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) Sgenerates 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

\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 independence.
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 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).  

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