# 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 dependent if there exists 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 independent if,
$\sum\limits_{i = 0}^n {{\alpha _i}} {u_i} = 0,\,\,\,{\alpha _i} \in F$

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

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

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 is linearly dependent.

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 the elements of $S$ is zero then the coefficient must be zero. $S$ is linearly independent.