*Notes to a video lecture on http://www.unizor.com*

__Vectors+ 07 - Abstract Vector Space__

This lecture represents a typical mathematical approach to move from a relatively simple and concrete object to some abstraction that allows to transfer properties of the original simple object to many others without proving these properties for each particular case.

Let's start with a simple prototype - a two-dimensional Euclidian plane with coordinates and vectors we studied before. Coordinates are real numbers, vectors can be added to each other and multiplied by a real numbers resulting in some other vector.

Our first step towards abstraction was to move to

*N*-dimensional space, which retains most of the properties of two-dimensional one. However, this is not the exact step towards abstraction, because we still considered a concrete object - an

*N*-dimensional space and demonstrated its properties.

The real abstraction is to choose any object that satisfies certain axioms and derive the properties from these axioms without referring to any concrete object.

Here is what can be done in this direction.

First of all, we introduce an abstract set

*, whose elements will be called*

**V****vectors**. It's modelled after an

*N*-dimensional Euclidean space as a prototype, but, in theory, it can be any set satisfying the rules below.

We also consider a set

*, whose elements will be called*

**S****scalars**. Its prototype is a set of all real numbers used for multiplication by vectors to change their magnitude, and for our purposes, to concentrate attention on vectors, we will consider this set of scalars to be exactly that, a set of all real numbers. However, it can be some other set, like all complex numbers.

We postulate that these sets satisfy the following rules.

(A) The operation of

**addition**is defined for each pair of vectors in

*that establishes a correspondence of this pair of vectors to a new vector called their*

**V****sum**. What's important, we don't define this operation, we don't establish any process it should follow. The only requirement is that this operation must satisfy the following axioms.

(A1) Addition of any two vectors

*and*

**a***is*

**b****commutative**

∀

*∈*

**a,b***:*

**V**

**a + b = b + a**(A2) Addition of any three vectors

*,*

**a***and*

**b***is*

**c****associative**

∀

*∈*

**a,b,c***:*

**V**

**(a + b) + c = a + (b + c)**(A3) There is one vector called

**null-vector**, denoted

*, with a property of not changing the value of any other vector*

**0***if added to it*

**a**∀

*∈*

**a***:*

**V**

**a + 0 = a**(A4) For any vector

*there is another vector called its opposite, denoted as*

**a***, such that the sum of a vector and its opposite equals to null-vector*

**−a**∀

*∈*

**a***∃(*

**V***)∈*

**−a***:*

**V**

**a + (−a) = 0***Note*: in many cases an expression

*a+(−b)*will be shortened to

*a−b*. Sign '−' here does not mean a new operation of subtraction, but just an indication of addition with an opposite element.

(B) The operation of

**multiplication of vector by scalar**is defined for each vector in

*and each scalar in*

**V***, taken in any order. This operation establishes a correspondence of this vector and this scalar to a new vector called the*

**S****product of a vector and a scalar**.

This operation must satisfy the following axioms.

(B1) Multiplication of any scalar

*α*by any vector

*is*

**a****commutative**

∀

*∈*

**a***, ∀*

**V***α*∈

*:*

**S***α*

**·a = a·**α(B2) Multiplication of any two scalars

*α*and

*β*by any vector

*is*

**a****associative**

∀

*∈*

**a***, ∀*

**V***α,β*∈

*:*

**S***(α·β)*

**·a =**α·(β**·a**)(B3) Multiplication of any vector by scalar

*0*results in null-vector

∀

*∈*

**a***:*

**V***0*

**·a = 0**(B4) Multiplication of any vector by scalar

*1*does not change the value of this vector

**a**∀

*∈*

**a***:*

**V***1*

**·a = a**(B5) Multiplication is

**distributive**relatively to addition of vectors. ∀

*∈*

**a,b***, ∀*

**V***α*∈

*:*

**S***α*

**·(a+b) =**α**·a+**α**·b**(B6) Multiplication is

**distributive**relatively to addition of scalars.

∀

*∈*

**a***, ∀*

**V***α,β*∈

*:*

**S***(α+β)*

**·a =**α**·a+**β**·a***Problem A*

Prove that there must be only one null-vector in vector space

*.*

**V***Proof A*

Assume there are two null-vectors

*and*

**0**_{1}*.*

**0**_{2}Since

*is null-vector, its addition to*

**0**_{2}*does not change*

**0**_{1}*.*

**0**_{1}

**0**_{1}+ 0_{2}= 0_{1}Since

*is null-vector, its addition to*

**0**_{1}*does not change*

**0**_{2}*.*

**0**_{2}

**0**_{2}+ 0_{1}= 0_{2}Since addition is commutative, left sides in the two equations above are equal.

**0**_{1}+ 0_{2}= 0_{2}+ 0_{1}Therefore, the right sides are equal:

**0**_{1}= 0_{2}End of proof.

*Problem B*

Prove that for any vector there must be only one opposite to it vector in vector space

*.*

**V***Proof B*

Assume that for some vector

*there are two opposite vectors*

**a***and*

**(−a)**_{1}*.*

**(−a)**_{2}Since

*is an opposite to vector*

**(−a)**_{1}*, its addition to*

**a***results in null-vector.*

**a**Therefore,

**((−a)**_{1}+a) + (−a)_{2}= 0 + (−a)_{2}= (−a)_{2}Since

*is an opposite to vector*

**(−a)**_{2}*, its addition to*

**a***results in null-vector.*

**a**Therefore,

**(−a)**_{1}+ (a+(−a)_{2}) = (−a)_{2}+ 0 = (−a)_{2}Since addition is associative, left sides in the two equations above are equal.

Therefore, the right sides are equal:

**(−a)**_{1}= (−a)_{2}End of proof.

*Problem C*

Prove that for any vector its product with scalar

*−1*results in an opposite vector.

*Proof C*

Since vector multiplication by scalar is distributive,

*(1+(−1))*

**·a =**1**·a +**(−1)**·a = a +**(−1)**·a**On the other hand, the same initial statement can be transformed differently

*(1+(−1))*

**·a =**0**·a = 0**Since left sides in the two equations above are equal, right sides are equal as well

**a +**(−1)**·a = 0**Therefore,

*(−1)*satisfies the definition of a vector opposite to

**·a***. But, as has been proven in*

**a***Problem B*, there must be only one vector opposite to

*, which we denoted as*

**a***.*

**(−a)**Hence,

*(−1)*

**·a = (−a)**End of proof.

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