vcla

1. AMIRAJ COLLEGE OF ENGINEERING AND
TECHNOLOGY
TOPIC:-VECTOR SPACE AND SUB SPACE
SUBJECT:-VC&LA(2110015)
BRANCH:-CIVIL DEPARTMENT
CREATED BY:-151080106007
:-151080106008
:-151080106009
:-151080106010
:-151080106011
:-151080106012
GUIDE BY :-HEENA MAM
Slide 4.1- 1© 2012 Pearson Education, Inc.

2. 4
4.1
4
4.1
VECTOR SPACES AND
SUBSPACES

3. Slide 4.1- 3© 2012 Pearson Education, Inc.
VECTOR SPACES AND SUBSPACES
 Definition: A vector space is a nonempty set V of
objects, called vectors, on which are defined two
operations, called addition and multiplication by
scalars (real numbers), subject to the ten axioms (or
rules) listed below. The axioms must hold for all
vectors u, v, and w in V and for all scalars c and d.
 The sum of u and v, denoted by , is in V.
 .
 .
 There is a zero vector 0 in V such that
.
u v
u v v u  
(u v) w u (v w)    
u ( u) 0  

4. Slide 4.1- 4© 2012 Pearson Education, Inc.
VECTOR SPACES AND SUBSPACES
 For each u in V, there is a vector in V such
that .
 The scalar multiple of u by c, denoted by cu, is
in V.
 .
 .
 .
 .
u
u ( u) 0  
(u v) u vc c c  
(c + d)u = cu + du
( u) ( )uc d cd
1u u

5. Slide 4.1- 5© 2012 Pearson Education, Inc.
VECTOR SPACES AND SUBSPACES
 Example 1. The space of all 3x3 matrices is a vector
space. The space of all matrices is not a vector space.

6. Slide 4.1- 6© 2012 Pearson Education, Inc.
SUBSPACES
 Definition: A subspace of a vector space V is a
subset H of V that has three properties:
 The zero vector of V is in H.
 H is closed under vector addition. That is, for
each u and v in H, the sum is in H.
 H is closed under multiplication by scalars.
That is, for each u in H and each scalar c, the
vector cu is in H.
u v

7. Slide 4.1- 7© 2012 Pearson Education, Inc.
SUBSPACES
 Properties (a), (b), and (c) guarantee that a subspace
H of V is itself a vector space, under the vector space
operations already defined in V.
 Every subspace is a vector space.
 Conversely, every vector space is a subspace (of itself
and possibly of other larger spaces).

8. Slide 4.1- 8© 2012 Pearson Education, Inc.
Example 1. The space of all 3x3 upper triangular
matrices is a subspace. The space of all matrices with
integer entries is not.

9. Slide 4.1- 9© 2012 Pearson Education, Inc.
A SUBSPACE SPANNED BY A SET
 The set consisting of only the zero vector in a vector
space V is a subspace of V, called the zero subspace
and written as {0}.
 As the term linear combination refers to any sum of
scalar multiples of vectors, and Span {v1,…,vp}
denotes the set of all vectors that can be written as
linear combinations of v1,…,vp.

10. Slide 4.1- 10© 2012 Pearson Education, Inc.
A SUBSPACE SPANNED BY A SET
 Example 2: Given v1 and v2 in a vector space V, let
. Show that H is a subspace of V.
 Solution: The zero vector is in H, since .
 To show that H is closed under vector addition, take
two arbitrary vectors in H, say,
and .
 By Axioms 2, 3, and 8 for the vector space V,
1 2
Span{v ,v }H 
1 2
0 0v 0v 
1 1 2 2
u v vs s  1 1 2 2
w v vt t 
1 1 2 2 1 1 2 2
1 1 1 2 2 2
u w ( v v ) ( v v )
( )v ( )v
s s t t
s t s t
    
   

11. Slide 4.1- 11© 2012 Pearson Education, Inc.
A SUBSPACE SPANNED BY A SET
 Theorem 1: If v1,…,vp are in a vector space V, then
Span {v1,…,vp} is a subspace of V.
 We call Span {v1,…,vp} the subspace spanned (or
generated) by {v1,…,vp}.
 Give any subspace H of V, a spanning (or
generating) set for H is a set {v1,…,vp} in H such
that
.1
Span{v ,...v }p
H 

12. Eigenfaces
Slide 4.1- 12© 2012 Pearson Education, Inc.

13. Slide 4.2- 13© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 Definition: The null space of an matrix A,
written as Nul A, is the set of all solutions of the
homogeneous equation . In set notation,
.
 Theorem 2: The null space of an matrix A is a
subspace of . Equivalently, the set of all solutions
to a system of m homogeneous linear
equations in n unknowns is a subspace of .
 Proof: Nul A is a subset of because A has n
columns.
 We need to show that Nul A satisfies the three
properties of a subspace.
m n
x 0A 
Nul A ={x : x is in n
and Ax = 0}
m n
n
x 0A 
n
n

14. Slide 4.2- 14© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 An Explicit Description of Nul A
 There is no obvious relation between vectors in Nul A
and the entries in A.
 We say that Nul A is defined implicitly, because it is
defined by a condition that must be checked.

