Beyond Fields: Vector Spaces and Algebras Bernd Schr oder logo1 - - PowerPoint PPT Presentation

logo1 Vector Spaces Bases Algebras

Beyond Fields: Vector Spaces and Algebras

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).
• 2. We have noted that C has all the properties we would like

when in comes to solving polynomial equations

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).
• 2. We have noted that C has all the properties we would like

when in comes to solving polynomial equations (Fundamental Theorem of Algebra).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).
• 2. We have noted that C has all the properties we would like

when in comes to solving polynomial equations (Fundamental Theorem of Algebra).

• 3. What other (useful) fields are there beyond R and C?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).
• 2. We have noted that C has all the properties we would like

when in comes to solving polynomial equations (Fundamental Theorem of Algebra).

• 3. What other (useful) fields are there beyond R and C?
• 4. To answer this question, we first need to talk about vector

spaces and algebras.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Introduction

• 1. We have seen that R is uniquely determined (up to

isomorphism) by its axioms (R is the complete, totally

• rdered field).
• 2. We have noted that C has all the properties we would like

when in comes to solving polynomial equations (Fundamental Theorem of Algebra).

• 3. What other (useful) fields are there beyond R and C?
• 4. To answer this question, we first need to talk about vector

spaces and algebras.

• 5. As we work with vector spaces, we prove a few results that

will be helpful when discussing the unsolvability of the quintic by radicals.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

• 4. For every x ∈ X there is a (−x) ∈ X so that x+(−x) = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

• 4. For every x ∈ X there is a (−x) ∈ X so that x+(−x) = 0.
• 5. For all x,y ∈ X and α ∈ F we have

α ·(x+y) = α ·x+α ·y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

• 4. For every x ∈ X there is a (−x) ∈ X so that x+(−x) = 0.
• 5. For all x,y ∈ X and α ∈ F we have

α ·(x+y) = α ·x+α ·y.

• 6. For all x ∈ X and α,β ∈ F we have

(α +β)·x = α ·x+β ·x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

• 4. For every x ∈ X there is a (−x) ∈ X so that x+(−x) = 0.
• 5. For all x,y ∈ X and α ∈ F we have

α ·(x+y) = α ·x+α ·y.

• 6. For all x ∈ X and α,β ∈ F we have

(α +β)·x = α ·x+β ·x.

• 7. For all x ∈ X and α,β ∈ F we have α ·(β ·x) = (αβ)·x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. A vector space over the field F is a triple (X,+,·)
• f a set X and two binary operations, vector addition

+ : X ×X → X and scalar multiplication · : F×X → X, so that the following hold.

• 1. For all x,y,z ∈ X we have (x+y)+z = x+(y+z).
• 2. For all x,y ∈ X we have x+y = y+x.
• 3. There is an element 0 ∈ X so that for all x ∈ X we have

x+0 = x.

• 4. For every x ∈ X there is a (−x) ∈ X so that x+(−x) = 0.
• 5. For all x,y ∈ X and α ∈ F we have

α ·(x+y) = α ·x+α ·y.

• 6. For all x ∈ X and α,β ∈ F we have

(α +β)·x = α ·x+β ·x.

• 7. For all x ∈ X and α,β ∈ F we have α ·(β ·x) = (αβ)·x.
• 8. For all x ∈ X we have 1·x = x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. An element of a vector space is also called a vector

and an element of F is also called a scalar.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. An element of a vector space is also called a vector

and an element of F is also called a scalar. We will usually refer to the set X as the vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. An element of a vector space is also called a vector

and an element of F is also called a scalar. We will usually refer to the set X as the vector space. As is customary for multiplications, the dot is usually omitted.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space. Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x) = f(x)+g(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x) = f(x)+g(x) = g(x)+f(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x) = f(x)+g(x) = g(x)+f(x) = (g+f)(x).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x) = f(x)+g(x) = g(x)+f(x) = (g+f)(x). The neutral element for addition is the function that is equal to 0 ∈ F for all x ∈ X.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. Let d ∈ N and let F be a field. The set

Fd :=

• (x1,...,xd) : xi ∈ F
• with componentwise addition and

scalar multiplication is a vector space.

