Editing Linear subspace (section)

===Equations for a subspace===
:'''Input''' A basis {'''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>} for a subspace ''S'' of ''K''<sup>''n''</sup>
:'''Output''' An (''n''&nbsp;−&nbsp;''k'')&nbsp;×&nbsp;''n'' matrix whose null space is ''S''.
:# Create a matrix ''A'' whose rows are {{nowrap| '''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>}}.
:# Use elementary row operations to put ''A'' into reduced row echelon form.
:# Let {{nowrap| '''c'''<sub>1</sub>, '''c'''<sub>2</sub>, ..., '''c'''<sub>''n''</sub> }} be the columns of the reduced row echelon form. For each column without a pivot, write an equation expressing the column as a linear combination of the columns with pivots.
:# This results in a homogeneous system of ''n'' − ''k'' linear equations involving the variables '''c'''<sub>1</sub>,...,'''c'''<sub>''n''</sub>. The {{nowrap| (''n'' − ''k'') × ''n''}} matrix corresponding to this system is the desired matrix with nullspace ''S''.
; Example
:If the reduced row echelon form of ''A'' is

::<math>\left[ \begin{alignat}{6}
1 && 0 && -3 && 0 &&  2 && 0 \\
0 && 1 &&  5 && 0 && -1 && 4 \\
0 && 0 &&  0 && 1 &&  7 && -9 \\
0 && \;\;\;\;\;0 &&  \;\;\;\;\;0 && \;\;\;\;\;0 &&  \;\;\;\;\;0 && \;\;\;\;\;0 \end{alignat} \,\right] </math>

:then the column vectors {{nowrap| '''c'''<sub>1</sub>, ..., '''c'''<sub>6</sub>}} satisfy the equations

::<math> \begin{alignat}{1}
\mathbf{c}_3 &= -3\mathbf{c}_1 + 5\mathbf{c}_2 \\
\mathbf{c}_5 &= 2\mathbf{c}_1 - \mathbf{c}_2 + 7\mathbf{c}_4 \\
\mathbf{c}_6 &= 4\mathbf{c}_2 - 9\mathbf{c}_4
\end{alignat}</math>

:It follows that the row vectors of ''A'' satisfy the equations

::<math> \begin{alignat}{1}
x_3 &= -3x_1 + 5x_2 \\
x_5 &= 2x_1 - x_2 + 7x_4 \\
x_6 &= 4x_2 - 9x_4.
\end{alignat}</math>

:In particular, the row vectors of ''A'' are a basis for the null space of the corresponding matrix.