Editing Row and column spaces (section)

===Basis===
The columns of {{mvar|A}} span the column space, but they may not form a [[basis (linear algebra)|basis]] if the column vectors are not [[linearly independent]].  Fortunately, [[elementary row operations]] do not affect the dependence relations between the column vectors.  This makes it possible to use [[row reduction]] to find a [[basis (linear algebra)|basis]] for the column space.

For example, consider the matrix
:<math>A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}.</math>
The columns of this matrix span the column space, but they may not be [[linearly independent]], in which case some subset of them will form a basis.  To find this basis, we reduce {{mvar|A}} to [[reduced row echelon form]]:
:<math>\begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}
\sim \begin{bmatrix} 1 & 3 & 1 & 4 \\ 0 & 1 & 1 & 1 \\ 0 & 2 & 2 & -3 \\ 0 & -1 & -1 & 4 \end{bmatrix}
\sim  \begin{bmatrix} 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 5 \end{bmatrix}
\sim  \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.</math><ref>This computation uses the [[Gaussian elimination|Gauss–Jordan]] row-reduction algorithm.  Each of the shown steps involves multiple elementary row operations.</ref>
At this point, it is clear that the first, second, and fourth columns are linearly independent, while the third column is a linear combination of the first two.  (Specifically, {{math|1='''v'''<sub>3</sub> = −2'''v'''<sub>1</sub> + '''v'''<sub>2</sub>}}.)  Therefore, the first, second, and fourth columns of the original matrix are a basis for the column space:
:<math>\begin{bmatrix} 1 \\ 2 \\ 1 \\ 1\end{bmatrix},\;\;
\begin{bmatrix} 3 \\ 7 \\ 5 \\ 2\end{bmatrix},\;\;
\begin{bmatrix} 4 \\ 9 \\ 1 \\ 8\end{bmatrix}.</math>
Note that the independent columns of the reduced row echelon form are precisely the columns with [[Pivot element|pivots]].  This makes it possible to determine which columns are linearly independent by reducing only to [[row echelon form|echelon form]].

The above algorithm can be used in general to find the dependence relations between any set of vectors, and to pick out a basis from any spanning set.  Also finding a basis for the column space of {{mvar|A}} is equivalent to finding a basis for the row space of the [[transpose]] matrix&nbsp;{{math|''A''<sup>T</sup>}}.

To find the basis in a practical setting (e.g., for large matrices), the [[singular-value decomposition]] is typically used.