Guided by: Prof.

L.D. College Of Engineering

Ahmedabad

TOPIC : VECTOR SPACES

BRANCH : CHEMICAL

DIVISION : F SEM : 2ND

ACADEMIC YEAR : 2014-15(EVEN)

Active Learning assignment

LINEAR ALIGEBRA AND VECTOR CALCULASACTIVE LEARNING ASSIGNMENT

1

NAME ENROLLMENT NO.

PATEL DHRUMIL P. 140280105035

SHAIKH AZIZUL AMINUL HAQUE 140280105048

SHAIKH MAHOMMED BILAL 140280105049

PANCHAL SAHIL C. 140280105031

SONI SWARUP J. 140280105056

RAMANI AJAY M. 140280105040

VAKIL JAYADEEP A. 140280105060

UMARALIYA KETAN B. 140280105057

MODI KATHAN J. 140280105027

SONI KEVIN R. 140280105054

PANCHAL SARTHAK G. 140280105032

2

CONTENTS

1) Real Vector Spaces2) Sub Spaces3) Linear combination4) Linear independence5) Span Of Set Of Vectors6) Basis7) Dimension8) Row Space, Column Space, Null Space9) Rank And Nullity10)Coordinate and change of basis

3

4

Vector Space Axioms

Real Vector Spaces

Let V be an arbitrary non empty set of objects on which two operations are defined, addition and multiplication by scalar (number). If the following axioms are satisfied by all objects u, v, w in V and all scalars k and l, then we call V a vector space and we call the objects in V vectors.

1) If u and v are objects in V, then u + v is in V2) u + v = v + u3) u + (v + w) = (u + v) + w4) There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V5) For each u in V, there is an object –u in V, called a negative of u, such that u + (-u) = (-u) + u = 06) If k is any scalar and u is any object in V then ku is in V7) k(u+v) = ku + kv8) (k+l)(u) = ku + lu9) k(lu) = (kl)u10) 1u = u

Example: Rn is a vector space

Let, V=Rn

u+v = (u1,u2,u3,. . . . . un)+(v1,v2,v3,. . . . vn)= (u1+v1,u2+v2,u3+v3, . . . .un+vn )

Ku =(ku1,ku2,ku3,. . . . . kun)

Step :1- identify the set V of

objects that will become vectors.

Step :2- identify the edition and scalar multiplication operations

on V.

Step :3- verify axioms 1 & 6, adding two

vectors in a V produces a vector in a V , and

multiplying a vector in a V by scalar

produces a vector in a V.

Step :4- confirm that axioms 2,3,4,5,7,8,9 and 10 hold.

5

Example: Set Is Not A Vector

6

Examples of vector spaces(1) n-tuple space: Rn

),,,(),,,(),,,( 22112121 nnnn vuvuvuvvvuuu

),,,(),,,( 2121 nn kukukuuuuk

(2) Matrix space: (the set of all m×n matrices with real values)

nmMV Ex: ：(m = n = 2)

22222121

12121111

2221

1211

2221

1211

vuvu

vuvu

vv

vv

uu

uu

2221

1211

2221

1211

kuku

kuku

uu

uuk

vector addition

scalar multiplication

vector addition

scalar multiplication

7

(3) n-th degree polynomial space: (the set of all real polynomials of degree n or less)

n

nn xbaxbabaxqxp )()()()()( 1100 n

n xkaxkakaxkp 10)(

(4) Function space:(the set of all real-valued continuous functions defined on the entire

real line.)

)()())(( xgxfxgf

),( cV

)())(( xkfxkf

vector addition

scalar multiplication

vector addition

scalar multiplication

8

9

PROPERTIES OF VECTORS

Let v be any element of a vector space V, and let c be any

scalar. Then the following properties are true.

vv

0v0v

00

0v

)1( (4)

or 0 then , If (3)

(2)

0 (1)

cc

c

10

Definition:

),,( V : a vector space

VW

W : a non empty subset

),,( W ：a vector space (under the operations of addition and

scalar multiplication defined in V)

W is a subspace of V

Subspaces

If W is a set of one or more vectors in a vector space V, then W is a sub space of V if

and only if the following condition hold;

a)If u,v are vectors in a W then u+v is in a W.

b)If k is any scalar and u is any vector In a W then ku is in W.

11

Every vector space V has at least two subspaces

(1)Zero vector space {0} is a subspace of V.

(2) V is a subspace of V.

