vcla ppt ch=vector space
TRANSCRIPT
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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