MAths PPT about the defination of vector Space

1. Prepared by
Tanuj Parikh

2. Chapter Two: Vector Spaces
Vector space ~ Linear combinations of vectors.
I. Definition of Vector Space
II. Linear Independence
III. Basis and Dimension
• Topic: Fields
• Topic: Crystals
• Topic: Voting Paradoxes
• Topic: Dimensional Analysis
Ref: T.M.Apostol, “Linear Algebra”, Chap 3, Wiley (97)

3. I. Definition of Vector Space
I.1. Definition and Examples
I.2. Subspaces and Spanning Sets

4. Algebraic Structures
Ref: Y.Choquet-Bruhat, et al, “Analysis, Manifolds & Physics”, Pt
I., North Holland (82)
Structure Internal Operations Scalar Multiplication
Group * No
Ring, Field * , + No
Module / Vector Space + Yes
Algebra + , * Yes
Field = Ring with idenity & all elements except 0 have inverses.
Vector space = Module over Field.

5. I.1. Definition and Examples
Definition 1.1: (Real) Vector Space ( V, § ;  )
A vector space (over ) consists of a set V along with 2 operations ‘§’ and ‘¨’ s.t.
(1) For the vector addition § :
" v, w, u Î V
a) v § w Î V ( Closure )
b) v § w = w § v ( Commutativity )
c) ( v § w ) § u = v § ( w § u ) ( Associativity )
d) $ 0 Î V s.t. v § 0 = v ( Zero element )
e) $ -v Î V s.t. v § (-v) = 0 ( Inverse )
(2) For the scalar multiplication ¨ :
" v, w Î V and a, b Î , [  is the real number field (,+,´)
f) a ¨ v Î V ( Closure )
g) ( a + b ) ¨ v = ( a ¨ v ) § (b ¨ v ) ( Distributivity )
h) a ¨ ( v § w ) = ( a ¨ v ) § ( a ¨ w )
i) ( a ´ b ) ¨ v = a ¨ ( b ¨ v ) ( Associativity )
j) 1 ¨ v = v
 § is always written as + so that one writes v + w instead of v § w
 ´ and ¨ are often omitted so that one writes a b v instead of ( a ´ b ) ¨ v

6. Definition in Conventional Notations
Definition 1.1: (Real) Vector Space ( V, + ;  )
A vector space (over ) consists of a set V along with 2 operations ‘+’ and ‘ ’ s.t.
(1) For the vector addition + :
" v, w, u Î V
a) v + w Î V ( Closure )
b) v + w = w + v ( Commutativity )
c) ( v + w ) + u = v + ( w + u ) ( Associativity )
d) $ 0 Î V s.t. v + 0 = v ( Zero element )
e) $ -v Î V s.t. v -v = 0 ( Inverse )
(2) For the scalar multiplication :
" v, w Î V and a, b Î , [  is the real number field (,+,´) ]
a) a v Î V ( Closure )
b) ( a + b ) v = a v + b v ( Distributivity )
c) a ( v + w ) = a v + a w
d) ( a ´ b ) v = a ( b v ) = a b v ( Associativity )
e) 1 v = v

7. Example 1.3: 2
x y
2 is a vector space if 1 1
æ ö æ ö
ax by
ax by
æ + ö
= ç ¸ è + ø
x y 1 1
a b a b
+ = ç ¸+ ç ¸
x y
è 2 ø è 2
ø
2 2
" a,bÎR
0
0
= æ ö ç ¸
è ø
with 0
Proof it yourself / see Hefferon, p.81.
Example 1.4: Plane in 3.
The plane through the origin 0
ì æ x
ö ü
= ï ç ¸ í ç ¸ + + = ï ý ï î çè ø¸ ï þ
P y x y z
z
is a vector space.
P is a subspace of 3.
Proof it yourself / see Hefferon,
p.82.

