We illustrate another method for finding a basis of a column space through the same example.

= [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. The five spanning vectors can be simplified by column operations. The computation in this example gives us a column echelon form.

= [B |
1 | 0 | 0 | 0 | 0 | ] |

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

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

1 | 0 | -1 | 0 | 0 |

Since zero columns do not contribute to the span, we have

`col A` =

On the other hand, the first three columns of ` B` are linearly independent.

`c`_{1}[col 1] + `c`_{2}[col 2] + `c`_{3}[col 3] = **0**

⇒ `c`_{1} = 0 (look at the 1st coordinate of `c`_{1}[col 1] + `c`_{2}[col 2] + `c`_{3}[col 3] = **0**)

⇒ `c`_{2} = 0 (look at the 2nd coordinate of `c`_{2}[col 2] + `c`_{3}[col 3] = **0**)

⇒ `c`_{3} = 0. (look at the 3rd coordinate of `c`_{3}[col 3] = **0**)

Thus the nonzero columns of `B`

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

[col 2] of

[col 3] of

form a basis of `col B` =

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

- If
is the column echelon form of`B`, then`A``col`=**B**`col`.**A** - The nonzero columns in
are linearly independent and span`B``col`.**B**

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

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

Example To find a basis of

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

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

we use the translation

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

c |
d |

to convert the problem to the basis of the column space of

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

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

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

1 | 3 | 1 | 2 |

Column operations give us a column echelon form.

[ | 1 | 0 | 0 | 0 | ] |

-1 | 3 | 0 | 0 | ||

1 | -3 | 0 | 0 | ||

1 | 1 | 1 | 0 |

The nonzero columns (1, -1, 1, 1), (0, 3, -3, 1), (0, 0, 0, 1) form a basis of the column space. The corresponding 2 by 2 matrices

[ | 1 | -1 | ], [ | 0 | 3 | ], [ | 0 | 0 | ] |

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

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

Example To find a basis for the range of the linear transformation

` T`(

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

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

1 | 1 | 1 | ||

0 | 2 | 2 |

By column operations, we get a column echelon form

[ | 1 | 0 | 0 | ], |

1 | 1 | 0 | ||

0 | 2 | 0 |

and the nonzero columns (1, 1, 0), (0, 1, 2) of the column echelon form is a basis of the column space. Correspondingly,
1 + `t` and `t` + 2`t`^{2} form a basis of `range T`.

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