15. Slide 4.2- 15© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 No explicit list or description of the elements in Nul
A is given.
 Solving the equation amounts to producing an
explicit description of Nul A.
 Example 1: Find a spanning set for the null space of
the matrix
.
x 0A 
3 6 1 1 7
1 2 2 3 1
2 4 5 8 4
A
   
   
 
   

16. Slide 4.2- 16© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 Solution: The first step is to find the general solution
of in terms of free variables.
 Row reduce the augmented matrix to reduce
echelon form in order to write the basic variables in
terms of the free variables:
,
x 0A 
 0A
1 2 0 1 3 0
0 0 1 2 2 0
0 0 0 0 0 0
  
 
 
  
1 2 4 5
3 4 5
2 3 0
2 2 0
0 0
x x x x
x x x
   
  


17. Slide 4.2- 17© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 The general solution is ,
, with x2, x4, and x5 free.
 Next, decompose the vector giving the general
solution into a linear combination of vectors where
the weights are the free variables. That is,
1 2 4 5
2 3x x x x  
3 4 5
2 2x x x  
1 2 4 5
2 2
3 4 5 2 4 5
4 4
5 5
2 3 2 1 3
1 0 0
2 2 0 2 2
0 1 0
0 0 1
x x x x
x x
x x x x x x
x x
x x
           
         
         
              
         
         
                 
u wv

18. Slide 4.2- 18© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX
 The spanning set produced by the method in
Example (1) is automatically linearly
independent because the free variables are the
weights on the spanning vectors.
 When Nul A contains nonzero vectors, the
number of vectors in the spanning set for Nul A
equals the number of free variables in the
equation .x 0A 

19. Slide 4.2- 19© 2012 Pearson Education, Inc.
COLUMN SPACE OF A MATRIX
 Definition: The column space of an matrix A,
written as Col A, is the set of all linear combinations
of the columns of A. If , then
.
 Theorem 3: The column space of an matrix A
is a subspace of .
m n
A = a1
an
é
ë
ù
û
1
Col Span{a ,...,a }n
A 
m n
m

20. Slide 4.2- 20© 2012 Pearson Education, Inc.
COLUMN SPACE OF A MATRIX
 The notation Ax for vectors in Col A also shows that Col
A is the range of the linear transformation .
 The column space of an matrix A is all of if
and only if the equation has a solution for each
b in .
 Example 2: Let ,
and .
x Ax
m n m
x bA 
m
2 4 2 1
2 5 7 3
3 7 8 6
A
 
   
 
  
3
2
u
1
0
 
 
 
 
 
 
3
v 1
3
 
  
 
  

21. Slide 4.2- 21© 2012 Pearson Education, Inc.
COLUMN SPACE OF A MATRIX
 Determine if u is in Nul A. Could u be in Col A?
 Determine if v is in Col A. Could v be in Nul A?
 Solution:
 An explicit description of Nul A is not needed
here. Simply compute the product Au.
3
2 4 2 1 0 0
2
u 2 5 7 3 3 0
1
3 7 8 6 3 0
0
A
 
                 
      
           
 

22. Slide 4.2- 22© 2012 Pearson Education, Inc.
COLUMN SPACE OF A MATRIX
 u is not a solution of , so u is not in Nul A.
 Also, with four entries, u could not possibly be in Col
A, since Col A is a subspace of .
 Reduce to an echelon form.
 The equation is consistent, so v is in
Col A.
x 0A 
3
 vA
 
2 4 2 1 3 2 4 2 1 3
v 2 5 7 3 1 0 1 5 4 2
3 7 8 6 3 0 0 0 17 1
A
    
         
   
      
:
x vA 

23. Slide 4.2- 23© 2012 Pearson Education, Inc.
KERNEL AND RANGE OF A LINEAR
TRANSFORMATION
 With only three entries, v could not possibly be
in Nul A, since Nul A is a subspace of .
 Subspaces of vector spaces other than are often
described in terms of a linear transformation instead of a
matrix.
 Definition: A linear transformation T from a vector
space V into a vector space W is a rule that assigns to each
vector x in V a unique vector T (x) in W, such that
 for all u, v in V, and
 for all u in V and all scalars c.
4
n
(u v) (u) (v)T T T  
( u) (u)T c cT

24. Slide 4.2- 24© 2012 Pearson Education, Inc.
KERNEL AND RANGE OF A LINEAR
TRANSFORMATION
 The kernel (or null space) of such a T is the set of all
u in V such that (the zero vector in W ).
 The range of T is the set of all vectors in W of the
form T (x) for some x in V.
 The kernel of T is a subspace of V.
 The range of T is a subspace of W.
(u) 0T 

25. Slide 4.2- 25© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
MATRIX Am n
Nul A Col A
Nul A is a subspace of
.
Col A is a subspace of
.
Nul A is implicitly
defined; i.e., you are given
only a condition
that vectors in
Nul A must satisfy.
Col A is explicitly
defined; i.e., you are told
how to build vectors in Col
A.
n m
( x 0)A 

26. Slide 4.2- 26© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
MATRIX Am n
It takes time to find
vectors in Nul A. Row
operations on are
required.
It is easy to find vectors
in Col A. The columns of a
are displayed; others are
formed from them.
There is no obvious
relation between Nul A and
the entries in A.
There is an obvious
relation between Col A and
the entries in A, since each
column of A is in Col A.
 0A

27. Slide 4.2- 27© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
MATRIX Am n
A typical vector v in Nul A
has the property that
.
A typical vector v in Col
A has the property that the
equation is
consistent.
Given a specific vector v,
it is easy to tell if v is in Nul
A. Just compare Av.
Given a specific vector v,
it may take time to tell if v is
in Col A. Row operations on
are required.
v 0A  x vA 
 vA

28. Slide 4.2- 28© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
MATRIX Am n
Nul if and only if
the equation has only
the trivial solution.
Col if and only if
the equation has a
solution for every b in .
Nul if and only if
the linear transformation
is one-to-one.
Col if and only if
the linear transformation
maps onto .
{0}A 
x 0A 
A = m
x bA 
m
{0}A 
x xAa
A = m
x xAa
n m