• Example. Let D be a set and let F be a field. The set F(D,F)
• f all functions f : D → F with addition defined pointwise by

(f +g)(x) := f(x)+g(x) and scalar multiplication defined pointwise by (α ·f)(x) := αf(x) is a vector space. All properties follow from the corresponding pointwise properties for F. For example, for all x ∈ D we have that (f +g)(x) = f(x)+g(x) = g(x)+f(x) = (g+f)(x). The neutral element for addition is the function that is equal to 0 ∈ F for all x ∈ X.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space. A subset S ⊆ X \{0} is

called linearly independent iff for all finite subsets {x1,...,xn} ⊆ S and all sets of scalars {α1,...,αn} ⊆ F the equation

n

∑

i=1

αixi = 0 implies α1 = α2 = ··· = αn = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space. A subset S ⊆ X \{0} is

called linearly independent iff for all finite subsets {x1,...,xn} ⊆ S and all sets of scalars {α1,...,αn} ⊆ F the equation

n

∑

i=1

αixi = 0 implies α1 = α2 = ··· = αn = 0. A sum

n

∑

i=1

αixi with αi ∈ F and xi ∈ X is also called a linear combination of x1,...,xn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space. A subset S ⊆ X \{0} is

called linearly independent iff for all finite subsets {x1,...,xn} ⊆ S and all sets of scalars {α1,...,αn} ⊆ F the equation

n

∑

i=1

αixi = 0 implies α1 = α2 = ··· = αn = 0. A sum

n

∑

i=1

αixi with αi ∈ F and xi ∈ X is also called a linear combination of x1,...,xn. Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space. A subset S ⊆ X \{0} is

called linearly independent iff for all finite subsets {x1,...,xn} ⊆ S and all sets of scalars {α1,...,αn} ⊆ F the equation

n

∑

i=1

αixi = 0 implies α1 = α2 = ··· = αn = 0. A sum

n

∑

i=1

αixi with αi ∈ F and xi ∈ X is also called a linear combination of x1,...,xn.

• Definition. A linearly independent set B ⊆ X such that for every

x ∈ X there are a finite subset {b1,...,bn} ⊆ B and a set of scalars {α1,...,αn} ⊆ F so that x =

n

∑

i=1

αibi is called a base of a vector space.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence:

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0, which for each j simply state that αj = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0, which for each j simply state that αj = 0, as was to be proved.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0, which for each j simply state that αj = 0, as was to be proved. Representation of elements:

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0, which for each j simply state that αj = 0, as was to be proved. Representation of elements: For each x = (x1,...,xd) ∈ Fd we have that x = (x1,...,xd) =

d

∑

i=1

xiei.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Example. In Fd, let ei denote the vector such that the ith

component is 1 and all other components are zero. Then {e1,...,ed} is a base of Fd. Linear independence: For each i = 1,...,d let e(j)

i

denote the jth component of ei. Then for any α1,...,αd the vector equation

d

∑

i=1

αiei = 0 leads to the scalar equations

d

∑

i=1

αie(j)

i

= 0, which for each j simply state that αj = 0, as was to be proved. Representation of elements: For each x = (x1,...,xd) ∈ Fd we have that x = (x1,...,xd) =

d

∑

i=1

xiei.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = / 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = /

• 0. Hence, C has more elements than F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = /

• 0. Hence, C has more elements than F. Because b ∈ F, we
• btain
• (L\{b})∪H
• ∩F
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = /

• 0. Hence, C has more elements than F. Because b ∈ F, we
• btain
• (L\{b})∪H
• ∩F
• =
• (L∩F)∪H
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = /

• 0. Hence, C has more elements than F. Because b ∈ F, we
• btain
• (L\{b})∪H
• ∩F
• =
• (L∩F)∪H
• > |L∩F|

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Theorem. Let X be a vector space with a finite base F. Then

every linearly independent subset L of X has at most as many elements as F. Moreover, all bases of X have as many elements as F.

