# theorem for the direct sum of finite dimensional vector spaces

Theorem
Let $S$ and $T$ be subspaces^{} of a finite dimensional vector space^{}
$V$. Then $V$ is the direct sum^{} of $S$ and $T$, i.e., $V=S\oplus T$,
if and only if $dimV=dimS+dimT$ and $S\cap T=\{0\}$.

*Proof.* Suppose that $V=S\oplus T$. Then, by definition,
$V=S+T$ and $S\cap T=\{0\}$.
The dimension theorem for subspaces states that

$$dim(S+T)+dimS\cap T=dimS+dimT.$$ |

Since the dimension^{} of the zero vector space $\{0\}$
is zero, we have that

$$dimV=dimS+dimT,$$ |

and the first direction of the claim follows.

For the other direction, suppose $dimV=dimS+dimT$ and $S\cap T=\{0\}$. Then the dimension theorem theorem for subspaces implies that

$$dim(S+T)=dimV.$$ |

Now $S+T$ is a subspace of $V$ with the same dimension as $V$ so, by Theorem 1 on this page (http://planetmath.org/VectorSubspace), $V=S+T$. This proves the second direction. $\mathrm{\square}$

Title | theorem for the direct sum of finite dimensional vector spaces |
---|---|

Canonical name | TheoremForTheDirectSumOfFiniteDimensionalVectorSpaces |

Date of creation | 2013-03-22 13:36:17 |

Last modified on | 2013-03-22 13:36:17 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 8 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 15A03 |