vector space

1. Presented By
150490106037 – Harekrishna Jariwala
150490106027 – Niraj Devaiya
SNPIT & RC
UMRAKH
VECTOR SPACE

2. CONTENTS
1) Real Vector Spaces
2) Sub Spaces
3) Linear combination
4) Linear independence
5) Span Of Set Of Vectors
6) Basis
7) Dimension
8)Coordinate and change of basis
9)Linear dependence and linear independence of
function
2

3.  Definition and Examples
Vector Space

4. 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.

5.  Addition conditions:-
1.If u and v are objects in V then u+v is in V.
2.u+v=v+u
3.u+(v+w) = (u+v)+w
4.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. 6. If K1 is any scalar and u is an object in V, then k1u is in V.
7.k1(u+v) = k1u +k1v
8.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:-

7. Determine whether the set R+ of all positive real numbers with
operations
x + y = xy
kx = xk.
Is a vector space.
Example:-

8. 8
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.

9. 9
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
   00,(1) 00
originhethrough tLines(2)
2
(3) R
• Ex: Subspace of R3
originhethrough tPlanes(3)
3
(4) R
   00,0,(1) 00
originhethrough tLines(2)
If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also
subspace of V.

10. 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,AAWA,A TT

)( 21212121 AAAAAAWAW,A TTT

)( kAkAkAWA,Rk TT

22ofsubspaceais  MW
)( 21 WAA 
)( WkA
Ex : (A subspace of M2×2)
10

11. WBA 






10
01
222 ofsubspaceanotis  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)
11

12. 12
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.

13. 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)

14. 14
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
23so,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

15. 15
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

16. dependent.linearlycalledisthen
zeros),allnot(i.e.,solutionnontrivialahasequationtheIf(2)
t.independenlinearlycalledisthen
)0(solutiontrivialonly thehasequationtheIf(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. 17
tindependenlinearlyis(1) 
dependent.linearlyis(2) SS 0
  tindependenlinearlyis(3) v0v 
21(4) SS 
dependentlinearlyisdependentlinearlyis 21 SS 
tindependenlinearlyistindependenlinearlyis 12 SS 
Notes

18.       10,2,,21,0,,32,1, S
023
02
02
321
21
31



ccc
cc
cc
 0vvv 332211 ccc
Sol:
Determine whether the following set of vectors in R
3
is L.I. or L.D.








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 

0123
0012
0201
  nEliminatioJordan-Gauss


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
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




0100
0010
0001
 solutiontrivialonly the0321  ccc
tindependenlinearlyisS
v1 v2 v3
Ex : Testing for linearly independent
18

19. SPAN

20. 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.

21. 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 ℜ³, and
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 ℜ³ consisting of all linear combinations of the vectors in S, then x ∈ W for
any x ∈ ℜ³.
Thus, W = span(S) = ℜ³.
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 ) 
r
x W ? r
x  A
r
k

22. 22
Ex: A spanning set for R3
  sapns)1,0,2(),2,1,0(),3,2,1(setthat theShow 3
RS 
.and,,ofncombinatiolinearaasbecanin
),,(vectorarbitraryanwhetherdeterminemustWe
321
3
321
vvv
u
R
uuu
Sol:
332211
3
vvvuu cccR 
3321
221
131
23
2
2
uccc
ucc
ucc



.and,,ofvaluesallforconsistentis
systemthiswhethergdeterminintoreducesthusproblemThe
321 uuu
0
123
012
201


A
u.everyforsolutiononeexactlyhasbx  A
3
)( RSspan 

23. 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.

24. 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.

25. 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.

26. 26
Basis
• Definition:
V：a vector space
Generating
Sets
Bases
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:

27. 27
(3) the standard basis for R
n
:
{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:








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
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


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





10
00
,
01
00
,
00
10
,
00
01
22
(4) the standard basis for mn matrix space:
{ Eij | 1im , 1jn }
(5) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x) {1, x, x2, x3}

28. 28
THEOREMS
Uniqueness 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

29. 29
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 vector
space V is called a BASIS for V.

30. 30
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).

31. 31
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.

32. 32
Dimensions of Some Familiar Vector Spaces
(1) Vector space Rn  basis {e1 , e2 ,  , en}
(2) Vector space Mm  basis {Eij | 1im , 1jn}
(3) Vector space Pn(x)  basis {1, x, x2,  , xn}
(4) Vector space P(x)  basis {1, x, x2, }
 dim(Rn) = n
 dim(Mmn)=mn
 dim(Pn(x)) = n+1
 dim(P(x)) = 

33. 33
Dimension of a Solution Space
EXAMPLE

34. 34
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

35. 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.













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








)5,3,2()2,1,0()1,0,1()1,2,1( 321332211
 cccccc uuux




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


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









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
















1
2
1
521
310
201
i.e.
152
23
12
3
2
1
321
32
31
c
c
c
ccc
cc
cc




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
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


 
2
8
5
][ B
x
Finding a coordinate matrix relative to a nonstandard basis

36. LINEAR DEPENDENCE AND
INDEPENDENCE OF FUNCTIONS
If 𝒇 𝟏=𝒇 𝟏 𝒙 , 𝒇 𝟐=𝒇 𝟐 𝒙 ,…, 𝒇 𝒏=𝒇 𝒏 𝒙
Are 𝒏 − 𝟏 times differentiable
Functions on the interval −∞, ∞
Then the wronskian of these function is……..

37. W=
𝒇 𝟏 𝒙 𝒇 𝟐 𝒙 … 𝒇 𝒏 𝒙
𝒇 𝟏
|
(𝒙) 𝒇 𝟐
|
(𝒙) … 𝒇 𝒏
|
(𝒙)
⋮ ⋮ ⋮
𝒇 𝟏
𝒏−𝟏
(𝒙) 𝒇 𝟐
𝒏−𝟏
(𝒙) … 𝒇 𝒏
𝒏−𝟏
(𝒙)
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.

38. • 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.

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

40. (2) 6,3𝑠𝑖𝑛2 𝑥,2𝑐𝑜𝑠2 𝑥
the wronskian of the function is
W=
6 3𝑠𝑖𝑛2 𝑥 2𝑐𝑜𝑠2 𝑥
0 6𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 −4𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥
0 6𝑐𝑜𝑠2𝑥 −4𝑐𝑜𝑠2𝑥
= 0
No conclusion can be made about the linear
independence of the functions.

41. 6=6𝑠𝑖𝑛2 𝑥+6𝑐𝑜𝑠2 𝑥
=2(3𝑠𝑖𝑛2 𝑥)+3(2𝑐𝑜𝑠2 𝑥)
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.

42. 42