• Proof. Let F = {f1,...,fn} be a finite base of X. Suppose for a

contradiction that there is a linearly independent set L ⊆ X that has more elements than F. WLOG assume that L is such that if ˜ L is another linearly independent subset of X with more elements than F, then

• ˜

L∩F

• ≤ |L∩F|. Let b ∈ L\F and

consider the linearly independent sets L\{b} and F. There is a subset H ⊆ F \L so that C :=

• L\{b}
• ∪H is a base of X (good

exercise). But L\{b} is not a base of X (good exercise), so H = /

• 0. Hence, C has more elements than F. Because b ∈ F, we
• btain
• (L\{b})∪H
• ∩F
• =
• (L∩F)∪H
• > |L∩F|, a

contradiction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X, then |B| ≤ |F|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X, then |B| ≤ |F|. Therefore B is finite.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X, then |B| ≤ |F|. Therefore B is finite. With the same argument as above, if L ⊆ X is linearly independent, then |L| ≤ |B|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X, then |B| ≤ |F|. Therefore B is finite. With the same argument as above, if L ⊆ X is linearly independent, then |L| ≤ |B|. Because F is linearly independent, we obtain |F| ≤ |B|, and hence |F| = |B|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Proof (concl.). Thus no linearly independent subset of X has more elements than F. Now, if B is another base of X, then |B| ≤ |F|. Therefore B is finite. With the same argument as above, if L ⊆ X is linearly independent, then |L| ≤ |B|. Because F is linearly independent, we obtain |F| ≤ |B|, and hence |F| = |B|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy). Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy).

• Example. Let D be a set and let F be a field.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy).

• Example. Let D be a set and let F be a field. The vector space

F(D,F) of all functions f : D → F is an algebra with the multiplication operation defined pointwise by (f ·g)(x) := f(x)·g(x).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy).

• Example. Let D be a set and let F be a field. The vector space

F(D,F) of all functions f : D → F is an algebra with the multiplication operation defined pointwise by (f ·g)(x) := f(x)·g(x). All properties follow from the corresponding pointwise properties for elements of fields.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

• Definition. Let X be a vector space over the field F and let

· : X ×X → X be a binary operation. Then (X,·) is called an algebra iff

• 1. The multiplication operation is associative, and
• 2. Multiplication is left- and right distributive over addition.
• 3. For all α ∈ F and all x,y ∈ X we have that

α(xy) = (αx)y = x(αy).

• Example. Let D be a set and let F be a field. The vector space

F(D,F) of all functions f : D → F is an algebra with the multiplication operation defined pointwise by (f ·g)(x) := f(x)·g(x). All properties follow from the corresponding pointwise properties for elements of fields.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

• 1. An algebra with a few more properties (commutativity,

unit element, multiplicative inverses) becomes a field.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

• 1. An algebra with a few more properties (commutativity,

unit element, multiplicative inverses) becomes a field.

• 2. So can we “go beyond C”?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

• 1. An algebra with a few more properties (commutativity,

unit element, multiplicative inverses) becomes a field.

• 2. So can we “go beyond C”?
• 3. It can be proved (with a lot of functional analysis) that any

complete (defined through analysis) normed algebra over C with a multiplicative unit element and multiplicative inverses for all nonzero elements is isomorphic to C.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”: Any complete normed field that contains C is an algebra over C

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”: Any complete normed field that contains C is an algebra over C, and hence it must be equal to C.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”: Any complete normed field that contains C is an algebra over C, and hence it must be equal to C. So we cannot expand without losing at least one useful property.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”: Any complete normed field that contains C is an algebra over C, and hence it must be equal to C. So we cannot expand without losing at least one useful property. (Quaternions lose the third property in the definition of an algebra as well as commutativity.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras

logo1 Vector Spaces Bases Algebras

Final Comments on Number Systems

So just like R, C is in some ways a “constant of nature”: Any complete normed field that contains C is an algebra over C, and hence it must be equal to C. So we cannot expand without losing at least one useful property. (Quaternions lose the third property in the definition of an algebra as well as commutativity.) On the other hand, because we want polynomial equations to be solvable, we cannot shrink to something smaller than C and expect an elegant theory.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Beyond Fields: Vector Spaces and Algebras