The system

x |
+ y |
+ z |
= | 1 |

2x |
+ 2y |
+ 2z |
= | 2 |

- x |
- y |
- z |
= | -1 |

x |
+ y |
+ z |
= | 1 |

consists of four equations and appears to be "large". However, we know the system *essentially* consists of just one equation: `x` + `y` + `z` = 1. The true size of the system is reflected in the fact that the augmented matrix

[ | 1 | 1 | 1 | 1 | ] |

2 | 2 | 2 | 2 | ||

-1 | -1 | -1 | -1 | ||

1 | 1 | 1 | 1 |

has only one pivot in the row echelon form.

[ | 1 | 1 | 1 | 1 | ] |

0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 |

Moreover, by this method of computing the basis of the column space, the number of pivots is the dimension of the column space of the matrix. The observation leads to the definition of the rank of a matrix.

[ | 1 | 3 | 2 | 0 | 1 | 0 | ]. |

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

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

1 | 3 | 2 | -2 | 0 | 0 |

In this example, we computed a row echelon form.

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |

0 | 2 | 1 | -1 | 0 | 1 | ||

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

0 | 0 | 0 | 0 | 0 | 1 |

Thus

rank[ |
1 | 3 | 2 | 0 | 1 | 0 | ] = 4. |

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

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

1 | 3 | 2 | -2 | 0 | 0 |

Moreover, if we take only the first four or five columns, then

rank[ |
1 | 3 | 2 | 0 | 1 | ] = rank[ |
1 | 3 | 2 | 0 | ] = 3. |

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

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

1 | 3 | 2 | -2 | 0 | 1 | 3 | 2 | -2 |

If we take only the first two or three columns, then

rank[ |
1 | 3 | ] = rank[ |
1 | 3 | 2 | ] = 2. |

-1 | -1 | -1 | -1 | -1 | |||

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

1 | 3 | 1 | 3 | 2 |

A matrix also has the null space. Its dimension is the nullity.

By this method of computing the basis of the null space, we have

`nullity A`

= number of free variables in the solution of

= number of nonpivot columns in (the row echelon form of)

where the second equality follows from this correspondence. Compared with the definition of `rank A, `we get the following equality.

Example By the computation in the earlier example and the relation between the rank and nullity, we have

nullity[ |
1 | 3 | 2 | 0 | 1 | 0 | ] = 6 - 4 = 2. |

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

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

1 | 3 | 2 | -2 | 0 | 0 |

nullity[ |
1 | 3 | 2 | 0 | 1 | ] = 5 - 3 = 2. |

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

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

1 | 3 | 2 | -2 | 0 |

nullity[ |
1 | 3 | 2 | 0 | ] = 4 - 3 = 1. |

-1 | -1 | -1 | 1 | ||

0 | 4 | 2 | 4 | ||

1 | 3 | 2 | -2 |