Basis
A basis of the null space of a matrix may be computed by Gaussian elimination.
For this purpose, given an m × n matrix A, we construct first the row augmented matrix where I is the n × n identity matrix.
Computing its column echelon form by Gaussian elimination (or any other available method), we get a matrix A basis of the null space of A consists in the non zero columns of C such that the corresponding column of B is a zero column.
In fact, the computation may be stopped as soon as the upper matrix is in column echelon form: the remainder of the computation consists in changing the basis of the vector space generated by the columns whose upper part is zero.
For example, suppose that
Then
Putting the upper part in column echelon form by column operations on the whole matrix gives
The last three columns of B are zero columns. Therefore, the three last vectors of C,
are a basis of the null space of A.
Read more about this topic: Kernel (matrix)
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![A=\left[ \begin{array}{cccccc}
1 & 0 & -3 & 0 & 2 & -8 \\
0 & 1 & 5 & 0 & -1 & 4 \\
0 & 0 & 0 & 1 & 7 & -9 \\
0 & 0 & 0 & 0 & 0 & 0 \end{array} \,\right].](http://upload.wikimedia.org/math/d/e/5/de56a2cf3e4a8a25494a76b8eaad5acd.png)
![\left=
\left[\begin{array}{cccccc}
1 & 0 & -3 & 0 & 2 & -8 \\
0 & 1 & 5 & 0 & -1 & 4 \\
0 & 0 & 0 & 1 & 7 & -9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\hline\\
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{array}\right].](http://upload.wikimedia.org/math/2/b/8/2b86db5ea4854c7f5097fae0baca1e07.png)
![\left=
\left[\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\hline\\
1 & 0 & 0 & 3 & -2 & 8 \\
0 & 1 & 0 & -5 & 1 & -4 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & -7 & 9 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{array}\right].](http://upload.wikimedia.org/math/0/9/d/09d8904b3c5d6dc1776e6f3583c0284f.png)
