Determine whether or not the set
\mathcal{S} = \begin{Bmatrix}\begin{pmatrix}1\\2\\1\end{pmatrix}, \begin{pmatrix}1\\0\\2\end{pmatrix}, \begin{pmatrix}5\\6\\ 7\end{pmatrix}\end{Bmatrix}is linearly independent.
Simply determine whether or not there exists a nontrivial solution for the α ’s in the homogeneous equation
α_{1}\begin{pmatrix}1\\2\\1\end{pmatrix}+α_{2} \begin{pmatrix}1\\0\\2 \end{pmatrix} + α_{3}\begin{pmatrix}5\\6\\7\end{pmatrix} = \begin{pmatrix}0\\0\\0 \end{pmatrix}
or, equivalently, if there is a nontrivial solution to the homogeneous system
\begin{pmatrix}1 &1& 5\\2 &0& 6\\1 &2& 7\end{pmatrix} \begin{pmatrix}α_{1}\\α_{2}\\α_{3}\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}.
If A = \begin{pmatrix}1 &1& 5\\2 &0& 6\\1 &2& 7\end{pmatrix}, then E_{A} = \begin{pmatrix}1 &0& 3\\0 &1& 2\\0 &0 &0\end{pmatrix}, and therefore there exist nontrivial solutions. Consequently, \mathcal{S} is a linearly dependent set. Notice that one particular dependence relationship in \mathcal{S} is revealed by E_{A} because it guarantees that A_{∗3} = 3A_{∗1}+2A_{∗2}. This example indicates why the question of whether or not a subset of ℜ^m is linearly independent is really a question about whether or not the nullspace of an associated matrix is trivial. The following is a more formal statement of this fact.