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 a 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 a 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 $$\,{a_1}\left( {1,2,1} \right) + {a_2}\left( {2,1,0} \right) + {a_3}\left( {1, – 1,2} \right) = 0$$
\[\begin{gathered} \Rightarrow \left( {{a_1} + 2{a_2} + {a_3},\,\,2{a_1} + {a_2} – {a_3},\,\,{a_1} + 2{a_3}} \right) = \left( {0,0,0} \right) \\ \Rightarrow {a_1} + 2{a_2} + {a_3} = 0,\,\,2{a_1} + {a_2} – {a_3} = 0,\,\,{a_1} + {3_3} = 0 \\ \Rightarrow {a_1} = {a_2} = {a_3} = 0 \\ \end{gathered} \]

Hence the given set is linearly independent.

Now let
\[\begin{gathered} \left( {1,0,0} \right) = x\left( {1,2,1} \right) + y\left( {2,1,0} \right) + z\left( {1, – 1,2} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x + 2y + z,\,2x + y – z,\,x + 2z} \right) \\ \end{gathered} \]

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 that every element 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 $${V_3}\left( F \right)$$.