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

Example The column 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 |

is the span of the five column vectors. By the row operations in this example, we get a row echelon form.

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

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

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

0 | 0 | 0 | 0 | 0 |

Since [col 3] and [col 5] are not pivot, [col 5] is a linear combination of [col 1] through [col 4], and [col 3] is a linear combination of [col 1] and [col 2]. Then by this property, we have

`col A`

=

=

=

On the other hand, if we pick [col 1], [col 2], [col 4] only, then

[ | 1 | 3 | 0 | ] |

-1 | -1 | 1 | ||

0 | 4 | 4 | ||

1 | 3 | -2 |

has the following row echelon form

[ | 1 | 3 | 0 | ], |

0 | 2 | -1 | ||

0 | 0 | 2 | ||

0 | 0 | 0 |

in which all columns are pivot. Thus the three columns are linearly independent. Since they also span `col A`, we conclude that

[col 1] of ** A** = (1, -1, 0, 1)

[col 2] of

[col 4] of

form a basis of `col A`.

The argument in the example can be applied to any matrix ** A**. We have

- The nonpivot columns of
are linear combinations of the preceeding columns. Therefore they can be deleted without affecting the columns space.`A` - Tautologically, the matrix formed by the pivot columns of
has all columns pivot. Therefore the pivot columns are linearly independent.`A`

This leads to the following method for finding a basis of the column space.

We emphasis that it is the columns of ** A**, not the columns of the row echelon form,
that form the basis.

Example We would like to find a basis of

= Hspan{ [ |
1 | -1 | ], [ | 2 | 1 | ], [ | 4 | 2 | ], [ | 3 | -3 | ] }. |

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

By the translation

= [A |
a |
b |
] ↔ [] = (Aa, b, c, d), |

c |
d |

the problem becomes the basis of the column space of

[ | 1 | 2 | 4 | 3 | ]. |

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

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

1 | 3 | 1 | 2 |

Row operations give us the row echelon form.

[ | 1 | 2 | 4 | 3 | ] |

0 | 1 | 2 | 0 | ||

0 | 0 | 5 | 1 | ||

0 | 0 | 0 | 0 |

Thus the first three columns of the matrix form a basis of the column space. Correspondingly, the first three 2 by 2 matrices

[ | 1 | -1 | ], [ | 2 | 1 | ], [ | 4 | 2 | ] |

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

form a basis of the span ** H**.

Example We would like to find a basis for the range of the linear transformation

` T`(

The range is spanned by

` T`(1) = 1 +

By the translation [`a` + `bt` + `ct`^{2}] =
(`a`, `b`, `c`), the problem becomes the basis of the column space of

[ [(1)] [T(Tt)] [(Tt^{2})] ] = [ |
1 | 0 | 0 | ]. |

1 | 1 | 1 | ||

0 | 2 | 2 |

By row operations, we get

[ | 1 | 0 | 0 | ]. |

0 | 1 | 1 | ||

0 | 0 | 0 |

Thus the first two columns form a basis of the column space. Correspondingly,
` T`(1) = 1 +