Vector SpacesVector SpacesA set V is called a vector space over a set K denoted V(K) if ,V is an Abelian group,

,,K is a field,

and

For every element vV and K there exists an element .v V called the “scalar multiple of v by ” satisfying

• (i)

• (ii)

• (iii)

• (iv)

vuvuKVvu ...,,

vvvVvK ...,,

vvVvK ...,,

vv .1

Notation : 0 K denotes the additive identity under +, K, - denotes the inverse of under +

V0 denotes the identity under Vv v denotes the inverse of v under

ExamplesExamplesExample 1 (Polynomials of degree n)

,,RKV = set of all polynomials of order n

The additive operation on vectors is defined as follows:

00111

11

011

1

011

1

baxbaxbaxba

bxbxbxb

axaxaxa

nnn

nnn

nn

nn

nn

nn

V(K) is a vector space

00000 1 xxx nn

Proof

v= 011

1 axaxaxa nn

nn

For 01

11 axaxaxav n

nn

n

Now vuuv by continuity of addition

on the real numbers

Examples:1Examples:1

Proof continued

Closure is trivial

Associativity of follows from associativity ofnormal addition

011

1 axaxaxavR nn

nn

.,

therefore

uv

baxba

xbaxba

baxbaxbaxba

uv

nnn

nnn

nnn

nnn

.

.

.

0011

111

00111

11

Similarly for properties (ii)-(iv)

Hence, ,V is an Abelian group

Examples:2Examples:2Example 2 (n dimensional vectors)

,,RK

RaaaV in ,,1

nn

nn

babauv

bbuaav

,,

,,,,,

11

11

then

Example 3 (Complex Numbers)

,,RK

RbabiaV ,

idbcauv

dicubiav

then

,

Example 4 (Matrices)

,,RK ,, ,mnMV

mnM , is the set of nm matrices

nM is the set of nn matrices

Properties of Vector SpacesProperties of Vector SpacesTheorem

0.0 v

Proof

vv .00.0 Identity under +

vv .0.0 by axiom (ii) of vector spaces

But vv .00.0 Identity under

Therefore

vvv .00.0.0

0.0 v by the cancellation law for ,V

K

0

V

v

0

Properties:2Properties:2Theorem

v. v.(i)

(ii) (-).( v) =.v

Proof

(i)

vv ..00 by previous theorem and inverse under +

Therefore,

vv ..0 by axiom (ii) of vector spaces

Also

= .( v)

0 (.v) v. by inverse under

Therefore

vv .. (.v) v.

v. (.v)

by the cancellation law for,V

Show v. v.

Properties:3Properties:3Show v. = .( v)

.(v ( v)) 0. by inverse under

Therefore,

.v . v 0. by axiom (i) vector spaces

Now

Vv vv .0.00. by previous theorem

and axiom (iii)

0.0 v by above theorem

.v . v 0

Also

.v (.v) 0 by inverse under

Therefore,

.v . v= .v (.v)

. v= (.v) By the cancellation law for ,V

(ii) proof ??

SubspacesSubspaces

Definition

Let WV such that W then W(K) is a subspace of V(K) if W is a vector space over K with the same definition of and scalar multiple as V

Clearly to show that W(K) is a subspace of V(K) we need only show that <W,> is a sub-group of <V, > and that

WuWuK .,

Characterisation Theorem

A non-empty subset W of V is a subspace of V iff .u vW for all K, u,vW

Proof

() trivial since .uW

Subspaces:2Subspaces:2

Proof (continued)

() Taking =1 then u,vW

WvuWvu .1 by axiom (iv)

Taking =-1 then u,vW

Wvu .1 and by a previous theorem

(-1).u= (1.u)= u by axiom (iv)

Therefore u uW W0 by inverse under

Therefore K, uW taking 0v

gives .uW

Therefore, taking =-1 gives WuWu .1

(1.u)W uW

by a previous theorem

Hence, ,W is a subgroup and

WuWuK ., as required

Examples of SubspacesExamples of Subspaces

Example 1

Let ,,RK

RaaaaW in ,,,, 320

RaaaV in ,,1

Then W(K) is a subspace of V(K)

W(K)

V(K)

