¡Descarga Lesson 1: Vector Spaces and Linear Maps y más Apuntes en PDF de Administración de Empresas solo en Docsity!
_______________________________________________________________________________
LESSON 1
VECTOR SPACES AND LINEAR MAPS
1. Vector Spaces
Definition
Let us consider:
- A set K with field structure and whose elements are called scalars , we denote
as: a, b, α, β,…
Remarks: a field is an algebraic structure with notions of addition, subtraction,
multiplication, and division, satisfying certain axioms. Example: real
numbers.
- A set V whose elements are called vectors and we denote as: u , v , w ...
It is said that V has got a structure of vector space onto the field K {V, K} if we can
define the two following operations and, besides, the following properties are
satisfied:
Internal means that it assigns to every couple of vector of V, another vector of V.
This operation is shown as (+) and is called “vector addition”.
V × V → V
( u , v )→ u + v
If u and v are any elements of V, then ( u + v ) is in V. (i.e. V is closed under the
operation +)
Set V and sum operation {V, +} have to verify the following properties:
a) Associative: ( u + v )+ w = u +( v + w )
b) Commutative: u + v = v + u
c) There is a unique element 0 in V such that: ∃! 0 ∈ V /∀ u ∈ V : 0 + u = u
(zero vector)
d) For each u ∈ V there is a unique element − u ∈ V such that
∀ u ∈ V / ∃ u ∈ V / u +(− u )= 0 (opposite of u )
_______________________________________________________________________________
Remark: Properties a , c and d define a group structure. If property b is satisfied as
well, we have an Abelian Group.
External means that it assigns to every element of V an element of K.
This operation is shown as (^) ( ⋅) and it will be called “scalar multiplication”
V × K → V
( u ,λ )→ λ u
If u is any element of V and λ is any real number, then λ ⋅ u is in V (i.e. V is closed
under the operation( ⋅) ).
Sets V, K, the addition and the scalar multiplication operations {V, +, ·}, must
satisfy the following properties
a) Distributivity of scalar multiplication over vector addition:
λ ( u + w )=λ u + λ w
b) Distributivity of scalar multiplication over field addition: (λ + μ) u =λ u + μ u
c) Compatibility of scalar multiplication with field multiplication:
( λμ ) u =λ( μ u )
d) Identity element of scalar multiplication: 1 u = u for all u ∈ V.
Examples of vector spaces {ℜn,ℜ}, n∈N
2
= {(a, b)/a, b∈ℜ } ⇒ a couple of real numbers: ℜxℜ
3
= {(a, b, c) / a, b, c ∈ℜ} ⇒ ℜxℜxℜ
ℜn= { ( a (^) 1 ,..., an )/ ai ∈ ℜ, i = 1 , 2 ,..., n } ⇒ ℜ ×ℜ×...×ℜ
Additional Properties
1. λ 0 = 0 ∀ λ ∈ℜ, 0 ∈ V
2. 0 v = 0
3. ( −λ ) v =λ(− v )=−( λ v )
0 , or 0 v
v
_______________________________________________________________________________
Geometrical interpretation.
space a couple of vectors will be L.I. if they are not proportional.
Three or more vectors, en ℜ^2 space, will be necessarily L.D.
- For ℜ^3 space three vectors will be L.I, if they are not proportional one with
each other. Four or more vectors will be necessarily L.D.
Theorem of uniqueness of coordinates
If { v 1 ,..., vp }is a set of L.I. vectors and the vector x is a linear combination of them,
then its coordinates λ 1 ,..., λ p ∈ ℜ are unique.
Definition: Linear Span (L.S.)
Let v = { v 1 , v 2 ,..., vn }be a set of vectors in a vector space V. The set v spans V, or V is
spanned by v if every vector u in V is a linear combination of the vectors in v , or in
the same way, { V , K }⇔∀ u ∈ V ∃λ 1 ,..... λ n ∈ K / u = λ 1 v 1 +λ 2 v 2 +...+ λ n vn
Definition : Basis of a vector space
B ={ v 1 ,..., vn } ⊂ V are basis of { V , K }⇔
(1) are L.S. (2) are L I.
Remark
Basis is equal to Reference system
Since a basis is a linear span , every vector in the vector space can be spanned by its
coordinates in connection with the basis, because of the following
1) unique coordinates
2) basis vectors are L.I.
(Exercise 2.3)
_______________________________________________________________________________
Properties of Basis
Theorem 1. If v = { v 1 , v 2 ,..., 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 v.
Theorem 2. Let v = { v 1 , v 2 ,..., vn }be a set of nonzero vectors in a vector space V and let
w span v. Then some subset of v is a basis for w.
Theorem 3. Let’s consider a vector space where exists a basis made up of “n”
vectors. The maximum number of L.I. vectors we might find in that space is “n”.
Theorem 4. If v = { v 1 , v 2 ,..., vn }is a basis for a vector space V and u = { u 1 ,...... u r }is a L.I.
set of vectors in V, then r ≤ n.
