Linear space

Linear space


	Linear space

Linear space is one of the fundamental concepts of algebra. Its formal definition can be found in numerous books on linear algebra, see also

http://en.wikipedia.org/wiki/Linear_space

Instead of copying this long definition we define linear space as follows.

Definition 1 ( Real Linear Space )

Consider a set of "n"-strings of real numbers.

$\begin{displaymath} {\rm R}^n = \{(x_1,\;x_2,\;\dots x_n),\;\;\mbox{ where } x_i \in {\rm R} \;\;\; (i=1,\;2,\; \dots n)\}, \end{displaymath}$

where

${\rm R}$

denotes the set of real numbers.

${\rm R}^n$

is called (real "n"-dimensional) linear space if it is equipped with the following operations.

Addition and Subtraction.

$\begin{eqnarray*} x &=& (x_1,\;x_2,\;\dots x_n) \ y &=& (y_1,\;y_2,\;\dots y... ...x \pm y &=& ( x_1 \pm y_1 ,\; x_2 \pm y_2,\; \dots x_n \pm y_n). \end{eqnarray*}$

Multiplication by a number.

$\begin{eqnarray*} x &=& (x_1,\;x_2,\;\dots x_n) \ \lambda x &=& (\lambda x_1 ,\;\lambda x_2 ,\; \dots \lambda x_n ). \end{eqnarray*}$

The length of the string in this definition is called the dimension of the linear space. That means

${\rm R}$

can be considered as "n"-dimensional linear space.

We illustrate the concept of real linear space by looking at zero-space of a matrix.

Definition 2 If

$A\in {\rm R}^{n\times m}$

is a real matrix with "n" rows and "m" columns then its zero-space is defined as

$\begin{displaymath} Ax=0 \end{displaymath}$

Example 1

Consider

$\begin{displaymath} \left( \begin{array}{cccc} 1 & -4 & 9 & -7 \ -1 & 3 & 8 ... ...n{array}{c} x_1 \ x_2 \ x_3 \ x_4 \end{array}\right) = 0 \end{displaymath}$

Gaussian elimination procedure leads us to the following system.

$\begin{displaymath} \left( \begin{array}{cccc} 1 & -4 & 9 & -7 \ 0 & -1 & 17... ...n{array}{c} x_1 \ x_2 \ x_3 \ x_4 \end{array}\right) = 0 \end{displaymath}$

Since we have four unknowns and three equations we can assign arbitrary values for

Therefore, we declare

as a parameter,

$\begin{displaymath} x_4 = t. \end{displaymath}$

Then the solution of the system is given by

$\begin{displaymath} x= t \cdot\left( \begin{array}{c} \frac{440}{9}\ \frac{110}{9}\ \frac{7}{9} \ 1 \end{array}\right) \end{displaymath}$

where "t" can take arbitrary real values.

One can calculate the number of independent parameters in the solution. This number is equal to the difference between the number of unknowns and number of nonzero equations obtained after Gaussian elimination procedure.

Next:ExercisesUp:Linear spacePrevious:Linear space

Sergey Nikitin 2004-10-05