# linearly independent

Let $V$ be a vector space^{} over a
field $F$. We say that ${v}_{1},\mathrm{\dots},{v}_{k}\in V$ are *linearly dependent* if there exist scalars ${\lambda}_{1},\mathrm{\dots},{\lambda}_{k}\in F$, not all zero, such that

$${\lambda}_{1}{v}_{1}+\mathrm{\cdots}+{\lambda}_{k}{v}_{k}=0.$$ |

If no such scalars exist, then we say that the vectors are *linearly independent*.
More generally, we say that a (possibly infinite) subset $S\subset V$ is linearly independent if all finite subsets of $S$ are linearly independent.

In the case of two vectors, linear dependence means that one of the vectors is a scalar multiple of the other. As an alternate characterization of dependence, we also have the following.

###### Proposition 1.

Let $S\mathrm{\subset}V$ be a subset of a vector space. Then, $S$ is
linearly dependent if and only if there exists a $v\mathrm{\in}S$ such that
$v$ can be expressed as a linear combination^{} of the vectors in the
set $S\mathrm{\backslash}\mathrm{\{}v\mathrm{\}}$ (all the vectors in $S$ other
than $v$ (http://planetmath.org/SetDifference)).

Remark. Linear independence can be defined more generally for modules over rings: if $M$ is a (left) module over a ring $R$. A subset $S$ of $M$ is linearly independent if whenever ${r}_{1}{m}_{1}+\mathrm{\cdots}+{r}_{n}{m}_{n}=0$ for ${r}_{i}\in R$ and ${m}_{i}\in M$, then ${r}_{1}=\mathrm{\cdots}={r}_{n}=0$.

Title | linearly independent |
---|---|

Canonical name | LinearlyIndependent |

Date of creation | 2013-03-22 11:58:40 |

Last modified on | 2013-03-22 11:58:40 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 30 |

Author | rmilson (146) |

Entry type | Definition |

Classification | msc 15A03 |

Synonym | linear independence |

Defines | linearly dependent |

Defines | linear dependence |