Theorem 5. If v = { v 1 , v 2 ,..., vn }and u = { u 1 ,...... u r }are bases for a vector space, then r=n.
Remark:
- A basis is a minimum linear span (the smallest possible number of vectors that
constitute a L.S.)
- A basis is a maximum set of L.I. vectors.
_______________________________________________________________________________
2. Matrices and Vector Spaces
(Matrix slides 56 – 96)
Rank of a matrix is defined to be the maximal number of L.I. columns (or L.I.
rows) of matrix A. Similarly, the rank of a matrix is defined to be the greatest order
of nonzero minors of matrix A.
Example
Work out the rank of
A
Remark: Rk A ( ) = Rk A ( t )
c 1 c 2
c 2 is a linear combination of c 1 : so we must put out c 2 and we replace with c 3.
It is easy to prove that c 2 = 2 c 1 Rk A ( ) ≥ 1
≠ 0 ⇒ Rk A ( ) ≥ 2 c 1 y c 3 are L.I.
c 3 c 4
c 1 c 3 c 4
c 4 is a linear combination of the c 1 y c 3. It is easy to prove c 4 = c 1 + c 3
so 1
Rk A Rk c c Rg
_______________________________________________________________________________
Practical applications of calculating the rank
1. How to know the maximum number of L.I. vectors in a set of vectors.
2. How to know if a vector is a L.C. of a given set of vectors.
Example
In a vector space in ℜ 3 we consider the following vectors v 1 = ( a ,− 6 , b ), v 2 =( 1 , 0 , 6 ),
v 3 =( 0 ,− 3 , 2 )
Try to find the relationship between a and b so that the three vectors are L.D.
(Exercise 2.11)
_______________________________________________________________________________
Example
{( 1 , 1 , 1 ),( 2 , 2 , 2 ) }is a linear span of^ S^ but it is not a basis.
Definition : Dimension of a vector subspace
Dim ( S ) =It is the greatest set of L.I. vectors we can find in a vector subspace S
= It is a set of vectors of any basis of S.
Remark
Dim ( S ) S ⊂ V ↓ 0 ≤ Dim ( S )≤ Dim ( V ) ↓ ↓
S = { 0 } S ≡ V
How to calculate Dim (S)
( 1 ,..., (^) p ) linear span
S = L v v 14243
If { v 1 , v 2 ,..., vp }is a linear span in S , then it can be proved that Dim S ( ) = Rk v ( 1 , v 2 ,... v p )
(Exercises 2.5, 2.8, 2.15)
What is the relationship between a vector subspace and a linear homogeneous
system
Theorem. The set of solutions, x = ( x 1 , x 2 ,..., xn )∈ℜ n , of an homogeneous system:
AX = 0 ( A ∈ Mm × m , x ∈ Mn × 1 , 0 ∈ Mm × 1 )is a vector subspace for ℜ n :
S = { x =( x 1 ,..., xn )∈ℜ n / AX = 0 }
How to find out the Dim(S) (through a homogeneous system)
{ /^0 }
S = x ∈ℜ n AX =
_______________________________________________________________________________
Dim ( S ) = It is the number of free components of a vector in S (there will be one
equation for each restricted variable or non-free variable)
Rk ( A ) = It is the number of non-redundant equations in the homogeneous system,
AX = 0
= either it is the number of L.I. columns in A
= or, it is the number of restricted variables
So, therefore, it will be: Dim ( S ) = n − Rk ( A )
3 1
3 2 1 x x
x x S x
1 1 0 1 0 ( ) ( ) 3 3 3 2 1 1 0 1 0 1
Dim S n rg A rk rk
Parametric and Cartesian Equations in a vector space system
1. Parametric Equations → Cartesian Equations
n p r r r i PARAMETRIC EQUATIONS
S = L v v = L v v = x ∈ℜ x = λ v + + λ v λ∈ℜ
Dim S ( ) = Rk v ( 1 − v (^) p )= r ≤ p
We can prove that ∀ x ∈ S , x is a linear combination of the basis ( v 1 ,..., vp ), so
Rk v ( 1 (^) ,..., vr , x ) = Rk v ( 1 ,..., vr )= r
2. Cartesian Equations → Parametric Equations
{ /^ 0 ,^ ,^1 , 0 1 }
n m n n m CARTESIAN EQUATIONS
S = x ∈ ℜ 1 AX 42 = 43 A ∈ M (^) × X ∈ M (^) × ∈ M ×
Dim S ( ) = n − rk A ( ) = n − h = r
If we solve the homogeneous system, there are " h " restricted variables with
respect to the " r " free variables.
