Thursday, August 12, 2021

Examining Fields and Vector Spaces

 The fundamental starting point for linear algebra is the concept of the vector space, and this in turn requires familiarity with fields.


By a field (F, +, ·) we mean a commutative ring with unity which satisfies the additional axiom: (A1): For every non-zero element a
  F there exists an element a-1   F such that: a· a-1 = 1. The element a-1 = 1/a is called the reciprocal or multiplicative inverse of a. Thus, the key thing about a field is that it comprises a set of elements which can be added or multiplied in such a way that addition and multiplications satisfy the ordinary rules of arithmetic, and in such a way one can divide by non-zero elements.

Of course, if one has a field, one can also have a sub-field. Thus, say P and L are fields and L is contained in P (e.g. L
  P ), then L is a sub-field of P.

A vector space V over the field K, say, is a set of objects that can be added and multiplied by elements of K in such a way that the sum of any 2 elements of V is again, an element of V.

One can also have some set W V , i.e. W is a subset of V, which satisfies specific conditions: i) the sum of any 2 elements of W is also an element of W, ii) the multiple m(w1) of an element of W is also an element of W, the element 0 of V is also an element of W, then we call W a subspace of V.

As an illustration, let V = R n and let W be the set of vectors in V whose last coordinate = 0. Then, W is a subset of V, which we could identify with R n-1 .

As another illustration, let V be an arbitrary vector space and let v 1, v2v3........v n be elements of V. Also let x 1, x 2, x 3...... x n  be numbers. Then it is possible to form an expression of the type:

x 1 v 1 + x 2  v 2 + x 3 v 3 +.............x n v n

which is called a linear combination of  v 1, v2v3........v n. The set of all linear combinations of v 1, v2v3........v n is a subspace of V,

Yet another example: let A be a vector in R 3. Let W be the set of all elements B in R 3 such that B · A = 0, i.e. such that B is perpendicular to A. Then W is a subspace of R 3.

An additional important consideration is whether elements of a vector space are linearly dependent or linearly independent. We say the elements v 1, v2v3........v n are linearly dependent over a field F if there exist elements in F not all equal to zero such that:

a 1 v 1 + a 2 v 2 + ..............a n v n = 0

If, on the other hand, there do not exist such numbers a1, a2 etc. we say that the elements v 1, v2v3........v n are linearly independent.

Now, if elements v 1, v2v3........v n of the vector space V generate V and also are linearly independent, then (v 1, v2v3........v n) is called a basis of V. One can also say that those elements v 1, v2v3........v n form a basis of V.

Example: Let W f  be a vector space of functions generated by the two functions: exp(t) and exp(2t), then {exp(t), exp(2t)} is a basis of W f

As a further illustration, let V be a vector space and let (v 1, v2v3........v n) be a basis of V. The elements of V can be represented by n-tuples relative to this basis, e.g. if an element v of V is written as a linear combination:

 v = x 1 v 1 + x 2  v 2 + x 3  v 3 +.............x n v n

then we call (x 1, x 2, .......x n) the coordinates of v with respect to our basis.

Let V be the vector space of functions generated by the two functions: exp(t) and exp(2t), then what are the coordinates for f(V) = 3 exp(t) + 5 exp(2t)?

Ans. The coordinates are (3, 5) with respect to the basis {exp(t), exp(2t)} .

Example Problem :

Show that the vectors (1, 1) and (-3, 2) are linearly independent.

Solution: Let a, b be two numbers associated with some vector space - call it W- such that:

a(1,1) + b(-3,2) = 0

Writing out the components as linear combinations:

a - 3b = 0 and a + 2b = 0

Then solve simultaneously:

a - 3b = 0
a + 2b = 0
----------
0 -5b = 0

or b = 0, so a = 0

Both a and b are equal to zero so the vectors are linearly independent.


Suggested Problem:

Show that the vectors (1, 1) and (-1, 2) form a basis of R 2.


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