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 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, $\ker S=\{0\}$, 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

$dimV$ $=$ $dim\ker T+dimimgT.$ (1)

Now $\ker T=W^{\omega }$ and $imgT=W^{*}$. To see the latter assertion, first note that from the definition of $T$, 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