Ex: Subspace of R2

0 0, (1) 00

origin he through tLines (2)

2 (3) R

• Ex: Subspace of R3

origin he through tPlanes (3)

3 (4) R

0 0, 0, (1) 00

origin he through tLines (2)

If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also subspace of V.

Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the

vector space M2×2, with the standard operations of matrix addition and scalar

multiplication.

sapcesvector : 2222 MMW

Sol:

) ( Let 221121 AA,AA WA,A TT

)( 21212121 AAAAAAWAW,A TTT

)( kAkAkAWA,Rk TT

22 of subspace a is MW

)( 21 WAA

)( WkA

Ex : (A subspace of M2×2)

12

WBA

10

01

222 of subspace anot is MW

Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×2 with

the standard operations.

WB,WA

10

00

00

01

Sol:

Ex : (The set of singular matrices is not a subspace of M2×2)

13

14

kkccc uuuv 2211

form in the written becan if in vectorsthe

ofn combinatiolinear a called is space vector ain A vector

21 vuuu

v

V,,,

V

k

Definition

scalars : 21 k,c,,cc

Linear combination

If S = (w1,w2,w3, . . . . , wr) is a nonempty set of vectors in a vector space V, then:

(a) The set W of all possible linear combinations of the vectors in S is a subspace of V.

(b) The set W in part (a) is the “smallest” subspace of V that contains all of the vectors in Sin the sense that any other subspace that contains those vectors contains W.

15

2123

2012

1101

nEliminatioJordan Guass

7000

4210

1101

)70(solution no has system this

332211 vvvw ccc

V1=(1,2,3) v2=(0,0,2) v3= (-1,0,1)Prove W=(1,-2,2) is not a linear combination of v1,v2,v3

w = c1v1 + c2v2 + c3v3

Finding a linear combination

Sol :

dependent.linearly called is then

zeros), allnot (i.e.,solution nontrivial a hasequation theIf (2)

t.independenlinearly called is then

)0(solution trivialonly the hasequation theIf (1) 21

S

S

ccc k

0vvv

vvv

kk

k

ccc

S

2211

21 ,,, : a set of vectors in a vector space V

Linear Independent (L.I.) and Linear Dependent (L.D.):

Definition:

16

Theorem

A set S with two or more vectors is(a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S.

17

tindependenlinearly is (1)

dependent.linearly is (2) SS 0

tindependenlinearly is (3) v0v

21 (4) SS

dependentlinearly is dependent linearly is 21 SS

t independenlinearly is t independenlinearly is 12 SS

Notes

1 0, 2,,2 1, 0,,3 2, 1, S

0 23

0 2

02

321

21

31

ccc

cc

cc

0vvv 332211 cccSol:

Determine whether the following set of vectors in R3

is L.I. or L.D.

0123

0012

0201

nEliminatioJordan - Gauss

0100

0010

0001

solution trivialonly the 0321 ccc

tindependenlinearly is S

v1 v2 v3

Ex : Testing for linearly independent

18

19

Span of set of vectors

If S={v1, v2,…, vk} is a set of vectors in a vector space V,

then the span of S is the set of all linear combinations of

the vectors in S.

)(Sspan

)in vectorsof nscombinatiolinear all ofset (the

2211

S

Rcccc ikk vvv

If every vector in a given vector space can be written as a linear combination of vectors in

a given set S, then S is called a spanning set of the vector space.

Definition:

If S={v1, v2,…, vk} is a set of vectors in a vector space

V, then the span of S is the set of all linear

combinations of the vectors in S.

20

(a)span (S) is a subspace of V.

(b)span (S) is the smallest subspace of V that contains S.

(Every other subspace of V that contains S must contain span (S).

If S={v1, v2,…, vk} is a set of vectors in a vector space V,

then

0)( (1) span

)( (2) SspanS

)()(

, (3)

2121

21

SspanSspanSS

VSS

Notes:

VS

SV

V S

VS

ofset spanning a is

by )(generated spanned is

)(generates spans

)(span

21

Ex: A spanning set for R3

sapns )1,0,2(),2,1,0(),3,2,1(set that theShow 3RS

. and ,, ofn combinatiolinear a as becan in

),,(vector arbitrary an whether determinemust We

321

3

321

vvv

u

R

uuu

Sol:

332211

3vvvuu cccR

3321

221

131

2 3

2

2

uccc

ucc

ucc

. and , , of valuesallfor consistent is

system this whether gdeterminin toreduces thusproblem The

321 uuu

0

123

012

201

Au.every for solution oneexactly has bx A

3)( RSspan

22

Basis • Definition:

