# Basis of a Vector Space

A subset $S$ of a vector space $V\left( F \right)$ is said to be a basis of $V\left( F \right)$, if

(i) $S$ consists of linearly independent vector, and

(ii) $S$generates $V\left( F \right)$, i.e. $L\left( S \right) = V$ i.e. each vector in $V$ is a linear combination of finite number of elements of $S$.

For example the set $\left\{ {\left( {1,0,0} \right),\left( {0,1,0} \right),\left( {0,0,1} \right)} \right\}$ is a basis of the vector space ${V_3}\left( R \right)$ over the field of real numbers.

Dimension:
The dimension of a vector space $V\left( F \right)$ is the number of elements in a basis of $V\left( F \right)$.

Example:
Show that the set $S = \left\{ {\left( {1,2,1} \right),\left( {2,1,0} \right),\left( {1, - 1,2} \right)} \right\}$ forms a basis for ${V_3}\left( F \right)$.

Solution:
For ${a_1},{a_2},{a_3} \in F$, then $\,{a_1}\left( {1,2,1} \right) + {a_2}\left( {2,1,0} \right) + {a_3}\left( {1, - 1,2} \right) = 0$

Hence the given set is linearly independence.

Now let

So that $x + 2y + z = 1,\,\,\,\,2x + y - z = 0,\,\,\,\,x + 2z = 0$
$\therefore \,\,\,x = - \frac{2}{9},\,\,\,y = \frac{5}{9},\,\,\,z = \frac{1}{9}$

Thus, the unit vector $\left( {1,0,0} \right)$ is a linear combination of the vectors of the given set, i.e.
$\left( {1,0,0} \right) = - \frac{2}{9}\left( {1,2,1} \right) + \frac{5}{9}\left( {2,1,0} \right) + \frac{1}{9}\left( {1, - 1,2} \right)$
$\left( {0,1,0} \right) = \frac{4}{9}\left( {1,2,1} \right) - \frac{1}{9}\left( {2,1,0} \right) - \frac{2}{9}\left( {1, - 1,2} \right)$
$\left( {0,0,1} \right) = \frac{1}{3}\left( {1,2,1} \right) - \frac{1}{3}\left( {2,1,0} \right) + \frac{1}{3}\left( {1, - 1,2} \right)$

Since ${V_3}\left( F \right)$ is generated by the unit vectors$\left( {1,0,0} \right)$, $\left( {0,1,0} \right)$, $\left( {0,0,1} \right)$ we see therefore that every elements of ${V_3}\left( F \right)$ is a linear combination of the given set $S$. Hence the vectors of this set form a basis of ${V_3}\left( F \right)$.