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 rank-nullity theorem to $S$ , which yields $\dim V = \dim \image S$ . We now have $\image S \subset V^*$ and $\dim \image S = \dim V^\ast$ . (The first assertion follows directly from the definition of $S$ .) Hence $\image S = V^\ast$ (see this page), 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 rank-nullity theorem to $T$ yields \begin{eqnarray} \label{eq0} \dim V &=& \dim \ker T + \dim \image T. \end{eqnarray}Now $\ker T = W^\omega$ and $ \image T = W^*$ . To see the latter assertion, first note that from the definition of $T$ , we have $\image T \subset W^*$ . Since $S$ is a linear isomorphism, we also have $\image T \supset W^*$ . Then, since $\dim W= \dim W^*$ , the result follows from equation .