# proof of the dimension theorem for subspaces

Let $S$ and $T$ be subspaces^{} of a vector space^{}.
By the rank-nullity theorem^{} and the second isomorphism theorem (for modules)
we have

$\mathrm{dim}(S+T)$ | $=\mathrm{dim}S+\mathrm{dim}((S+T)/S)$ | ||

$=\mathrm{dim}S+\mathrm{dim}(T/(S\cap T)).$ |

Therefore

$\mathrm{dim}(S+T)+\mathrm{dim}(S\cap T)$ | $=\mathrm{dim}S+\mathrm{dim}(T/(S\cap T))+\mathrm{dim}(S\cap T)$ | ||

$=\mathrm{dim}S+\mathrm{dim}T,$ |

by the rank-nullity theorem again.

Title | proof of the dimension theorem for subspaces |
---|---|

Canonical name | ProofOfTheDimensionTheoremForSubspaces |

Date of creation | 2013-03-22 16:35:17 |

Last modified on | 2013-03-22 16:35:17 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 5 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 15A03 |