# dimension theorem for symplectic complement (proof)

We denote by ${V}^{\star}$ the dual space^{} of $V$, i.e.,
linear mappings from $V$ to $\mathbb{R}$. Moreover, we assume known that
$dimV=dim{V}^{\ast}$ for any vector space^{} $V$.

We begin by showing that the
mapping $S:V\to {V}^{*}$, $a\mapsto \omega (a,\cdot )$
is an linear isomorphism. First, linearity is clear, and since
$\omega $ is non-degenerate, $\mathrm{ker}S=\{0\}$, so $S$ is injective.
To show that $S$ is surjective, we apply the
http://planetmath.org/node/2238rank-nullity theorem^{} to
$S$, which yields $dimV=dimimgS$.
We now have
$imgS\subset {V}^{*}$
and
$dimimgS=dim{V}^{\ast}$.
(The first assertion follows directly from the definition of $S$.)
Hence $imgS={V}^{\ast}$ (see this page (http://planetmath.org/VectorSubspace)),
and $S$ is a surjection. We have shown that
$S$ is a linear isomorphism.

Let us next define the mapping $T:V\to {W}^{*}$, $a\mapsto \omega (a,\cdot )$. Applying the http://planetmath.org/node/2238rank-nullity theorem to $T$ yields

$dimV$ | $=$ | $dim\mathrm{ker}T+dimimgT.$ | (1) |

Now $\mathrm{ker}T={W}^{\omega}$ and $imgT={W}^{*}$. To see the latter assertion, first note that from the definition of $T$, we have $imgT\subset {W}^{*}$. Since $S$ is a linear isomorphism, we also have $imgT\supset {W}^{*}$. Then, since $dimW=dim{W}^{*}$, the result follows from equation 1. $\mathrm{\square}$

Title | dimension^{} theorem for symplectic complement (proof) |
---|---|

Canonical name | DimensionTheoremForSymplecticComplementproof |

Date of creation | 2013-03-22 13:32:52 |

Last modified on | 2013-03-22 13:32:52 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 5 |

Author | matte (1858) |

Entry type | Proof |

Classification | msc 15A04 |