Editing Linear independence (section)

=== Vectors in R<sup>2</sup> ===
'''Three vectors:'''  Consider the set of vectors <math>\mathbf{v}_1 = (1, 1),</math> <math>\mathbf{v}_2 = (-3, 2),</math> and <math>\mathbf{v}_3 = (2, 4),</math> then the condition for linear dependence seeks a set of non-zero scalars, such that
:<math>a_1 \begin{bmatrix} 1\\1\end{bmatrix} + a_2 \begin{bmatrix} -3\\2\end{bmatrix} + a_3 \begin{bmatrix} 2\\4\end{bmatrix} =\begin{bmatrix} 0\\0\end{bmatrix},</math>
or
:<math>\begin{bmatrix} 1 & -3 & 2 \\ 1 & 2 & 4 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}.</math>

[[Row reduction|Row reduce]] this matrix equation by subtracting the first row from the second to obtain,
:<math>\begin{bmatrix} 1 & -3 & 2 \\ 0 & 5 & 2 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}.</math>
Continue the row reduction by (i) dividing the second row by 5, and then (ii) multiplying by 3 and adding to the first row, that is
:<math>\begin{bmatrix} 1 & 0 & 16/5 \\ 0 & 1 & 2/5 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}.</math>

Rearranging this equation allows us to obtain
:<math>\begin{bmatrix} 1 & 0  \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}= \begin{bmatrix} a_1\\ a_2 \end{bmatrix}=-a_3\begin{bmatrix} 16/5\\2/5\end{bmatrix}.</math>
which shows that non-zero ''a''<sub>''i''</sub> exist such that <math>\mathbf{v}_3 = (2, 4)</math> can be defined in terms of <math>\mathbf{v}_1 = (1, 1)</math> and <math>\mathbf{v}_2 = (-3, 2).</math>  Thus, the three vectors are linearly dependent.

'''Two vectors:'''  Now consider the linear dependence of the two vectors <math>\mathbf{v}_1 = (1, 1)</math> and <math>\mathbf{v}_2 = (-3, 2),</math> and check, 
:<math>a_1 \begin{bmatrix} 1\\1\end{bmatrix} + a_2 \begin{bmatrix} -3\\2\end{bmatrix}  =\begin{bmatrix} 0\\0\end{bmatrix},</math>
or
:<math>\begin{bmatrix} 1 & -3  \\ 1 & 2  \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}.</math>

The same row reduction presented above yields,
:<math>\begin{bmatrix} 1 & 0  \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}.</math>
This shows that <math>a_i = 0,</math> which means that the vectors <math>\mathbf{v}_1 = (1, 1)</math> and <math>\mathbf{v}_2 = (-3, 2)</math> are linearly independent.