# 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 $\dim V=\dim V^{\ast}$ 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 $\dim V=\dim\mathop{\mathrm{img}}S$. We now have $\mathop{\mathrm{img}}S\subset V^{*}$ and $\dim\mathop{\mathrm{img}}S=\dim V^{\ast}$. (The first assertion follows directly from the definition of $S$.) Hence $\mathop{\mathrm{img}}S=V^{\ast}$ (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

 $\displaystyle\dim V$ $\displaystyle=$ $\displaystyle\dim\ker T+\dim\mathop{\mathrm{img}}T.$ (1)

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

Title dimension   theorem for symplectic complement (proof) DimensionTheoremForSymplecticComplementproof 2013-03-22 13:32:52 2013-03-22 13:32:52 matte (1858) matte (1858) 5 matte (1858) Proof msc 15A04