Each bilinear form induces a unique bilinear form from the dual space

Question

Let $V$ be a finite dimensional vector space over a field $K$.

Let $\gamma: V \times V \to K$ be a nondegenerate bilinear form.

I now want to show that there exists one and only one bilinear form $\overline{\gamma}: V^* \times V^* \to K$ that satisfies the property $\gamma(x, y) = \overline{\gamma}(\Gamma_\gamma(x), \Gamma_\gamma(y))$ where $\Gamma_\gamma$ is defined as the linear transformation: $V \to V^*, w \mapsto (v \mapsto \gamma(v, w))$. Also, why is $\overline{\gamma}$ nondegenerate?

Thanks in advance. I'm not very used to working with these constructions yet.

Answer 1

It seems to me that $\bar{\gamma}$ is defined by the inverse of $\Gamma_{\gamma}$ as follows: $$ \bar{\gamma}(\alpha,\beta) := \beta(\Gamma^{-1}_{\gamma}(\alpha))$$ where $\alpha,\beta \in V^*$.