Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism

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

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

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

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

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

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

Then:
 * $M^{\circ \circ} = J \sqbrk M$

where $J \sqbrk M$ denotes the image of $M$ under $J$.

Proof
Let Dimension of Annihilator on Algebraic Dual be applied to $M^\circ$ instead of $M$.

It is seen that the annihilator $M^{\circ \circ}$ of $M^\circ$ has dimension $n - \paren {n - m} = m$.

But clearly:
 * $J \sqbrk M \subseteq M^{\circ \circ}$.

As $J$ is an isomorphism, $J \sqbrk M$ has dimension $m$.

So by Dimension of Proper Subspace is Less Than its Superspace:
 * $J \sqbrk M = M^{\circ \circ}$.

As a consequence:
 * $\map {J^\gets} {M^{\circ \circ} } = M$

Hence the result:
 * $M^{\circ \circ} = J \sqbrk M$