Prepared by

Tanuj Parikh

Chapter Two: Vector Spaces

I. Definition of Vector Space

II. Linear Independence

III. Basis and Dimension• Topic: Fields

• Topic: Crystals

• Topic: Voting Paradoxes

• Topic: Dimensional Analysis

Vector space ~ Linear combinations of vectors.

Ref: T.M.Apostol, “Linear Algebra”, Chap 3, Wiley (97)

I. Definition of Vector Space

I.1. Definition and Examples

I.2. Subspaces and Spanning Sets

Algebraic Structures

Ref: Y.Choquet-Bruhat, et al, “Analysis, Manifolds & Physics”, Pt I., North Holland (82)

Structure Internal Operations Scalar Multiplication

Group * No

Ring, Field * , + No

Module / Vector Space + Yes

Algebra + , * Yes

Field = Ring with idenity & all elements except 0 have inverses.

Vector space = Module over Field.

I.1. Definition and Examples

Definition 1.1: (Real) Vector Space ( V, ♣ ; )A vector space (over ) consists of a set V along with 2 operations ‘♣’ and ‘♦’ s.t.(1) For the vector addition ♣ :

∀ v, w, u ∈ V a) v ♣ w ∈ V ( Closure )b) v ♣ w = w ♣ v ( Commutativity )c) ( v ♣ w ) ♣ u = v ♣ ( w ♣ u ) ( Associativity )d) ∃ 0 ∈ V s.t. v ♣ 0 = v ( Zero element )e) ∃ −v ∈ V s.t. v ♣ (−v) = 0 ( Inverse )

(2) For the scalar multiplication ♦ :

∀ v, w ∈ V and a, b ∈ , [ is the real number field (,+,×) f) a ♦ v ∈ V ( Closure ) g) ( a + b ) ♦ v = ( a ♦ v ) ♣ (b ♦ v ) ( Distributivity )h) a ♦ ( v ♣ w ) = ( a ♦ v ) ♣ ( a ♦ w )i) ( a × b ) ♦ v = a ♦ ( b ♦ v ) ( Associativity )j) 1 ♦ v = v

♣ is always written as + so that one writes v + w instead of v ♣ w

× and ♦ are often omitted so that one writes a b v instead of ( a × b ) ♦ v

Definition 1.1: (Real) Vector Space ( V, + ; )A vector space (over ) consists of a set V along with 2 operations ‘+’ and ‘ ’ s.t.(1) For the vector addition + :

∀ v, w, u ∈ V a) v + w ∈ V ( Closure )b) v + w = w + v ( Commutativity )c) ( v + w ) + u = v + ( w + u ) ( Associativity )d) ∃ 0 ∈ V s.t. v + 0 = v ( Zero element )e) ∃ −v ∈ V s.t. v −v = 0 ( Inverse )

(2) For the scalar multiplication : ∀ v, w ∈ V and a, b ∈ , [ is the real number field (,+,×) ]a) a v ∈ V ( Closure )b) ( a + b ) v = a v + b v ( Distributivity )c) a ( v + w ) = a v + a wd) ( a × b ) v = a ( b v ) = a b v ( Associativity )e) 1 v = v

Definition in Conventional Notations

Example 1.3: 2

2 is a vector space if1 1

2 2

x ya b a b

x y

+ = + ÷ ÷

x y 1 1

2 2

ax by

ax by

+ = ÷+

,a b∀ ∈R

0

0

= ÷

0with

Example 1.4: Plane in 3.

The plane through the origin 0

x

P y x y z

z

÷= + + = ÷ ÷

is a vector space.

P is a subspace of 3.

Proof it yourself / see Hefferon, p.81.

Proof it yourself / see Hefferon, p.82.

Example 1.5:

Let ♣ & ♦ be the (column) matrix addition & scalar multiplication, resp., then

( n, + ; ) is a vector space.

( n, + ; ) is not a vector space since closure is violated under scalar multiplication.

Example 1.6:

0

0

0

0

V

÷ ÷= ÷ ÷

Let then (V, + ; ) is a vector space.

Definition 1.7: A one-element vector space is a trivial space.

Example 1.8: Space of Real Polynomials of Degree n or less, n

0

nk

n k kk

a x a=

= ∈

∑P R { }2 33 0 1 2 3 ka a x a x a x a= + + + ∈P R

Vector addition: 0 0

n nk k

k kk k

a x b x= =

+ = +∑ ∑a b ( ) k kka b+ = +a b

Scalar multiplication:0

nk

kk

b b a x=

= ÷

∑a

Zero element:0

0n

k

k

x=

= ∑0 ( ) 0k

k= ∀0i.e.,

n is a vector space with vectors 0

nk

kk

a x=

≡ ∑a

( )0

nk

k kk

a b x=

= +∑

0

nk

kk

ba x=

= ∑

i.e.,

( ) kkb ba=ai.e.

