FACULTY :Prof. D.K. KatariyaProf. S.J. VoraProf. D.H. Parmar

SHANTILAL SHAH ENGINEERING COLLEGE, BHAVNAGAR

(INSTRUMENTATION & CONTROL DEPARTMENT)

Roll no.

Enrollment no. Name

3017 140430117019 JITIN J PILLAI

3007 140430117008 CHAUHAN ASHISH MUKESHBHAI

3010 140430117012 DODIYA PARTH KISHORBHAI

3016 140430117018 ISHAN JATINKUMAR BHATT

3002 140430117002 BAVISHI BIRJU DINESHBHAI

Group Members:

BATCH B1

VECTOR SPACES

Vector Spaces

I. Definition of Vector Space

II. Subspaces

III. Basis and Dimension

Vector space ~ Linear combinations of vectors.

I. Definition of Vector Space

I.1. Definition and ExamplesI.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.

Definition and ExamplesDefinition : (Real) Vector Space ( V, ; R )A vector space (over R) 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 R, [ R is the real number field (R,+,)

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 : (Real) Vector Space ( V, + ; R )A vector space (over R) 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 R, [ R is the real number field (R,+,) ]

f) a v V ( Closure )g) ( a + b ) v = a v + b v ( Distributivity )h) a ( v + w ) = a v + a wi) ( a b ) v = a ( b v ) = a b v ( Associativity )j) 1 v = v

Definition in Conventional Notations

Example : R2

R2 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 : Plane in R3.

The plane through the origin 0

x

P y x y z

z

is a vector space.

P is a subspace of R3.

Example :

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

( Zn, + ; Z ) is a vector space.

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

Example :0

0

0

0

V

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

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

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

0

nk

n k kk

a x a

P R 2 3

3 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.,

Pn 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 baai.e.

,

E.g.,

Pn is isomorphic to Rn+1 with 10

0

~ , ,n

k nk n n

k

a x a a

P R

Inverse: 0

nk

kk

a x

a kka ai.e.

,

kkaa

The kth component of a is

i.e.,

Subspaces and Spanning Sets

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

0

x

P y x y z

z

is a subspace of R3.

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 : Plane in R3 0

x

P y x y z

z

is a subspace of R3.

Proof: Let

→ 1 1 1 2 2 20 , 0x y z x y z

with

→ 1 2a b P r r ,a b R

0

Example : The x-axis in Rn is a subspace.

,0, ,0 -axisT

x x r Proof follows directly from the fact that

Example : • { 0 } is a trivial subspace of Rn.

• Rn is a subspace of Rn.

Both are improper subspaces.

All other subspaces are proper.

Example : Subspace is only defined wrt inherited operations.

({1}, ; R) is a vector space if we define 11 = 1 and a1=1 aR.

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

Definition : Span

Let S = { s1 , …, sn | sk ( V,+,R ) } 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 : The span of any subset of a vector space is a subspace.

Proof:

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

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 :

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

Example :

Proof:

The problem is tantamount to showing that for all x, y R, unique a,b R 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

21 1,

1 1span

R

Example : P2

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 P2 ?

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

Example 2.19: All Possible Subspaces of R3

Planes thru 0

Lines thru 0

Vector Spaces

Definition: A subspace of a vector space V is a subset H of V that has three properties:

a. The zero vector of V is in H.

b. H is closed under vector addition. That is, for each u and v in H,

the sum is in H.

c. H is closed under multiplication by scalars. That is, for each u in

H and each scalar c, the vector cu is in H.

u v

II. Subspaces

Properties (a), (b), and (c) guarantee that a subspace H of V is itself a vector space, under the vector space operations already defined in V.

Every subspace is a vector space.

Conversely, every vector space is a subspace (of itself and possibly of other larger spaces).

A Subspace Spanned By A Set

The set consisting of only the zero vector in a vector space V is a subspace of V, called the zero subspace and written as {0}.

As the term linear combination refers to any sum of scalar multiples of vectors, and Span {v1,…,vp} denotes the set of all vectors that can be written as linear combinations of v1,…,vp.

Example 2: Given v1 and v2 in a vector space V, let

. Show that H is a subspace of V.

Solution: The zero vector is in H, since . To show that H is closed under vector addition, take two arbitrary vectors in

H, say,

and . By Axioms 2, 3, and 8 for the vector space V,

So is in H. Furthermore, if c is any scalar, then by Axioms 7 and 9,

which shows that cu is in H and H is closed under scalar multiplication. Thus H is a subspace of V.

u w

1 1 2 2 1 1 2 2u ( v v ) ( )v ( )vc c s s cs cs

Theorem 1: If v1,…,vp are in a vector space V, then Span {v1,…,vp} is a subspace of V.

We call Span {v1,…,vp} the subspace spanned (or generated) by {v1,…,vp}.

Give any subspace H of V, a spanning (or generating) set for H is a set {v1,…,vp} in H such that

1Span{v ,...v }pH

Subspace

Basis

Definition 1.1: Basis

A basis of a vector space V is an ordered set of linearly independent (non-zero) vectors that spans V.

Notation:

1 , , nβ β

Example 1.2:

2 1,

4 1B

is a basis for R2

B is L.I. :

2 1 0

4 1 0a b

→2 0

4 0

a b

a b

→0

0

a

b

B spans R2:

2 1

4 1

xa b

y

→

2

4

a b x

a b y

→ 1

22

a y x

b x y

L.I. → Minimal

Span → Complete

Example 1.3:

1 2,

1 4B

is a basis for R2 that differs from B only in order.

