The columns space `col A` of a matrix

The span of vectors `v`_{1}, `v`_{2}, ..., ` v_{k}` is

`span`{`v`_{1}, `v`_{2}, ..., ` v_{k}`} =
{

Example Any vector in **R**^{3} can be expressed as follows.

` x` = (

Thus any vector is a linear combination of the
standard basis
**e**_{1}, **e**_{2}, **e**_{3}.
In other words, the standard basis vectors span the euclidean space **R**^{3}. By the same reason,
the standard basis **e**_{1}, **e**_{2}, ..., ` e_{n}`
span

Example Since any polynomial `a`_{0} + `a`_{1}`t`
+ `a`_{2}`t`^{2} + ... + `a _{n}t^{n}` of degree ≤

Example By the equality
1 = cos^{2}`t` + sin^{2}, the constant value 1 function is in the span of
cos^{2}`t` and sin^{2}`t`.
Similarly, the function cos2`t` = cos^{2}`t` - sin^{2}`t`
∈ `span`{cos^{2}`t`, sin^{2}`t`}.

The fact that `col A` is a subspace can also be extended.

Proof We only show that `span`{`v`_{1}, `v`_{2}, `v`_{3}} is a subspace. The general case is similar.

For ` u`,

` u` +

For ` u` ∈

`d u`
=

Example (see this example) By (`a` + `b`, 3`a` - `b`, 2`a` + `b`) = `a`(1, 3, 2) + `b`(1, -1, 1), the subset

{(`a` + `b`, 3`a` - `b`, 2`a` + `b`): `a`, `b` ∈ **R**} ⊂ **R**^{3}

is the span of (1, 3, 2) and (1, -1, 1) and is a subspace.

The subset

{`a` cos`t` + `b` sin`t`: `a`, `b` ∈ **R**} ⊂ `F`(**R**)

is a subspace because it is the span of the vectors cos`t` and sin`t`.

The question of whether a vector ** b** ∈

`c`_{1}`v`_{1} + `c`_{2}`v`_{2} + ... +
`c _{k}v_{k}` =

where `c`_{1}, `c`_{2}, ..., `c _{k}` are the unknown variables,
has solutions.

Example If cos`t` ∈ `span`{cos^{2}`t`, sin^{2}`t`}, then

`a` cos^{2}`t` + `b` sin^{2}`t` = cos`t`

for some numbers `a` and `b`. By taking `t` = 0, we get `a` = 1. By taking `t` = π, we get `a` = - 1. The contradiction shows that cos`t` ∉ `span`{cos^{2}`t`, sin^{2}`t`}.

Example Consider polynomials
`p`_{1} = 1 - `t` + `t`^{3},
`p`_{2} = 3 - `t` + 4`t`^{2} + 3`t`^{3},
`p`_{3} = 2 - `t` + 2`t`^{2} + 2`t`^{3},
`p`_{4} = `t` + 4`t`^{2} - 2`t`^{3},
`p`_{5} = 1 + 3`t`^{2} in `P`_{3}.
We would like to determine whether `q` = `t` + 3`t`^{2} ∈
`span`{`p`_{1}, `p`_{2}, `p`_{3}, `p`_{4}, `p`_{5}}.
The question is the same as whether we can find
`c`_{1}, `c`_{2}, `c`_{3}, `c`_{4}, `c`_{5} such that

`c`_{1}(1 - `t` + `t`^{3}) +
`c`_{2}(3 - `t` + 4`t`^{2} + 3`t`^{3}) +
`c`_{3}(2 - `t` + 2`t`^{2} + 2`t`^{3}) +
`c`_{4}(`t` + 4`t`^{2} - 2`t`^{3}) +
`c`_{5}(1 + 3`t`^{2}) =
`t` + 3`t`^{2}.

By comparing the coefficients of 1, `t`, `t`^{2}, and `t`^{3}, we get a system of four equations.

x_{1} |
+ 3x_{2} |
+ 2x_{3} |
+ x_{5} |
= | 0 | |

- x_{1} |
- x_{2} |
- x_{3} |
+ x_{4} |
= | 1 | |

4x_{2} |
+ 2x_{3} |
+ 4x_{4} |
+ 3x_{5} |
= | 3 | |

x_{1} |
+ 3x_{2} |
+ 2x_{3} |
- 2x_{4} |
= | 0 |

By this example, the system has no solution. Therefore `q` is not in the span.

Observe that the augmented matrix for the system is

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] =
[ [p_{1}]
[p_{2}]
[p_{3}]
[p_{4}]
[p_{5}]
[q] ], |

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

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

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

where we used the translation

`p` = `a` + `bt` + `ct`^{2} + `dt`^{3} ∈ `P`_{3}
↔ [`p`] = (`a`, `b`, `c`, `d`) ∈ **R**^{4}.

Thus problems about spans in general vector spaces can be translated into existence problems about usual systems of linear equations.

If we modify `q` to become `r` = `t` + 2`t`^{2}, then the augmented matrix of the corresponding existence problem is changed to

[ [p_{1}]
[p_{2}]
[p_{3}]
[p_{4}]
[p_{5}]
[r] ] = [ |
1 | 3 | 2 | 0 | 1 | 0 | ]. |

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

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

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

By an earlier example,
the system has solutions, so that `t` + 2`t`^{2} ∈
`span`{`p`_{1}, `p`_{2}, `p`_{3}, `p`_{4}, `p`_{5}}.

In the earlier example, we also saw that the system with the augmented matrix

[ [p_{1}]
[p_{2}]
[p_{4}]
[r] ] = [ |
1 | 3 | 0 | 0 | ] |

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

0 | 4 | 4 | 2 | ||

1 | 3 | -2 | 0 |

has solutions. Correspondingly, we have `t` + 2`t`^{2} ∈
`span`{`p`_{1}, `p`_{2}, `p`_{4}}.