dimension theorem for symplectic complement (proof)

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

We begin by showing that themapping $S:V\to V^{*}$, $a\mapsto \omega (a,\cdot )$is an linear isomorphism. First, linearity is clear, and since$\omega$ is non-degenerate, $\ker S=\{0\}$, so $S$ is injective.To show that $S$ is surjective, we apply thehttp://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\ker T+dimimgT.$(1)

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