# span

The *span* of a set of vectors ${\overrightarrow{\mathbf{v}}}_{1},\mathrm{\dots},\overrightarrow{{\mathbf{v}}_{\mathbf{n}}}$ of a vector space^{} $V$ over a field $K$ is the set of linear combinations^{} ${a}_{1}{\overrightarrow{\mathbf{v}}}_{1}+\mathrm{\dots}+{a}_{n}{\overrightarrow{\mathbf{v}}}_{n}$ with ${a}_{i}\in K$. It is denoted $\mathrm{Sp}({\overrightarrow{\mathbf{v}}}_{1},\mathrm{\dots},{\overrightarrow{\mathbf{v}}}_{n})$. More generally, the span of a set $S$ (not necessarily finite) of vectors is the collection of all (finite) linear combinations of elements of $S$. The span of the empty set is defined to be the singleton consisting of the zero vector $\overrightarrow{\mathrm{\U0001d7ce}}$.

For example, the standard basis vectors $\widehat{\u0131}$ and $\widehat{\u0237}$ span ${\mathbb{R}}^{2}$ because every vector of ${\mathbb{R}}^{2}$ can be represented as a linear combination of $\widehat{\u0131}$ and $\widehat{\u0237}$.

$\mathrm{Sp}({\overrightarrow{\mathbf{v}}}_{1},\mathrm{\dots},{\overrightarrow{\mathbf{v}}}_{n})$ is a subspace^{} of $V$ and is the smallest subspace containing ${\overrightarrow{\mathbf{v}}}_{1},\mathrm{\dots},{\overrightarrow{\mathbf{v}}}_{n}$.

*Span* is both a noun and a verb; a *set of vectors* can span a vector space, and a vector can be *in the span* of a set of vectors.

Checking span: To see whether a vector is *in the span* of other vectors, one can set up an augmented matrix, since if $\overrightarrow{\mathbf{u}}$ is in the span of ${\overrightarrow{\mathbf{v}}}_{1},{\overrightarrow{\mathbf{v}}}_{2}$, then $\overrightarrow{\mathbf{u}}={x}_{1}{\overrightarrow{\mathbf{v}}}_{1}+{x}_{2}{\overrightarrow{\mathbf{v}}}_{2}$. This is a system of linear equations. Thus, if it has a solution, $\overrightarrow{\mathbf{u}}$ is in the span of ${\overrightarrow{\mathbf{v}}}_{1},{\overrightarrow{\mathbf{v}}}_{2}$. Note that the solution does not have to be unique for $\overrightarrow{\mathbf{u}}$ to be in the span.

To see whether a set of vectors *spans* a vector space, you need to check that there are at least as many linearly independent^{} vectors as the dimension^{} of the space. For example, it can be shown that in ${\mathbb{R}}^{n}$, $n+1$ vectors are never linearly independent, and $n-1$ vectors never span.

Remark. We can define the concept of span also for a module $M$ over a ring $R$. Given a subset $X\subset M$ we define the module generated by $X$ as the set of all finite linear combinations of elements of $X$. Be aware that in general there does not exist a linearly independent subset which generates the whole module, i.e. there does not have to exist a basis. Also, even if $M$ is generated by $n$ elements, it is in general not true that any other set of $n$ linearly independent elements of $M$ spans $M$. For example $\mathbb{Z}$ is generated by $1$ as a $\mathbb{Z}$-module but not by $2$.

Title | span |

Canonical name | Span |

Date of creation | 2013-03-22 11:58:18 |

Last modified on | 2013-03-22 11:58:18 |

Owner | mathwizard (128) |

Last modified by | mathwizard (128) |

Numerical id | 22 |

Author | mathwizard (128) |

Entry type | Definition |

Classification | msc 16D10 |

Classification | msc 15A03 |

Synonym | linear span |

Related topic | LinearCombination |

Related topic | Basis |

Related topic | ProofOfGramSchmidtOrthogonalizationProcedure |

Related topic | FinitelyGeneratedRModule |

Defines | spanning set |