Examples of Subspaces:2Examples of Subspaces:2

Example 1 (continued)proof

(i) Clearly W and WV

(ii) For R and u,vW such that

naau ,,, 20 and nbbv ,,, 20

then

Wbabavu nn ,,,. 220

Example 2

RaaaV in ,,1

,,RK

RaaaaW in ,,,1 22

(i) Clearly W and WV

W0,,0,1,2 and 2R but

W 0,,0,2,40,,0,1,22 W(K) is not a subspace of V(K)

Linear CombinationsLinear Combinations

Definition

If VvvS k ,,1 then

k

iiiv

1

where Ki

is a linear combination of S

1v

2v

Linear Combinations:2Linear Combinations:2Theorem

For VvvS k ,,1 then

kiKvU i

k

iii ,,1,

1

is a subspace of V

Proof

U0 since Uvv k .0.0 1 and 0.0 iv

by a previous theorem and hence U

If UvuK ,, then kkvvu 11

and kkvvv 11 for some Kii ,

Then

kkkk

kk

kk

vvvv

vv

vvvu

1111

11

11

..

..

by axiom (i) kkkk vvvv 1111 ..

by axiom (iii) kkk vv .. 111 by axiom (ii)U

Linear IndependenceLinear Independence

Definition

A subset kvv ,,1 of V is linearly independentiff

kivv ikk ,,1,0011 Otherwise if there exist one 0i such that

ikkvv 011

then kvv ,,1 are linearly dependent

Example 1

1,0,0,2,1,2,1,1,1,1,0,1

is linearly dependent over R since

0,0,01,0,002,1,21,1,11,0,1

Example 2

1,0,0,1,0,1 is linearly independent since

00,0

0,0,01,0,01,0,1

Linear Dependence in Linear Dependence in MatricesMatrices

Theorem

KMaA nji ,If and jnjjj aaac ,,2,1 ,,,

nj ,,1 are the n column vectors of A then

ncc ,,1 is linearly dependent over K if and

only if 0det A

Proof

If ncc ,,1 is linearly dependent over K, then

there exists Kn ,,, 21 (not all zero)

such that0,,0,02211 nnccc

Without loss of generality assume 01 then

0,,0,01

21

21 n

n ccc

Hence, performing a column operation where

nn cc1

21

2

is added to column 1 gives

a matrix with zero first column. Hence, 0det A

Matrices:2Matrices:2Proof (continued)

If 0det A then the system of equations

nixa j

n

jji ,,1:0

1,

has a non-trivial solution, nxx ,,1

But this is the same as saying that there exist

Kxx n ,,1 (not all zero) such that

0,,0,02211 nncxcxcx

By considering the transpose of A we obtain

Corollary

The n rows of a matrix KMA n are linearly

dependent if and only if 0det A

BasisBasis

A set

Definition

kvvS ,,1 is a basis for V iff

(i) S is linearly independent over K

VkiKv i

k

iii

,,1,.1

(ii)

Condition (ii) means that S is a spanning set for V

Definition (Finitely Generated)

A vector space V is said to be finitely generated if it has a Basis S with a finite number of elements

S V

i

k

iiv

1

Examples of BasisExamples of Basis

Let ,,RK

Example 1

RaaaaV i 321 ,,

Then 1,0,0,0,1,0,0,0,1 is a basis for V

Linear independence:

0

0,0,01,0,00,1,00,0,1

321

321

Spanning:

1,0,00,1,00,0,1,, 321321 aaaaaa

1,1,4,1,1,2,1,0,1 is also a basis for V

Proof ??

Examples of Basis:2Examples of Basis:2

Example 2

RMV 2Let then 4321 ,,, vvvv is a basis

where

00

011v

00

102v

01

003v

10

004v

Example 3

RaaxaxaV i 012

2then

1,,2 xx is a basis

DimensionDimension

Theorem Every Basis of a finitely generated vector space has the same number of elements.

Definition (Dimension)

The number of elements in a basis for a finitely generated vector space V is called the dimension of V and denoted dim V.

Examples

RaaxaxaV i 012

2then

1,,2 xx is a basis

dim(V) = 3 RMV 2Let then

is a basis and dim(V)=4

00

011v

00

102v

01

003v

10

004v