Linear Dependence and Linear Independence Vectors

Linear Dependence:
Let V\left( F \right) be a vector space and let S = \left\{ {{u_1},{u_2}, \ldots ,{u_n}} \right\} be a finite subset of V. Then S is said to be linearly dependence if there exist scalar{\alpha _1},{\alpha _2}, \ldots ,{\alpha _n} \in F, not all zero, such that

{\alpha _1}{u_1} + {\alpha _2}{u_2} + \cdots + {\alpha _n}{u_n} = 0

Linear Independence:
Let V\left( F \right) be a vector space and let S = \left\{ {{u_1},{u_2}, \ldots ,{u_n}} \right\} be a finite subset of V. Then S is said to be linearly independence if,

\sum\limits_{i = 0}^n {{\alpha _i}} {u_i} = 0,\,\,\,{\alpha _i} \in F

Holds only when {\alpha _i} = 0,\,\,\,i = 1,2,3, \ldots ,n.

An infinite subset S of V is said to be linearly independence if every finite subset S is linearly independent, otherwise it is linearly dependence.

Example 1: Show that the system of three vectors \left( {1,3,2} \right), \left( {1, - 7, - 8} \right), \left( {2,1, - 1} \right) of {V_3}\left( R \right) is linearly dependent.

Solution: For {\alpha _1},{\alpha _2},{\alpha _3} \in R.

\begin{gathered} \,\,\,\,\,\,\,{\alpha _1}\left( {1,3,2} \right) + {\alpha _2}\left( {1, - 7, - 8} \right) + {\alpha _3}\left( {2,1, - 1} \right) \\ \Leftrightarrow \left( {{\alpha _1} + {\alpha _2} + 3{\alpha _3},\,\,3{\alpha _1} - 7{\alpha _2} + {\alpha _3},\,\,2{\alpha _1} - 8{\alpha _2} - {\alpha _3}} \right) = 0 \\ \Leftrightarrow {\alpha _1} + {\alpha _2} + 3{\alpha _3} = 0,\,\,3{\alpha _1} - 7{\alpha _2} + {\alpha _3} = 0,\,\,2{\alpha _1} - 8{\alpha _2} - {\alpha _3} = 0 \\ \Leftrightarrow {\alpha _1} = 3,\,\,\,{\alpha _2} = 1,\,\,\,{\alpha _3} = - 2 \\ \end{gathered}

Therefore, the given system of vectors in linearly dependence.

Example 2: Consider the vector space {\mathbb{R}^3}\left( R \right) and the subset S = \left\{ {\left( {1,0,0} \right),\,\left( {0,1,0} \right),\left( {0,0,1} \right)} \right\} of {\mathbb{R}^3}. Prove that S is linearly independent.

Solution: For {\alpha _1},{\alpha _2},{\alpha _3} \in R.

\begin{gathered} \,\,\,\,\,\,\,{\alpha _1}\left( {1,0,0} \right) + {\alpha _2}\left( {0,1,0} \right) + {\alpha _3}\left( {0,0,1} \right) = \left( {0,0,0} \right) \\ \Leftrightarrow \left( {{\alpha _1},{\alpha _2},{\alpha _3}} \right) = \left( {0,0,0} \right) \\ \Leftrightarrow {\alpha _1} = 0,\,\,\,{\alpha _2} = 0,\,\,\,{\alpha _3} = 0 \\ \end{gathered}

This shows that if any linear combination of elements of S is zero then the coefficient must be zero. S is linearly independent.