Non-degenerate bilinear map induces isomorphism

Question

Let $V$ be a finite dimensional vector space. Let $\alpha :V\times V \longrightarrow \mathbb{F}$ be a non-degenerate bilinear form.

Let $$f:V \longrightarrow V^*$$ $$f(u)(v) = \alpha(u,v)$$

clearly $f$ is linear and injective, but how do I show that $f$ is surjective to show that it is an isomorphism

Answer 1

For maps between isomorphic spaces of finite dimension, injective $\implies$ surjective. In particular, it suffices to take any basis $e_1,e_2,\dots,e_n$ of $V$, and note that the dual vectors $f(e_1),f(e_2),\dots,f(e_n)$ are necessarily linearly independent.