this is the ppt on vector spaces of linear algebra and vector calculus (VCLA) contents : Real Vector Spaces Sub Spaces Linear combination Linear independence Span Of Set Of Vectors Basis Dimension Row Space, Column Space, Null Space Rank And Nullity Coordinate and change of basis this is made by dhrumil patel which is in chemical branch in ld college of engineering (2014-18) i think he is the best ppt maker,dhrumil patel,harshid panchal

1. 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 CALCULAS
ACTIVE LEARNING ASSIGNMENT
1

2. 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

3. 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) Row Space, Column Space, Null Space
9) Rank And Nullity
10)Coordinate and change of basis
3

4. 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 V
2) u + v = v + u
3) u + (v + w) = (u + v) + w
4) There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V
5) For each u in V, there is an object –u in V, called a negative of u, such that u + (-u) = (-u) + u = 0
6) If k is any scalar and u is any object in V then ku is in V
7) k(u+v) = ku + kv
8) (k+l)(u) = ku + lu
9) k(lu) = (kl)u
10) 1u = u

5. 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

6. Example: Set Is Not A Vector
6

7. 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
uu
k
vector addition
scalar multiplication
vector addition
scalar multiplication
7

8. (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. 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)
or0then,If(3)
(2)
0(1)
cc
c

10. 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. 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
   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.

12. 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)
12

13. 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)
13

14. 14
kkccc uuuv  2211
formin thewrittenbecanifinvectorsthe
ofncombinatiolinearacalledisspacevectorainA 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 S
in the sense that any other subspace that contains those vectors contains W.

15. 15













2123
2012
1101
   nEliminatioJordanGuass












7000
4210
1101
)70(solutionnohassystemthis  
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 :

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.









 

0123
0012
0201
  nEliminatioJordan-Gauss










0100
0010
0001
 solutiontrivialonly the0321  ccc
tindependenlinearlyisS
v1 v2 v3
Ex : Testing for linearly independent
18

19. 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  
)invectorsofnscombinatiolinearallofset(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. 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
VS
VS
ofsetspanningais
by)(generatedspannedis
)(generatesspans
)(span



21. 21
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 

22. 22
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:

23. 23
(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:






























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}

24. 24
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

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

26. 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. 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. 28
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)) = 

29. 29
Dimension of a Solution Space
EXAMPLE

30. 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.

31.  
 
 



























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. 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. 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. 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. 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. 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. 37
Finding the solution space of a homogeneous system
Ex:
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
.. EJG
A 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. 38
Rank and Nullity
If A is an mn 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 of
A 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. 39
THEOREM
If A is an mn 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. 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. 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

42. 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. 43
Change of basis problem
You 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
spacevectorafor
basestwobe}...,,{nda},...,,{etL 2121 uuuuuu 
BB P  ][][hent vv         BBnBB
v,...,,  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. 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 R
n
. 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   1
PIn 

45. B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R
2
(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
23
P (the transition matrix from B' to B)
(a)
Sol:










22
21
22
43

 G.J.E.
B B'
(a)

46. 

























 
0
1
2
1
12
23
][][
2
1
][ BBB
P 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
211
P
 Check: 2
1
10
01
32
21
12
23
IPP 






















)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)

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