Let $V$ and $W$ be finite dimensional vector spaces. Let $b:V \times W \to \mathbb{R}$ a bilinear map satisfying:
- $\forall v \in V. (\forall w \in W. b(v,w)=0) \implies v=0$,
- $\forall w \in W. (\forall v \in V. b(v,w)=0) \implies w=0$.
Show an induced $\phi:V \to W^*, \phi(v)=b(v,.)$ is surjective.
The results I have established so far in solving this problem are:
a) A bilinear map $b$ can be split into linear maps. $b(v,w)=\gamma(v).\delta(w), \gamma:V \to \mathbb{R}, \delta:W \to \mathbb{R}$.
b) Let $B_V=\{v_1, ..., v_n\}$ be the basis of $V$. Let $B_W=\{w_1, ..., w_m\}$ be the basis of $W$. Then $1$ and $2$ hold on $b$ iff $\forall i \forall j. b(v_i, w_j)\neq 0$.
c) $\phi$ surjective iff $\forall \beta \in W^* \exists v \in V \forall w_i \forall w_j \frac{\beta(w_i)}{b(v,w_i)}=\frac{\beta(w_j)}{b(v,w_j)}$.
Point 1 shows that the induced map $V \to W^{*}$ is injective. Therefore $$\dim V \leq \dim W^{*} = \dim W.$$
Point 2 shows that the induced map $W \to V^{*}$ is injective, and similarly, $$\dim W \leq \dim V^{*} = \dim V.$$
Thus $\dim V = \dim W = \dim W^{*}$. Since the map $V \to W^{*}$ is injective, it must also be surjective.