Consider the set of $n=4$ vectors from $\complex{5}$, \begin{equation*} S=\set{ \colvector{2\\-1\\3\\1\\2},\, \colvector{1\\2\\-1\\5\\2},\, \colvector{2\\1\\-3\\6\\1},\, \colvector{-6\\7\\-1\\0\\1} } \end{equation*}
To determine linear independence we first form a relation of linear dependence, \begin{equation*} \alpha_1\colvector{2\\-1\\3\\1\\2}+ \alpha_2\colvector{1\\2\\-1\\5\\2}+ \alpha_3\colvector{2\\1\\-3\\6\\1}+ \alpha_4\colvector{-6\\7\\-1\\0\\1} =\zerovector \end{equation*}
We know that $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$ is a solution to this equation, but that is of no interest whatsoever. That is always the case, no matter what four vectors we might have chosen. We are curious to know if there are other, nontrivial, solutions. Theorem SLSLC tells us that we can find such solutions as solutions to the homogeneous system $\homosystem{A}$ where the coefficient matrix has these four vectors as columns, which we then row-reduce \begin{equation*} A= \begin{bmatrix} 2&1&2&-6\\ -1&2&1&7\\ 3&-1&-3&-1\\ 1&5&6&0\\ 2&2&1&1 \end{bmatrix} \rref \begin{bmatrix} \leading{1}&0&0&-2\\ 0&\leading{1}&0&4\\ 0&0&\leading{1}&-3\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix} \end{equation*}
We could solve this homogeneous system completely, but for this example all we need is one nontrivial solution. Setting the lone free variable to any nonzero value, such as $x_4=1$, yields the nontrivial solution \begin{equation*} \vect{x}=\colvector{2\\-4\\3\\1} \end{equation*} completing our application of Theorem SLSLC, we have \begin{equation*} 2\colvector{2\\-1\\3\\1\\2}+ (-4)\colvector{1\\2\\-1\\5\\2}+ 3\colvector{2\\1\\-3\\6\\1}+ 1\colvector{-6\\7\\-1\\0\\1} =\zerovector \end{equation*}
This is a relation of linear dependence on $S$ that is not trivial, so we conclude that $S$ is linearly dependent.