8. Example 1.5:
Let § & ¨ be the (column) matrix addition & scalar multiplication, resp., then
( n, + ;  ) is a vector space.
( n, + ;  ) is not a vector space since closure is violated under scalar multiplication.
Example 1.6:
0
0
0
0
V
ì æ ö ü
ï ç ¸ ï = ï ç ¸ ï í ç ¸ ý ï ç ¸ ï îï è ø ïþ
Let then (V, + ;  ) is a vector space.
Definition 1.7: A one-element vector space is a trivial space.

9. Example 1.8: Space of Real Polynomials of Degree n or less, n
ìï ïü = í Î ý
îï ïþ
P å R { 2 3 }
0
n
k
n k k
k
a x a
=
3 0 1 2 3 k P = a + a x + a x + a x a ÎR
n is a vector space with vectors
a º å
n n
The kth component of a is
a + b = å +å ( ) k k k a + b = a + b
Vector
addition: 0 0
k k
a x b x
k k
k k
= =
Scalar multiplication:
æ ö
a å
b b a x
= ç ¸
è 0
ø
n
k
k
k
=
Zero element:
= å +
= å
0 = å ( ) 0 k i.e., 0 = "k
0
0
n
k
k
x
=
0
n
k
k
k
a x
=
( )
0
n
k
k k
k
a b x
=
0
n
k
k
k
ba x
=
i.e.
,
( ) k k i.e. b a = ba
,
E.g.
,
-a º å - ( ) k k i.e. -a = -a
å ÎP L ÎR
is isomorphic to n+1 with ( ) 1
n 0
0
~ , ,
n
k n
k n n
k
a x a a +
=
Inverse: ( )
0
n
k
k
k
a x
=
,
( ) k k a = a

10. Example 1.9: Function Space
The set { f | f :  →  } of all real valued functions of natural numbers is
a vector space if
Vector addition: ( f1 + f2 ) ( n) º f1 ( n) + f2 ( n)
Scalar multiplication: ( a f ) ( n) º a f ( n)
nÎN
aÎR
Zero
element:
zero(n) = 0
Inverse: ( - f ) (n) º - f ( n)
f ( n ) is a vector of countably infinite dimensions: f = ( f(0), f(1), f(2), f(3), … )
E.g.,
f ( n) = n2 +1 ~ f = (1, 2, 5,10,L)

11. Example 1.10: Space of All Real Polynomials, 
ìï ïü = í Î Î ý
îï ïþ
P å R N
0
,
n
k
k k
k
a x a n
=
 is a vector space of countably infinite dimensions.
å ÎP ( L ) ÎR
0 1 2
0
k ~ , , ,
k
k
a x a a a
¥
¥
=
Example 1.11: Function Space
The set { f | f :  →  } of all real valued functions of real numbers is a
vector space of uncountably infinite dimensions.

12. Example 13: Solution Space of a Linear Homogeneous Differential Equation
ì ü
= í ® + = ý
î þ
2
2 S f : d f f 0
R R is a vector space with
d x
Vector addition: ( f + g) ( x) º f ( x) + g ( x)
Scalar multiplication: ( a f ) ( x) º a f ( x)
Zero
element:
zero(x) = 0
Inverse: ( - f ) (x) º - f ( x)
Closure:
aÎR
+ = + = ( ) ( )
2 2
2 2 d f f 0 & d g g 0
d x d x
2
d a f bg
a f bg
2 0
d x
+
→ + + =
Example 14: Solution Space of a System of Linear Homogeneous Equations

13. Remarks:
• Definition of a mathematical structure is not unique.
• The accepted version is time-tested to be most concise & elegant.
Lemma 1.16: Lose Ends
In any vector space V,
1. 0 v = 0 .
2. ( -1 ) v + v = 0 .
3. a 0 = 0 .
" v ÎV and a Î .
Proof:
1. 0 = v - v = (1+ 0) v - v = v + 0v - v = 0v
2.
( -1) v + v = ( -1+1) v = 0v = 0
3.
a 0 = a ( 0 v) = ( a 0) v = 0 v = 0

14. Exercises 2.I.1.
1. At this point “the same” is only an intuition, but nonetheless for
each vector space identify the k for which the space is “the same” as k.
(a) The 2´3 matrices under the usual operations
(b) The n ´ m matrices (under their usual operations)
(c) This set of 2 ´ 2 matrices
ìï æ a
0
ö ïü í ç ¸ a + b + c
= 0
ý
îï è b c
ø ïþ
2.
(a) Prove that every point, line, or plane thru the origin in 3 is a
vector space under the inherited operations.
(b) What if it doesn’t contain the origin?

