relation of 12n

v ,v ,,v

Suppose that

12n

v ,v ,,v are linearly dependent, then there exist two

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,

§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 12n

v ,v ,,v form a basis for a vector space V

if and only if (i) 12n

v ,v ,,v are linearly independent (ii) 12n

v ,v ,,v span V .

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

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 {}12n v ,v ,,v 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 {}12m u ,u ,,u be a collection of m vectors in V . Then each u i

can be written as a linear combination of

12n

v ,v ,,v .

i 1122n

u =v +v ++v i i in a a a

A linear combination

1122m

u + u u m c c c ++ can be written in the form

n

n

n

11j 22j j j=1

j=1

j=1

v + v v j j m nj c a c a c a ++∑∑∑

Rearranging the terms, we see that

1122m j 1

1

u + u u ()v n

m

m ij

i j i c c c a

c ==++=

∑∑

Then we consider the equation 1122m m c u + c u c u 0++= to see if we can

find a nontrivial solution (12n c ,c ,,c ). The left-hand side of the equation can be written as a linear combination of

12n

v ,v ,,v . We show that there

are scalars

12n c ,c ,,c , not all zero, such that 1122m m c u + c u c u 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 {}12n v ,v ,,v and {}12m u ,u ,,u are both bases for a vector space V , then n=m. (all the bases must have the same number of vectors.)

Proof Since

12n

v ,v ,,v span V , if m>n, then {}12m u ,u ,,u must be

linearly dependent. This contradicts the hypothesis that {}12m u ,u ,,u is linearly independent. Hence m n

≤. By the same reasoning,

n m

≤. 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.

★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.

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

12n

v ,v ,,v are linearly independent and v is

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

12n v ,v ,,v ,v

must be linearly dependent. Then we show that v can be

expressed in terms of 12n

v ,v ,,v .

Proof of II: If

12n

v ,v ,,v 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: