Sunday, June 1, 2014

Linear Algebra: #2 Basic Definitions

  • Linear Algebra: #2 Basic Definitions

Let X and Y be sets. The Cartesian product X × Y , of X with Y is the set of all possible pairs (x, y) such that x ∈ X and y ∈ Y .

A group is a non-empty set G, together with an operation (The operation is usually called “multiplication” in abstract group theory, but the sets we will deal with are also groups under “addition”.), which is a mapping ‘ · ’ : G × G → G, such that the following conditions are satisfied.

  1. For all a, b, c ∈ G, we have (a · b) · c = a · (b · c),
  2. There exists a particular element (the “neutral” element), often called e in group theory, such that e · g = g · e = g, for all g ∈ G.
  3. For each g ∈ G, there exists an inverse element g-1 ∈ G such that g · g-1 = g-1· g = e. 

If, in addition, we have a · b = b · a for all a, b ∈ G, then G is called an “Abelian” group.

A field is a non-empty set F, having two arithmetical operations, denoted by ‘+’ and ‘·’, that is, addition and multiplication (Of course, when writing a multiplication, it is usual to simply leave the ‘·’ out, so that the expression a · b is simplified to ab). Under addition, F is an Abelian group with a neutral element denoted by ‘0’. Furthermore, there is another element, denoted by ‘1’, with 1 ≠ 0, such that F \ {0} (that is, the set F, with the single element 0 removed) is an Abelian group, with neutral element 1, under multiplication. In addition, the distributive property holds:

 a · (b + c) = a · b + a · c and (a + b) · c = a · c + b · c,
 for all a, b, c ∈ F.

The simplest example of a field is the set consisting of just two elements {0, 1} with the obvious multiplication. This is the field ℤ/2ℤ (Here ℤ is the set of integers). Also, as we have seen in the analysis lectures, for any prime number p ∈ N, the set ℤ/pℤ of residues modulo p is a field. The following theorem, which should be familiar from the analysis lectures, gives some elementary general properties of fields.

Theorem 1
Let F be a field. Then for all a, b ∈ F, we have:
  1. a · 0 = 0 · a = 0,
  2. a · (−b) = −(a · b) = (−a) · b,
  3. −(−a) = a,
  4. (a−1)−1 = a, if a ≠ 0,
  5. (−1) · a = −a,
  6. (−a) · (−b) = a · b,
  7. a · b = 0 ⇒ a = 0 or b = 0. 

Proof. An exercise (dealt with in the analysis lectures).

So the theory of abstract vector spaces starts with the idea of a field as the underlying arithmetical system. But in physics, and in most of mathematics (at least the analysis part of it), we do not get carried away with such generalities. Instead we will usually be confining our attention to one of two very particular fields, namely either the field of real numbers ℜ, or else the field of complex numbers ℂ.

Despite this, let us adopt the usual generality in the definition of a vector space.

A vector space V over a field F is an Abelian group — with vector addition denoted by v + w, for vectors v, wV. The neutral element is the “zero vector” 0. Furthermore, there is a scalar multiplication F × VV satisfying (for arbitrary a, b ∈ F and v, wV):

  1. a · (v + w) = a · v + a · w
  2. (a + b) · v = a · v + b · v
  3. (a · b) · v = a · (b · v), and 
  4. 1 · v = v for all vV

Given any field F, then we can say that F is a vector space over itself. The vectors are just the elements of F. Vector addition is the addition in the field. Scalar multiplication is multiplication in the field.

Let ℜn be the set of n-tuples, for some n ∈ N. That is, the set of ordered lists of n real numbers. One can also say that this is
n = ℜ × ℜ × · · · × ℜ (n times), 

the Cartesian product, defined recursively. Given two elements

 (x1, · · · , xn) and (y1, · · · , yn)

in ℜn, then the vector sum is simply the new vector

(x1+ y1, . . . , xn+ yn

Scalar multiplication is
a ·  (x1, · · · , xn) =  (a ·x1, · · · , a ·xn)

It is a trivial matter to verify that ℜn, with these operations, is a vector space over ℜ.

Let C0([0, 1], ℜ) be the set of all continuous functions f : [0, 1] → ℜ. This is a vector space with vector addition

(f + g)(x) = f(x) + g(x), 

for all x ∈ [0, 1], defining the new function (f + g) ∈ C0([0, 1], ℜ), for all f, g ∈ C0([0, 1], ℜ). Scalar multiplication is given by

(a · f)(x) = a · f(x) 

for all x ∈ [0, 1].

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