L.D. College Of EngineeringAhmedabad

TOPIC : VECTOR SPACES

BRANCH : MECHANICAL

DIVISION : B SEM : 2ND

ACADEMIC YEAR : 2014-15

LINEAR ALIGEBRA AND VECTOR CALCULASACTIVE LEARNING ASSIGNMENT

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NAME ENROLLMENT NO.

TEJAS 140280119081

TRUPAL 140280119082

VATSAL 140280119083

JUGAL 140280119084

DHARMANSHU 140280119085

PRATIK 140280119086

-------- 140280119087

KISHAN 140280119088

MAHARSH 140280119089

NIRAV 140280119090

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CONTENTS

1) Real Vector Spaces2) Sub Spaces3) Linear combination4) Linear independence5) Span Of Set Of Vectors6) Basis7) Dimension8)Coordinate and change of basis9)Linear dependence and linear independence of function

Definition and Examples

Vector Space

Vector space is a system consisting of a set of generalized vectors and a field of scalars,having the same rules for vector addition and scalar multiplication as physical vectors and scalars.

What is Vector Space?

Let V be a non empty set of objects on which the operations of addition and multiplication by scalars are defined. If the following axioms are satisfied by all objects u,v,w in V and all scalars k1,k2 then V is called a vector space and the objects in V are called vectors.

Addition conditions:-

1.If u and v are objects in V then u+v is in V.2.u+v=v+u3.u+(v+w) = (u+v)+w4.There is an object 0 in V, called zero vector , such that 0+u=u+0 for all u in V.5.For each object u in V, there exists an object -u in V called a negative of u.

6. If K1 is any scalar and u is an object in V, then k1u is in V.7.k1(u+v) = k1u +k1v8.If k1,k2 are scalars and u is an object in V, then (k1+k2)u = k1u+k2u.9.k1(k2u) = (k1k2)u.10. 1u=u .

Scalar conditions:-

Determine whether the set R+ of all positive real numbers with operations

x + y = xy kx = xk. Is a vector space.

Example:-

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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.

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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.

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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

01Sol:

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

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Linear Combination

• A vector V is called a linear combination of the vectors v1,v2,..., vr if it can be expressed in the form as V = k1v1 + k2v2 + ... + krvr where k1, k2, ...., kr are scalars.

• Note: If r=1, then V = k1v1. This shows that a vector V is a linear combination of a single vector v1 if it is a scalar multiple of v1.

Example:

Every vector v = (a, b, c) in R3 is expressible as a linear combination of the standard basis vectors i = (1,0,0), j = (0,1,0), k=(0,0,1)

since v = (a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)

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Linear Combination• Example: Consider the vectors u=(1,2,-1) and v=(6,4,2) in R3. Show that

w=(9,2,7) is a linear combination of u and v and that w’=(4,-1,8) is not a linear combination of u and v.

vuw

vuw

23 so ,2,3

72

242

96

)2,42,6()7,2,9(

)2,4,6()1,2,1()7,2,9(

21

21

21

21

212121

21

21

kk

kk

kk

kk

kkkkkk

kk

kk

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Linear Combination

System of equations is inconsistent, so no such scalar k1 and k2 exist. w’ is not a linear combination of u and v.

822

142

46

)22,42,6()8,1,4(

)2,4,6()1,2,1()8,1,4(

21

21

21

212121

21

21

kk

kk

kk

kkkkkk

kk

kk vuw

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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:

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.

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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

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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

SPAN

What is the Spanning Set?

Let S = {v1, v2,…, vr } be a set of vectors in a vector space V, then there exists a subspace W of V consisting of all linear combinations of the vectors in S.

W is called the space spanned by v1, v2,…, vr. Alternatively, we say that the vectors v1, v2,…, vr span W.

Thus, W = span(S) = span {v1, v2,…, vr } and the set S is the spanning set of the subspace W.

In short, if every vector in V can be expressed as a linear combinations of the vectors in S, then S is the spanning set of the vector space V.

How to Find the Space Spanned by a Set of Vectors?

S = {u, v, w } = {(1,1,2),(-1,3,0),(0,1,2)} is a set of vectors in the vector space ³, and ℜ

Is Or can we solve for any x?Yes, if A-1 exists. Find det(A) to see if there is a unique solution?If we let W be the subspace of ³ℜ consisting of all linear combinations of the vectors in S, then x W for ∈

any x ³.∈ ℜ Thus, W = span(S) = ³ℜ .

The span of any subset of a vector space is a subspace

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

(x1, x2 , x 3) rx W ?

x A

rk

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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

3213

321

vvv

u

R

uuuSol:

3322113 vvvuu 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

Consistency of a system of linear equations:

Theorem: The set of equation Ax=B are consistence if and only if the coefficient matrix A and augmented matrix [A|B] have the same rank.

Conditions for consistency of non homogeneous linear equation Ax=B :

If rank of [A|B]=rank of (A)=no. of variables , the equation are consistent and have unique solution.

If rank of [A|B]=rank of(A)<no. of variables, the equation are consistent and have infinite solution.

If rank of [A|B]≠rank of (A),the equation are inconsistent and have no solution.

Condition for consistency of homogeneous linear equation Ax=0 :

x=0 is always solution . This solution in which each x1=0,x2=0,x3=0…..xn=0 is called null solution or the trivial solution.

If rank of (A)=number of variable ,the system has only trivial solution.If rank of (A)<no. of variable ,the system has an infinite non-trivial

solution.

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Basis • Definition:

V： a vector space

GeneratingSets

BasesLinearly

IndependentSets

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:

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(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}

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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

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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.

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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).

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Theorems for dimentionTHEOREM 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.

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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)) =

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Dimension of a Solution Space EXAMPLE

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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

LINEAR DEPENDENCE AND INDEPENDENCE OF FUNCTIONS

If =, =,…, =Are times differentiableFunctions on the interval

Then the wronskian of these function is……..

W=

THEOREM: if the wronskian of (n-1) times differentiable functions on the interval is not identically zero on this interval then these functions are linearly independent.

• NOTE: If the Wronskian of the functions is identically zero on the interval , then no conclusion can be made about the linear dependence or independence of the functions.

EXAMPLES

• Which of following set of the function F are linearly independent ?(1) x,sinx the wronskian of the functions is W= = xcosx-sinx Since, the function is not zero for all values of x in the interval , the given function are linearly independent.

(2) 6,3,2 the wronskian of the function is

W=

No conclusion can be made about the linear independence of the functions.

6=6+6 =2(3)+3(2 this shows that 6 can be expressed as a linear combination to given functions, hence the given functions are linearly dependent.NOTE : appropriate can be used directly to show linear dependence without using Wronskian.

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