In the next example, we find a basis of the null space of a matrix.

Example The null space of the matrix

= [A |
1 | 3 | 2 | 0 | 1 | ] |

-1 | -1 | -1 | 1 | 0 | ||

0 | 4 | 2 | 4 | 3 | ||

1 | 3 | 2 | -2 | 0 |

consists of all the solutions of the homogeneous system ** Ax** =

[ | 1 | 0 | 1/2 | 0 | -1/4 | ] |

0 | 1 | 1/2 | 0 | 1/4 | ||

0 | 0 | 0 | 1 | 1/2 | ||

0 | 0 | 0 | 0 | 0 |

Then we derive the solutions of the homogeneous system.

`x`_{1} = - (1/2)`x`_{3} - (1/4)`x`_{5}

`x`_{2} = - (1/2)`x`_{3} - (1/4)`x`_{5}

`x`_{3} is arbitrary

`x`_{4} = - (1/2)`x`_{5}

`x`_{5} is arbitrary

Thus `nul A` consists of vectors of the form

= [x |
x_{1} |
] = [ | - (1/2)x_{3} - (1/4)x_{5} |
] = x_{3}[ |
-1/2 | ] + x_{5}[ |
-1/4 | ] = x_{3} + ux_{5}.v |

x_{2} |
- (1/2)x_{3} - (1/4)x_{5} |
-1/2 | -1/4 | |||||

x_{3} |
x_{3} |
1 | 0 | |||||

x_{4} |
- (1/2)x_{5} |
0 | -1/2 | |||||

x_{5} |
x_{5} |
0 | 1 |

Since `x`_{3} and `x`_{5} are arbitrary, we have `nul A` =

`x`_{3}` u` +

⇒ The solution

⇒ All coordinates of the solution

⇒ The 3rd and 5th coordinates

Consequently, ** u** and

The discussion in the example applies to the general case. Thus we have the following method for finding bases of the null spaces.

If the general solution of ** Ax** =

` x` =

where `c`_{1}, `c`_{2}, ..., `c _{k}` are the free variables. Then

The bases of kernels of linear transformations can be found in a similar way.

Example Consider the linear transformation

(T) = X - AX: XAM(2, 2) → M(2, 2), where = [A |
1 | 2 | ]. |

3 | 4 |

Denote

= [X |
x |
y |
]. |

z |
w |

Then by looking at the four entries, ` T`(

3y |
- 2z |
= | 0 | ||

2x |
+ 3y |
- 2w |
= | 0 | |

- 3x |
- 3z |
+ 3w |
= | 0 | |

- 3y |
+ 2z |
= | 0 |

The general solution is `x` = - `z` + `w`, `y` = (2/3)`z`, with `z` and `w` arbitrary. In the matrix form, the solution is

= [X |
- z + w |
(2/3)z |
] = z[ |
-1 | 2/3 | ] + w[ |
1 | 0 | ]. |

z |
w |
1 | 0 | 0 | 1 |

Thus a basis of `kernel T` is given by the matrices

[ | -1 | 2/3 | ], [ | 1 | 0 | ]. |

1 | 0 | 0 | 1 |