# Definition of Vector Space and its Theorem

Scalar addition and multiplication operations are used in
diverse contexts in mathematics. Regardless of the context, however, these
operations usually fulfill the same rules of arithmetic. So the general theory
of mathematical systems involving scalar addition and multiplication will be
applicable to various fields of mathematics. Mathematical systems with this
form are called Vector Rooms or Linear Space. In this article we explain the
definition of vector space and develop some general theories about vector
space.

###
**Vector Space**

We can review R

^{n}as a set of all n x 1 matrix with real number entries. Scalar addition and multiplication for vectors in R^{n}are the usual sum and scalar multiplication of the matrix. More generally, suppose R^{m}x R^{n}expresses the set of all matrix**m x n**with real number entries.
If A = (a

_{ij}) and B = (b_{ij}), then the number of A + B is defined as the matrix C = (c_{ij}) which has the order m x n, where c_{ij}= a_{ij}+ b_{ij}.
If given scalar

**α**, then we can define**αA**as the matrix**m x n**where the**ij**entry is**αaij**.###
**Axiom of the Vector Space**

Let

**V**be the set where the scalar addition and multiplication operations are defined. By this we mean that for each pair of elements**x**and**y**in**V**, we can associate it with a single**x + y**element which is also at**V**, and with each element**x**in**V**and every scalar**α**, we can associate it with single αx element inside**V**.
The set

**V**together with scalar addition and multiplication operations is said to form a vector space if the following axioms are fulfilled.
A1 x + y = y
+ x for every x and y in V.

A2 (x + y) +
z = x + (y + z) for every x, y, z in V.

A3 There is
an element 0 at V so x + 0 = x for every x ε V.

A4 For each
x ε V there is a -x element in V so x + (-x) = 0.

A5 α (x + y)
= αx + αy for each α scalar and every x and y in V.

A6 (α + β) x
= αx + βx for each scalar α and β and every x ε V.

A7 (α β) x =
α (β x) for each scalar α and β and every x ε V.

A8 x = x for
each x ε V

###
**Vector Space C | a, b |**

Suppose C [a, b] expresses
the set of all real-valued functions that are defined and continuous at the
closing interval [a, b]. In this case the set should be a set of functions. So
the vectors are functions in C [a, b]. The number of

*f*+*g*of the two functions in C [a, b] is defined by
(

*f*+*g*) (x) = f(x) + g(x)
For all x in [a, b]. The
new function

*f*+*g*is an element of C [a, b], because the sum of two continuous functions is continuous.
If f is a function in C
[a, b] and α a real number, then α f is defined by

(α

*f*) (x) = α*f*(x)
For all

**x**in [a, b]. It is clear that αf is in C [a, b] because if a constant multiplied by a continuous function is always continuous.
So in C [a, b] we have
defined operations — scalar addition and multiplication operations. To show
that the first axiom is

*f*+

*g*=

*g*+

*f*

fulfilled, we must show
that

(

*f*+*g*) (x) = (*g*+*f*) (x) for each x ϵ [a, b]
This equation is correct
because

(

*f*+*g*) (x) =*f*(x) +*g*(x) =*g*(x) +*f*(x) = (*g*+*f*) (x) for each x in [a, b]
Additional Properties of
the Vector Room

From the explanation above
can be summarized with a theorem which states three more basic properties for
vector space, which will be explained by the following theorem,

**Theorem 1.**

If

**V**is a vector space and**x**is any element of**V**, then
(i) 0x = 0

(ii) x + y = 0 results in y = -x (that is, the
inverse of x is singular)

(iii) (-1) x = -x

Proof:

Based on the A6 and A8
axioms then

x = 1x = (1 + 0) x = 1x +
0x = x + 0x

so

- x + x = - x + (x + 0x) =
(-x + x) + 0x ...... (A2)

0 = 0 + 0x = 0x
........................................ (A1, A3 and A4)

To prove (ii), suppose
that x + y = 0, then

- x = - x + 0 = - x + (x +
y)

Therefore,

- x = (-x + x) + y = 0 + y
= y ...................... (A1, A2, A3 and A4)

Finally, to prove (iii),
note that

0 = 0x = (1 + (-1)) x = 1x
+ (-1) x ................ [(i) and A6]

So

X + (-1) x = 0

And from section (ii) is
obtained

(-) x = -x

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