1 1 ( ) ( )
( ,..., (^) h , (^) h ,..., (^) h r n ) n Rg A h n h r Dim S VARIABLESRESTRICTED DEGREES OF FREEDOMFREE VARIABLES OR
x x x (^) + x + = = = − = = = = =
(Exercises 2.10, 2.14, 2.17, 2.18)
_______________________________________________________________________________
Definition: The Kernel of a Linear Transformation
Let f : V → V ′be a linear transformation. The kernel of f , ker(f) is the subset of V
consisting of all the vectors such that f ( v )= 0 V ′, that is:
Ker f = { v^ ∈ V / f ( v )= 0 V ′}
Definition: The Image of a Linear Transformation
Let f : V → V ′be a linear transformation. The Image of f , Im(f) , is the set of all
vectors in V’ that are images, under f , of vectors in V. Thus a vector v ' is in the
image f if we can find some vector v in V such that f v ( ) = v ', or
f ( V ) = Im ( f ) = { v ′∈ V ′/ ∃ v ∈ V , f ( v )= v ′}
Properties
1. If f : V → V 'is a linear transformation, then Im ( f )is a subspace of V’. Its
dimension is called as linear dimension of a linear map or transformation :
dim ( f ) =dim (Im f )
2. Besides, the images of a basis in V constitute a linear span of the vector space
V’ :
{ 1 ,...,^ }^ { (^1 ) ,...,^ (^ )}
f n v Basis in V Linear Span in V
V V
B u u f u f u G ′
3. Ker f ≠Ø since, at least, f ( 0 V )= 0 V ′
4. Ker f is a vector subspace in V.
5. If f : V → V 'is a linear transformation, then
dim ( V ) = dim( Ker f ) +dim(Im f )
Example
Given a linear transformation
xy z f x y z x y z x y z
f → = − − − +
a) Compute the Kerf
_______________________________________________________________________________
b) Compute the Im( f )
a) Ker f = { x ∈ℜ^3 / f ( x )= 0 ∈ℜ^2 } ⇒( x − y − 5 z , 2 x − 3 y + 5 z )=( 0 , 0 )⇒
x y z
x y z x y z
x y z 2 3 5
so,
FREE
Ker f = x y z ∈ℜ x = z y = z z
Dim ( Ker f ) = n − Rk A ( ) = 3 − Rk
= 3 − 2 = 1 = number of free variables.
To get a basis of Ker f ( z = 1 ) ⇒ B ={( 20 , 15 , 1 )}
b) Im( f ) = { x ∈ ℜ 2 / ∃ x ∈ ℜ^3 , f ( ) x = x ′
x x y z
x x y z 2 3 5
2
1
If we take the natural basis from the original vector space C ={( 1 , 0 , 0 ),( 0 , 1 , 0 ),( 0 , 0 , 1 )}
its images make up a linear span of the vector space Im( f )in ℜ^2 :
f f f
A basis of Im( f ) will be, for instance, B ′
rk rk
−^ −^ −
=^ =
−^ −
B ′= {( 1 , 2 ),(− 1 ,− 3 )}
2
Im( f ) = x ′ ∈ℜ / x ′= λ 1 (1, 2) + λ 2 ( 1,− − 3), λ 1 ,λ 2 ∈ℜ
Dim (Im f ) = 2
It is easy to check that
Dim ( ℜ^3 ) = Dim ( Ker f ) + Dim (Im f )
_______________________________________________________________________________
Y
y
y
y
3
2
A
123 { X
x
x
2
1 2
1 2
( ) ( )
f u f u
x x
↑----- L.S.----↑
so, {( 1 , 1 , 0 ),( 0 , 1 , 0 )} (columns of A ), makes up Im( f ), and as the Image is a vector
subspace, if we want to know which the Dim ( f )is, it suffices to know the number of
L.I. vectors (columns of A ), that is, Rk A ( )
( ) 1 1 2 (Im( )) 0 1
Rk A Rk Dim f
In this particular case, the linear span of Im( f ) : (1, 1, 0), (0, 1, 1){ }is a basis of Im(^ f )
Example
Set out in a matrix way, find out the kernel and the image.
f : ℜ^3 →ℜ^3 /
1 2 3 2 2 3 1 3
f v v v f v v f v v v
Matrix expression
x y f x AX
f → = =
3 1 3
2 1 2
1 3
3
2
1
3
2
1 2 1
y x x
y x x
y x
x
x
x
y
y
y
_______________________________________________________________________________
Im( f )
{ }^ { }
1 3 3 2 1 2 3 3
Im( ) / ( ) / 1 2 0 1 0 1
y f y y f x y y x x x y
= ∈ℜ = = ∈ ℜ ^ ^ = ^ ^ + ^ ^ +^
^ ^ ^ ^ ^
3 3
( ) 1 2 0 3 (Im( )) 3 ( ) Im( ) 1 0 1
Rk A Rk Dim f Dim f
= ^ = ⇒ = = ℜ ⇒ = ℜ
⇒ {( 0 , 1 , 1 ),( 0 , 2 , 0 ),( 1 , 0 , 1 )}is a basis of Im( f )
Ker f
3 3 1 2 1 3
x Ker f x AX x x Rk A x x E D x
^ =
Ker f = { 0 } Dim Ker f ( ) = n − Rk ( A ) = 3 − 3 = 0
(Exercise 4.5)
4.3. Linear Map Classification
Let V and V’ denote vector spaces over a field, K. Let f : V → V’ be a linear
transformation.
- f is said to be injective or a monomorphism if f is one-to-one as a map of sets.
