*.*

**fields**As blogged about earlier (with groups), a field is defined:

By a field (

**F, +, ·)**we mean a commutative ring with unity which satisfies the additional axiom: (A1): For every non-zero element a Î F there exists an element a^-1 C 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 comrpises 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 C P ), then L is a sub-field of P.

Now, a vector space

**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.**

*V*One can also have some set W C

**, 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 muliple 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**

*V**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 v1, v2, v3........vn be elements of V. Also let x1, x2, x3......xn be numbers. Then it is possible to form an expression of the type:

x1 v1 + x2 v2 + x3 v3 +.............xn vn

which is called a linear combination of v1, v2, v3........vn. The set of all linear combinations of v1, v2, v3........vn 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 v1, v2, v3........vn are linearly dependent over a field F if there exist elements in F not all equal to zero such that:

a1 v1 + a2 v2 + ..............an vn = 0

If, on the other hand, there do not exist such numbers a1, a2 etc. we say that the elements v1, v2, v3........vn are

**linearly independent**.

Now, if elements v1, v2, v3........vn of the vector space V generate V and also are linearly independent, then (v1, v2, v3........vn) is called a

**basis**of V. One can also say that those elements v1, v2, v3........vn 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 fW.

As a further illustration, let V be a vector space and let (v1, v2, v3........vn) 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 = x1v1 + x2 v2 +.........xn vn, then we call (x1, x2, .......xn) the coordiantes of v with respect to our basis.

Example: 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)} .

Sample Problem (1):

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.

Sample Problem (2):

Find the coordinates of (1, 0) with respect to the two vectors (1,1) and (-1, 2)

Solution:

We must find numbers a and b which meet the condition:

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

This can be rewritten:

a - b = 1 and a + 2b = 0

Solve simultaneously, by subtracting the 2nd from the 1st:

a - b = 1

a + 2b = 0

-----------

0 - 3b = 1

and 3b = -1, so b = - 1/3, then a = 1 + b = 1 - 1/3 = 2/3

Then the coordinates of (1, 0) with respect to (1, 1) and (-1,2) are: (2/3, -1/3)

Sample Problem (3):

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

**R^2**.

Solution: This requires showing; a) the vectors are linearly independent, and b) they generate

**R^2**.

As before (earlier problems), we set out the condition via expression for linear independence:

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

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

solve simultaneusly by subtracting the 2nd from the 1st:

a - b = 0

a + 2b = 0

----------

0 - 3b = 0 so that b = 0 and a = 0

Thus the vectors are linearly independent.

(b) To show generation of

**R^2**, let (a,b) be an arbitrary element of of

**R^2**and write out:

x (1, 1) + y(-1, 2) = (a, b)

which leads to the pair of simultaenous equations:

x - y = a and x + 2y = b

As before, subtracting the 2nd from the 1st eqn.

x - y = a

x + 2y = b

----------

0 - 3y = a - b or y = (b - a)/ 3

Therefore, (x,y) are the coordinates of (a,b) with respect to the basis {(1,1), (-1,2)}.

Problems:

1) Show the following vectors are linearly independent:

a) (π, 0) and (0, 1)

b) (1, 1, 0), (1, 1, 1) and (0, 1, -1)

2) Express X as a linear combination of the given vectors A, B and find the coordinates of X with respect to A, B:

a) X = (1, 0), A = (1, 1), B = (0, 1)

b) X = (1,1), A = (2, 1), B = (-1, 0)

Solutions to these will be given along with those from the previous problem set in the next instalment of linear algebra!

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