V：a vector space

Generating

SetsBases

Linearly

Independent

Sets

S is called a basis for V

S ={v1, v2, …, vn}V

• S spans V (i.e., span(S) = V )

• S is linearly independent

(1) Ø is a basis for {0}

(2) the standard basis for R3:

{i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

Notes:

23

(3) the standard basis for Rn :

{e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1)

Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}

Ex: matrix space:

10

00,

01

00,

00

10,

00

01

22

(4) the standard basis for mn matrix space:

{ Eij | 1im , 1jn }

(5) the standard basis for Pn(x):

{1, x, x2, …, xn}

Ex: P3(x) {1, x, x2, x3}

24

THEOREMSUniqueness of basis representation

If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be

written in one and only one way as a linear combination of vectors in S.

If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more

than n vectors in V is linearly dependent.

Bases and linear dependence

25

If a vector space V has one basis with n vectors, then every basis for V has n

vectors. (All bases for a finite-dimensional vector space has the same number of

vectors.)

Number of vectors in a basis

An INDEPENDENT set of vectors that SPANS a vectorspace V is called a BASIS for V.

26

Dimension

Definition:

The dimension of a finite dimensional vector space V is defined to be the number of vectors

in a basis for V. V: a vector space S: a basis for V

Finite dimensional

A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements

Infinite dimensional

If a vector space V is not finite dimensional,then it is called infinite dimensional.

• Dimension of vector space V is denoted by dim(V).

27

Theorems for dimention

THEOREM 1

All bases for a finite-dimensional vector space have the same number of vectors.

THEOREM 2

Let V be a finite-dimensional vector space, and let be any basis.

(a) If a set has more than n vectors, then it is linearly dependent.(b) If a set has fewer than n vectors, then it does not span V.

28

Dimensions of Some Familiar Vector Spaces

(1) Vector space Rn basis {e1 , e2 , , en}

(2) Vector space Mm basis {Eij | 1im , 1jn}

(3) Vector space Pn(x) basis {1, x, x2, , xn}

(4) Vector space P(x) basis {1, x, x2, }

dim(Rn) = n

dim(Mmn)=mn

dim(Pn(x)) = n+1

dim(P(x)) =

29

Dimension of a Solution SpaceEXAMPLE

30

Row space:

The row space of A is the subspace of Rn spanned by the row vectors of A.

Column space:

The column space of A is the subspace of Rm spanned by the column vectors of A.

},,{ 21

)((2)

2

(1)

1 RAAAACS n

n

n

}|{)( 0xx ARANS n

Null space:

The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn.

Let A be an m×n matrix

Raw space ,column space and null space

The null space of A is also called the solution space of the homogeneous system Ax = 0.

mmnmm

n

n

A

A

A

aaa

aaa

aaa

A

2

1

21

22221

11211

)(

(2)

(1)

] ,, ,[

] ,, ,[

] ,, ,[

nmnm2m1

2n2221

1n1211

Aaaa

Aaaa

Aaaa

Row vectors of A Row Vectors:

n

mnmm

n

n

AAA

aaa

aaa

aaa

A

21

21

22221

11211

mn

n

n

mm a

a

a

a

a

a

a

a

a

2

1

2

22

12

1

21

11

Column vectors of A Column Vectors:

|| || ||

A(1)

A(2)

A(n)

32

THEOREM 1

Elementary row operations do not change the null space of a matrix.

THEOREM 2

The row space of a matrix is not changed by elementary row operations.

RS(r(A)) = RS(A) r: elementary row operations

THEOREM 3

If a matrix R is in row echelon form, then the row vectors with the leading 1′s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1′s of the row vectors form a basis for the column space of R.

33

Find a basis of row space of A =

2402

1243

1603

0110

3131

Sol:

2402

1243

1603

0110

3131

A=

0000

0000

1000

0110

3131

3

2

1

w

w

w

B = .E.G

bbbbaaaa 43214321

Finding a basis for a row space

EXAMPLE

A basis for RS(A) = {the nonzero row vectors of B} (Thm 3)

= {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}

34

THEOREM 4

If A and B are row equivalent matrices, then:

(a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent.(b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B.

If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero

row vectors of B form a basis for the row space of A.

THEOREM 5

35

Finding a basis for the column space of a matrix

2402

1243

1603

0110

3131

A

Sol:

3

2

1

..