15. I.2. Subspaces and Spanning Sets
Definition 2.1: Subspaces
For any vector space, a subspace is a subset that is itself a vector space,
under the inherited operations.
Note: A subset of a vector space is a subspace iff it is closed under § & ¨.
ì æ x
ö ü
= ï ç ¸ í ç ¸ + + = ï ý ï î çè ø¸ ï þ
Example 2.2: Plane in 3 P y x y z
0
z
is a subspace of 3.
→ It must contain 0. (c.f. Lemma 2.9.)
Proof: Let ( ) ( ) 1 1 1 1 2 2 2 2 , , , , , T T r = x y z r = x y z ÎP
→ 1 1 1 2 2 2 x + y + z = 0 , x + y + z = 0
( ) 1 2 1 2 1 2 1 2 , , T ar + br = ax + bx ay + by az + bz
with ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 1 1 2 2 2 ax + bx + ay + by + az + bz = a x + y + z + b x + y + z
→ 1 2 ar + br ÎP " a,bÎR QED
= 0

16. Example 2.3: The x-axis in n is a subspace.
( ,0, ,0) -axis T Proof follows directly from the fact that r = x L Î x
Example 2.4:
• { 0 } is a trivial subspace of n.
• n is a subspace of n.
Both are improper subspaces.
All other subspaces are proper.
Example 2.5: Subspace is only defined wrt inherited operations.
({1},§ ; ) is a vector space if we define 1§1 = 1 and a¨1=1 "aÎ.
However, neither ({1},§ ; ) nor ({1},+ ; ) is a subspace of the vector space
(,+ ; ).

17. Example 2.6: Polynomial Spaces.
n is a proper subspace of m if n < m.
Example 2.7: Solution Spaces.
The solution space of any real linear homogeneous ordinary
differential equation,  f = 0,
is a subspace of the function space of 1 variable { f :  →  }.
Example 2.8: Violation of Closure.
+ is not a subspace of  since (-1) v Ï + " vÎ +.

18. Lemma 2.9:
Let S be a non-empty subset of a vector space ( V, + ;  ).
W.r.t. the inherited operations, the following statements are equivalent:
1. S is a subspace of V.
2. S is closed under all linear combinations of pairs of vectors.
3. S is closed under arbitrary linear combinations.
Proof: See Hefferon, p.93.
Remark: Vector space = Collection of linear combinations of vectors.

19. Example 2.11: Parametrization of a Plane in 3
ì æ x
ö ü
= ï ç ¸ í ç ¸ - 2 + = 0
ï ý ï î çè ø¸ ï þ
S y x y z
z
is a 2-D subspace of 3.
ì æ 2
y - z
ö ü
= ï ç ¸ í ç y ¸ y ,
z
Î R
ï ý ï î çè z
ø¸ ï þ
ì æ 2 ö æ - 1
ö ü
= ï ç í y 1 ¸+ ç ¸ z ç 0 ¸ ç ¸ y ,
z
Î R
ï ý ï î çè 0 ø¸ èç 1
ø¸ ï þ
i.e., S is the set of all linear combinations of 2 vectors (2,1,0)T, &
(-1,0,1)T.
Example 2.12: Parametrization of a Matrix Subspace.
ìï æ a
0
ö ïü = í ç ¸ + + = 0
ý
îï è ø ïþ
L a b c
b c
is a subspace of the space of 2´2 matrices.
ìï æ- b - c
0
= ö í ç ¸ b ,
c
Î R
ïü b c
ý
îï è ø ïþ
ìï æ - 1 0 ö æ - 1 0
= b ¸+ c ö í ç ç ¸ b ,
c
Î R
ïü 1 0 0 1
ý
îï è ø è ø ïþ

