Consider the set of $n=9$ vectors from $\complex{4}$, \begin{equation*} R=\set{ \colvector{-1\\3\\1\\2},\, \colvector{7\\1\\-3\\6},\, \colvector{1\\2\\-1\\-2},\, \colvector{0\\4\\2\\9},\, \colvector{5\\-2\\4\\3},\, \colvector{2\\1\\-6\\4},\, \colvector{3\\0\\-3\\1},\, \colvector{1\\1\\5\\3},\, \colvector{-6\\-1\\1\\1} }. \end{equation*}
To employ Theorem LIVHS, we form a $4\times 9$ matrix, $C$, whose columns are the vectors in $R$ \begin{equation*} C= \begin{bmatrix} -1&7&1&0&5&2&3&1&-6\\ 3&1&2&4&-2&1&0&1&-1\\ 1&-3&-1&2&4&-6&-3&5&1\\ 2&6&-2&9&3&4&1&3&1 \end{bmatrix}. \end{equation*}
To determine if the homogeneous system $\homosystem{C}$ has a unique solution or not, we would normally row-reduce this matrix. But in this particular example, we can do better. Theorem HMVEI tells us that since the system is homogeneous with $n=9$ variables in $m=4$ equations, and $n>m$, there must be infinitely many solutions. Since there is not a unique solution, Theorem LIVHS says the set is linearly dependent.