Definition 1.5: Standard / Natural Basis for Rn

1 0 0

0 1 0, , ,

0 0 1

n

E 1 2, , , n e e e

kth component of ei =1

0

i kfor

i k

Example 1.6:

For the function space cos sin ,a b a b R

a natural basis is cos , sin

Another basis is cos sin , 2cos 3sin

Proof is straightforward.

Example 1.7:

For the function space of cubic polynomials P3 ,

a natural basis is

Other choices can be

Proof is again straightforward.

Rule: Set of L.C.’s of a L.I. set is L.I. if each L.C. contains a different vector.

Example 1.8:

The trivial space { 0 } has only one basis, the empty one .

Note: By convention, 0 does not count as a basis vector. ( Any set of vectors containing 0 is linearly dependent. )

Example 1.9:The space of all finite degree polynomials has a basis with infinitely many elements 1, x, x2, … .

Example 1.10: Solution Set of Homogeneous SystemsThe solution set of

0

0

x y w

z w

1 1

1 0,

0 1

0 1

y w y w

Ris

1 1

1 0,

0 1

0 1

Span

( Proof of L.I. is left as exercise )

Example 1.11: MatricesFind a basis for this subspace of M22 : 2 0

0

a ba b c

c

S

2,

0

b c bb c

c

S R

Solution:

1 1 2 0,

0 0 1 0b c b c

R

∴ Basis is1 1 2 0

,0 0 1 0

( Proof of L.I. is left as exercise )

Theorem 1.12:In any vector space, a subset is a basis if and only if each vector in the space can be expressed as a linear combination of elements of the subset in a unique way.

Proof: A basis is by definition spanning

→ every vector can be expressed as a linear combination of the basis vectors.

Let i i i ii i

c d βv β then i i ii

c d β 0

∴ L.I. uniqueness

Definition 1.13: Representation wrt a Basis

Let B = β1 , …, βn be a basis of vector space V and

Then the representation of v wrt B is

1i i

n

i

c

βv

1

2Rep

n

c

c

c

vB

B

cj are called the coordinates (components) of v wrt B.

Subscript B is often omitted

Example 1.14: P3

Let 2 31, 2 , 2 , 2x x xB 2 31 ,1 , ,x x x x x x D

Then

2

0

1/ 2Rep

1/ 2

0

x x

B

B

2

0

0Rep

1

0

x x

D

D

vB

Dimension

Definition 2.1 A vector space is finite-dimensional if it has a basis with only finitely many vectors.

Lemma 2.2: Exchange Lemma

Assume that B = β1 , …, βn is a basis for a vector space, and that for the vector v the relationship holds:

To be proved:

All bases for a vector space have the same number of elements.

→ Dimension Number of vectors in basis.

→ Basis = Minimal spanning = L.I. set = Smallest set.

Then exchanging βj for v yields another basis for the space.

1 1 j j n nc c c v β β β where cj 0.

Proof: See Hefferon p.120.

Theorem 2.3: In any finite-dimensional vector space, all of the bases have the same number of elements.

Proof:

Let B = β1 , …, βn be a basis of n elements.

Any other basis D = δ1 , …, δm must have m n.

1 1 1 k k n nc c c δ β β β Let with ck 0.

By lemma 2.2, D1 = β1 , …, βk 1 ,δ1 ,βk + 1 , …, βn is a basis.

Next, replacing βj in D1 begets

D2 = β1 , …, βk 1 ,δ1 ,βk + 1 , …, βj 1 ,δ2 ,βj + 1 , …, βn

Repeating the process n times results in a basis Dn = δ1 , …, δn that spans V.

Which contradicts with the assumption that D is L.I.

1 1 1n n nc c δ δ δ with at least one ck 0.If m > n, then we can write

Hence m = n.

Definition 2.4: DimensionThe dimension of a vector space is the number of vectors in any of its bases.

Example 2.5: Rn Any basis for Rn has n vectors since the standard basis En has n vectors. → Rn is n-D.

Example 2.6: Pn

dim Pn = n+1. since its natural basis, 1, x, x2, …, xn , has n+1 elements.

Example 2.7: A trivial space is 0-D since its basis is empty.

Comments:

All results in this book are applicable to finite-D vectors spaces.

Most of them are also applicable to countably infinite-D vectors spaces.

For uncountably infinite-D vectors spaces, e.g., Hilbert spaces, convergence most be taken into account.

Corollary 2.8:No L.I. set can have a size greater than the dimension of the enclosing space.

Example 2.9 : Only subspaces in R3.

2-D: Planes thru 0

1-D: Lines thru 0

0-D: {0}

Corollary 2.10:Any L.I. set can be expanded to make a basis.

Corollary 2.11: Any spanning set can be shrunk to a basis.

Corollary 2.12: In an n-D space, a set of n vectors is L.I. iff it spans the space.

Remark 2.13:The statement ‘any infinite-dimensional vector space has a basis’ is known to be equivalent to a statement called the Axiom of Choice.

Mathematicians differ philosophically on whether to accept or reject this statement as an axiom on which to base mathematics (although, the great majority seem to accept it).

References

• http://math.kennesaw.edu/~plaval/math3260/basis.pdf

• https://www.google.co.in/search?q=basis+and+dimension

• https://www.google.co.in/search?q=vector+spaces

• www2.gsu.edu/~matixb/sec4_1.pdf

• en.wikipedia.org/wiki/Vector_space

• https://www.khanacademy.org/math/linear-algebra/vectors_and_

spaces

• LINEAR ALGEBRA AND VECTOR CALCULUS by Ravish R Singh and Mukul Bhatt (McGraw Hill Education)

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