20. Definition 2.13: Span
Let S = { s1 , …, sn | sk Î ( V,+, ) } be a set of n vectors in vector space V.
The span of S is the set of all linear combinations of the vectors in S, i.e.,
ì ü
= í Î Î ý
î þ
å s s R with span Æ = { 0 }
span S c S c
1
,
n
k k k k
k
=
Lemma 2.15: The span of any subset of a vector space is a subspace.
Proof:
n n
Let S = { s, …, s| sÎ ( V,+,) }
1 n k  and u =å s v =å s Î
u ,
v span S
k k k k
= =
k k
1 1
w = u + v =å + s
® a b ( au bv
)
1
n
k k k
k
=
=å s Î " a,bÎR
1
n
k k
k
w span S
=
QED
Converse: Any vector subspace is the span of a subset of its members.
Also: span S is the smallest vector space containing all members of
S.

21. Example 2.16:
For any vÎV, span{v} = { a v | a Î } is a 1-D subspace.
Example 2.17:
Proof:
The problem is tantamount to showing that for all x, y Î, $ unique a,b Î s.t.
1 1
1 1
x
æ ö = a æ ö + b
æ ö ç è y
¸ ç ¸ ç ¸ ø è ø è - ø
i.e., a b x
+ =
- =
a b y
has a unique solution for arbitrary x & y.
Since 1 ( )
a = x + y 1 ( )
2
b = x - y " x, yÎR QED
2
2 1 1
,
1 1
span
ì æ ö æ ö ü í ç ¸ ç ¸ ý = î è ø è - ø þ
R

22. Example 2.18: 2
Let S = span{ 3x - x2 , 2x } = { a ( 3x - x2 ) + 2bx a,bÎR }
Question:
0 2 0
?
c S = =P
Answer is yes since
ì ü
= í ý
î þ
å = subspace of 2 ?
1 c = 3a + 2b 2 c = -a
1 3
2
a = -c b = ( c - a
) 2 1
and
1 3
2
( ) 1 2
= c + c
2
1
k
k
k
c x
=
Lesson: A vector space can be spanned by different sets of vectors.
Definition: Completeness
A subset S of a vector space V is complete if span S = V.

23. Example 2.19: All Possible Subspaces of 3
See next section for proof.
Planes thru
0
Lines thru 0

24. Exercises 2.I.2
ì æ x
ö ü
ï ç í ç y ¸ ï çè ø¸ ¸ x + y + z
= ý ï z
ï î þ
1. Consider the set 1
under these operations.
x x x x 1
y y y y
z z z z
æ ö æ + - ç 1 ¸+ ç 2 ö æ 1 2
ö
¸= ç ¸ çç 1 ¸¸ çç + 2 ¸¸ çç 1 2
¸¸ è 1 ø è + 2 ø è 1 2
ø
x rx r 1
r y r y
z rz
æ ö æ - + ö
ç ¸= ç ¸ çç ¸¸ çç ¸¸ è ø è ø
(a) Show that it is not a subspace of 3. (Hint. See Example 2.5).
(b) Show that it is a vector space.
( To save time, you need only prove axioms (d) & (j), and closure
under all linear combinations of 2 vectors.)
(c) Show that any subspace of 3 must pass thru the origin, and so any
subspace of 3 must involve zero in its description.
Does the converse hold?
Does any subset of 3 that contains the origin become a subspace
when given the inherited operations?

25. 2. Because ‘span of’ is an operation on sets we naturally consider how it
interacts with the usual set operations. Let [S] º Span S.
(a) If S Í T are subsets of a vector space, is [S] Í [T] ?
Always? Sometimes? Never?
(b) If S, T are subsets of a vector space, is [ S È T ] = [S] È [T] ?
(c) If S, T are subsets of a vector space, is [ S Ç T ] = [S] Ç [T] ?
(d) Is the span of the complement equal to the complement of the span?
3. Find a structure that is closed under linear combinations, and yet is not
a vector space. (Remark. This is a bit of a trick question.)