,

E.g.,

n is isomorphic to n+1 with ( ) 10

0

~ , ,n

k nk n n

k

a x a a +

=

∈ ∈∑ P L R

Inverse: ( )0

nk

kk

a x=

− −≡ ∑a ( ) kka− = −ai.e.

,

( ) kka=a

The kth component of a is

Example 1.9: Function Space

The set { f | f : → } of all real valued functions of natural numbers is a vector space if

( ) ( ) ( ) ( )1 2 1 2f f n f n f n≡+ +Vector addition:

( ) ( ) ( )a f n na f≡Scalar multiplication:

n ∈N

a ∈R

f ( n ) is a vector of countably infinite dimensions: f = ( f(0), f(1), f(2), f(3), … )

E.g.,

( ) 2 1f n n= + ~ ( )1, 2, 5,10,=f L

Zero element:

( ) 0zero n =

Inverse: ( ) ( )( )f n f n− ≡ −

Example 1.10: Space of All Real Polynomials,

0

, n

kk k

k

a x a n=

= ∈ ∈

∑P R N

is a vector space of countably infinite dimensions.

( )0 1 20

~ , , ,kk

k

a x a a a∞

∞

=

∈ ∈∑ P L R

Example 1.11: Function Space

The set { f | f : → } of all real valued functions of real numbers is a vector space of uncountably infinite dimensions.

Example 13: Solution Space of a Linear Homogeneous Differential Equation

2

2: 0

d fS f f

d x

= → + =

R R is a vector space with

( ) ( ) ( ) ( )f g x f x g x≡+ +Vector addition:

( ) ( ) ( )a f x xa f≡Scalar multiplication:

Zero element:

( ) 0zero x =

Inverse: ( ) ( )( )f x f x− ≡ −

Closure:2 2

2 20 & 0

d f d gf g

d x d x+ = + = ( ) ( )

2

20

d a f bga f bg

d x

++ + =→

a ∈R

Example 14: Solution Space of a System of Linear Homogeneous Equations

Remarks:

• Definition of a mathematical structure is not unique.

• The accepted version is time-tested to be most concise & elegant.

Lemma 1.16: Lose Ends

In any vector space V,

1. 0 v = 0 .

2. ( −1 ) v + v = 0 .

3. a 0 = 0 .

∀ v ∈V and a ∈ .

Proof:

( )1 0= − = + −0 v v v v 0= + −v v v 0= v1.

2.

( ) ( )1 1 1− + = − +v v v 0= v = 0

3.

( )0a a=0 v ( )0a= v 0= v = 0

Exercises 2.I.1.

1. At this point “the same” is only an intuition, but nonetheless for each vector space identify the k for which the space is “the same” as k.(a) The 2×3 matrices under the usual operations(b) The n × m matrices (under their usual operations)(c) This set of 2 × 2 matrices

2.

(a) Prove that every point, line, or plane thru the origin in 3 is a vector space under the inherited operations.(b) What if it doesn’t contain the origin?

00

aa b c

b c

+ + = ÷

I.2. Subspaces and Spanning Sets

Definition 2.1: SubspacesFor any vector space, a subspace is a subset that is itself a vector space, under the inherited operations.

Example 2.2: Plane in 3 0

x

P y x y z

z

÷= + + = ÷ ÷

is a subspace of 3.

Note: A subset of a vector space is a subspace iff it is closed under ♣ & ♦.

→ It must contain 0. (c.f. Lemma 2.9.)

Proof: Let ( ) ( )1 1 1 1 2 2 2 2, , , , ,T T

x y z x y z P= = ∈r r

→ 1 1 1 2 2 20 , 0x y z x y z+ + = + + =

∴ ( )1 2 1 2 1 2 1 2, ,T

a b ax bx ay by az bz+ = + + +r r

with ( ) ( ) ( ) ( ) ( )1 2 1 2 1 2 1 1 1 2 2 2ax bx ay by az bz a x y z b x y z+ + + + + = + + + + +

→ 1 2a b P+ ∈r r QED,a b∀ ∈R

0=

Example 2.3: The x-axis in n is a subspace.

( ),0, ,0 -axisT

x x= ∈r LProof follows directly from the fact that

Example 2.4:

• { 0 } is a trivial subspace of n.

• n is a subspace of n.

Both are improper subspaces.

All other subspaces are proper.

Example 2.5: Subspace is only defined wrt inherited operations.

({1},♣ ; ) is a vector space if we define 1♣1 = 1 and a♦1=1 ∀a∈.

However, neither ({1},♣ ; ) nor ({1},+ ; ) is a subspace of the vector space (,+ ; ).

Example 2.6: Polynomial Spaces.

n is a proper subspace of m if n < m.

Example 2.7: Solution Spaces.

The solution space of any real linear homogeneous ordinary differential equation, f = 0,

is a subspace of the function space of 1 variable { f : → }.