00000

11100

65910

23301

21103

42611

04013

23301

w

w

w

BA EGT

EXAMPLE

36

CS(A)=RS(AT)

(a basis for the column space of A)

A Basis For CS(A) = a basis for RS(AT)

= {the nonzero vectors of B}

= {w1, w2, w3}

1

1

1

0

0

,

6

5

9

1

0

,

2

3

3

0

1

37

Finding the solution space of a homogeneous systemEx:

Find the null space of the matrix A.

Sol: The null space of A is the solution space of Ax = 0.

3021

4563

1221

A

0000

1100

3021

3021

4563

1221.. EJGA x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t

21 vvx tsts

t

t

s

ts

x

x

x

x

1

1

0

3

0

0

1

232

4

3

2

1

},|{)( 21 RtstsANS vv

38

Rank and Nullity

If A is an mn matrix, then the row space and the column space of A have the

same dimension.

dim(RS(A)) = dim(CS(A))

THEOREM

The dimension of the row (or column) space of a matrix A is called the rank ofA and is denoted by rank(A).

rank(A) = dim(RS(A)) = dim(CS(A))

Rank:

Nullity: The dimension of the null space of A is called the nullity of A.

nullity(A) = dim(NS(A))

39

THEOREM

If A is an mn matrix of rank r, then the dimension of the solution space of Ax = 0 is n – r.

That is

n = rank(A) + nullity(A)

• Notes:

(1) rank(A): The number of leading variables in the solution of Ax=0.

(The number of nonzero rows in the row-echelon form of A)

(2) nullity (A): The number of free variables in the solution of Ax = 0.

40

Rank and nullity of a matrix

Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5.

120930

31112

31310

01201

A

a1 a2 a3 a4 a5

EXAMPLE

Find the Rank and nullity of the matrix.

Sol: Let B be the reduced row-echelon form of A.

00000

11000

40310

10201

120930

31112

31310

01201

BA

a1 a2 a3 a4 a5 b1 b2 b3 b4 b5

235)(rank)(nuillity AnA

G.E.

rank(A) = 3 (the number

of nonzero rows in B)

235)(rank)(nuillity AnA

41

Coordinates and change of basis

• Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a

vector space V and let x be a vector in V such that .2211 nnccc vvvx

The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The

coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn

whose components are the coordinates of x.

n

B

c

c

c

2

1

x

Find the coordinate matrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis

B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)}

Sol:

2100

8010

5001

1521

2310

1201

E. G.J.

)5 ,3 ,2()2 ,1 ,0()1 ,0 ,1()1 ,2 ,1( 321332211

cccccc uuux

1

2

1

521

310

201

i.e.

152

23

12

3

2

1

321

32

31

c

c

c

ccc

cc

cc

2

8

5

][ B

x

Finding a coordinate matrix relative to a nonstandard basis

43

Change of basis problemYou were given the coordinates of a vector relative to one basis B and were asked to find

the coordinates relative to another basis B'.

Transition matrix from B' to B:

V

BB nn

space vector afor

bases twobe }...,,{ nda },...,,{et L 2121 uuuuuu

BB P ][][hen t vv BBnBBv,...,, 1 uuu 2

BnBB

P uuu 2 ..., , ,1

where

is called the transition matrix from B' to B

If [v]B is the coordinate matrix of v relative to B

[v]B‘ is the coordinate matrix of v relative to B'

44

If P is the transition matrix from a basis B' to a basis B in Rn, then

(1) P is invertible

(2) The transition matrix from B to B' is P–1

THEOREM 1 The inverse of a transition matrix

THEOREM 2

Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases for Rn. Then the transition

matrix P–1 from B to B' can be found by using Gauss-Jordan elimination on the n×2n

matrix as follows.

Transition matrix from B to B'

BB 1PIn

B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R2

(a) Find the transition matrix from B' to B.

(b)

(c) Find the transition matrix from B to B' .

BB ][ find ,2

1][Let ' vv

Ex : (Finding a transition matrix)

Sol:

22

21

22

43

12

23

10

01

G.J.E.

B B' I P

12

23P (the transition matrix from B' to B)

(a)

Sol:

22

21

22

43

G.J.E.

B B'

(a)

0

1

2

1

12

23][ ][

2

1][

BBBP vvv

(b)

22

43

22

21

32

21

10

01

G.J.E.

B' B I P-1

(the transition matrix from B to B')

32

211P

Check:2

1

10

01

32

21

12

23IPP

)2,3()2,4)(0()2,3)(1(0

1][

)2,3()2,2)(2()2,1)(1( 2

1][

:Check

vv

vv

B

B

(c)

References:

Anton-linear algebra book& . .. . Dhrumil Patel(L.D. College of engineering)