Mapping to Annihilator on Algebraic Dual is Bijection

Theorem
Let $G$ be an $n$-dimensional vector space over a field.

Let $J: G \to G^{**}$ be the evaluation isomorphism.

Let $G^*$ be the algebraic dual of $G$.

Let $M$ be an $m$-dimensional subspace of $G$.

Let $M^\circ$ be the annihilator of $M$.

Let $G_m$ denote the set of all $m$-dimensional subspaces of $G$.

Let ${G^*}_{n - m}$ denote the set of all $n - m$-dimensional subspaces of $G^*$.

Let $\phi: G_m \to {G^*}_{n - m}$ be the mapping from $G_m$ to the power set of ${G^*}_{n - m}$ defined as:
 * $\forall M \in \powerset G: \map \phi M = M^\circ$

Then $\phi$ is a bijection.

Proof
From Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism, we have that:
 * $M^{\circ \circ} = J \sqbrk M$

From Evaluation Isomorphism is Isomorphism, $J: G \to \G^{**}$ is a bijection.