Example 2.8: Violation of Closure.

+ is not a subspace of since (−1) v ∉ + ∀ v∈ +.

Lemma 2.9:

Let S be a non-empty subset of a vector space ( V, + ; ).

W.r.t. the inherited operations, the following statements are equivalent:

1. S is a subspace of V.

2. S is closed under all linear combinations of pairs of vectors.

3. S is closed under arbitrary linear combinations.

Proof: See Hefferon, p.93.

Remark: Vector space = Collection of linear combinations of vectors.

Example 2.11: Parametrization of a Plane in 3

2 0

x

S y x y z

z

÷= − + = ÷ ÷

is a 2-D subspace of 3.

2

,

y z

y y z

z

− ÷= ∈ ÷ ÷

R2 1

1 0 ,

0 1

y z y z

− ÷ ÷= + ∈ ÷ ÷ ÷ ÷

R

i.e., S is the set of all linear combinations of 2 vectors (2,1,0)T, & (−1,0,1)T.

Example 2.12: Parametrization of a Matrix Subspace.

00

aL a b c

b c

= + + = ÷

is a subspace of the space of 2×2 matrices.

0,

b cb c

b c

− − = ∈ ÷

R1 0 1 0

,1 0 0 1

b c b c − − = + ∈ ÷ ÷

R

Definition 2.13: Span

Let S = { s1 , …, sn | sk ∈ ( V,+, ) } be a set of n vectors in vector space V.

The span of S is the set of all linear combinations of the vectors in S, i.e.,

1

,n

k k k kk

span S c S c=

= ∈ ∈

∑ s s R { }span ∅ = 0with

Lemma 2.15: The span of any subset of a vector space is a subspace.

Proof:

Let S = { s1 , …, sn | sk ∈ ( V,+, ) }1 1

,n n

k k k kk k

u v span S= =

= = ∈∑ ∑u s v sand

→ ( )1

n

k k kk

a b au bv=

= + = +∑w u v s1

n

k kk

w span S=

= ∈∑ s ,a b∀ ∈RQED

Converse: Any vector subspace is the span of a subset of its members.

Also: span S is the smallest vector space containing all members of S.

Example 2.16:

For any v∈V, span{v} = { a v | a ∈ } is a 1-D subspace.

Example 2.17:

Proof:

The problem is tantamount to showing that for all x, y ∈, ∃ unique a,b ∈ s.t.

1 1

1 1

xa b

y

= + ÷ ÷ ÷−

i.e.,a b x

a b y

+ =− =

has a unique solution for arbitrary x & y.

Since ( )1

2a x y= + ( )1

2b x y= − ,x y∀ ∈R QED

21 1,

1 1span

= ÷ ÷− R

Example 2.18: 2

Let { }23 , 2S span x x x= − ( ){ }23 2 ,a x x bx a b= − + ∈R

Question:0

2 0

?c

S=

=P

Answer is yes since

1 3 2c a b= + 2c a= −

2a c= − ( )1

13

2b c a= −

and

( )1 2

13

2c c= +

2

1

kk

k

c x=

=

∑ = subspace of 2 ?

Lesson: A vector space can be spanned by different sets of vectors.

Definition: Completeness

A subset S of a vector space V is complete if span S = V.

Example 2.19: All Possible Subspaces of 3

See next section for proof.

Planes thru 0

Lines thru 0

Exercises 2.I.2

(a) Show that it is not a subspace of 3. (Hint. See Example 2.5).(b) Show that it is a vector space.

( To save time, you need only prove axioms (d) & (j), and closure under all linear combinations of 2 vectors.)

(c) Show that any subspace of 3 must pass thru the origin, and so any subspace of 3 must involve zero in its description. Does the converse hold?Does any subset of 3 that contains the origin become a subspace when given the inherited operations?

1. Consider the set 1

x

y x y z

z

÷ + + = ÷ ÷

under these operations.

1 2 1 2

1 2 1 2

1 2 1 2

1x x x x

y y y y

z z z z

+ − ÷ ÷ ÷+ = + ÷ ÷ ÷ ÷ ÷ ÷+

1x rx r

r y r y

z rz

− + ÷ ÷= ÷ ÷ ÷ ÷

2. Because ‘span of’ is an operation on sets we naturally consider how it interacts with the usual set operations. Let [S] ≡ Span S.(a) If S ⊆ T are subsets of a vector space, is [S] ⊆ [T] ?

Always? Sometimes? Never?(b) If S, T are subsets of a vector space, is [ S ∪ T ] = [S] ∪ [T] ?(c) If S, T are subsets of a vector space, is [ S ∩ T ] = [S] ∩ [T] ?(d) Is the span of the complement equal to the complement of the span?

3. Find a structure that is closed under linear combinations, and yet is not a vector space. (Remark. This is a bit of a trick question.)