Results Concerning Annihilator of Vector Subspace

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

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

Let $$N$$ be a $p$-dimensional subspace of $$G^*$$.

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

Then:
 * $$(1) \quad M^\circ$$ is an $$\left({n - m}\right)$$-dimensional subspace of $$G^*$$, and $$M^{\circ \circ} = J \left({M}\right)$$


 * $$(2) \quad J^{-1} \left({N^\circ}\right)$$ is an $$\left({n - p}\right)$$-dimensional subspace of $$G$$


 * $$(3)$$    The mapping $$M \to M^\circ$$ is a bijection from the set of all $$m$$-dimensional subspaces of $$G$$ onto the set of all $$\left({n - m}\right)$$-dimensional subspaces of $$G^*$$


 * $$(4)$$    Its inverse is the bijection $$N \to J^{-1} \left({N^\circ}\right)$$.

Proof

 * $$(1) \quad M^\circ$$ is an $$\left({n - m}\right)$$-dimensional subspace of $$G^*$$, and $$M^{\circ \circ} = J \left({M}\right)$$

Let $$\left \langle {a_n} \right \rangle$$ be an ordered basis of $$G$$ such that $$\left \langle {a_m} \right \rangle$$ is an ordered basis of $$M$$.

Let $$\left \langle {a'_n} \right \rangle$$ be the ordered dual basis of $$G^*$$.

Let $$t' = \sum_{k=1}^n \lambda_k a'_k \in M^\circ$$.

Then:

$$ $$ $$ $$

So $$t'$$ is a linear combination of $$\left\{{a'_k: m + 1 \le k \le n}\right\}$$.

But $$a'_k$$ clearly belongs to $$M^\circ$$ for each $$k \in \left[{{m + 1} \,. \, . \, n}\right]$$.

Therefore $$M^\circ$$ has dimension $$n - m$$.

When we apply this result to $$M^\circ$$ instead of $$M$$, it is seen that the annihilator $$M^{\circ \circ}$$ of $$M^\circ$$ has dimension $$n - \left({n - m}\right) = m$$.

But clearly $$J \left({M}\right) \subseteq M^{\circ \circ}$$.

As $$J$$ is an isomorphism, $$J \left({M}\right)$$ has dimension $$m$$.

So by Dimension of Proper Subspace Less Than its Superspace, $$J \left({M}\right) = M^{\circ \circ}$$.

As a consequence, $$J^{-1} \left({M^{\circ \circ}}\right) = M$$.

Hence the result: $$M^{\circ \circ} = J \left({M}\right)$$


 * $$(2) \quad J^{-1} \left({N^\circ}\right)$$ is an $$\left({n - p}\right)$$-dimensional subspace of $$G$$

If $$N$$ is a $$p$$-dimensional subspace of $$G$$, then $$N^\circ$$ and hence also $$J^{-1} \left({N^\circ}\right)$$ have dimension $$n - p$$ by what has just been proved.

By definition: $$\left({J^{-1} \left({N^\circ}\right)}\right)^\circ = \left\{{z' \in G: \forall x \in G: \forall t' \in N: t' \left({x}\right) = 0: z' \left({x}\right) = 0}\right\}$$

Thus $$N \subseteq \left({J^{-1} \left({N^\circ}\right)}\right)^\circ$$.

But as $$\left({J^{-1} \left({N^\circ}\right)}\right)^\circ$$ has dimension $$n - \left({n - p}\right) = p$$, it follows that $$N = \left({J^{-1} \left({N^\circ}\right)}\right)^\circ$$ by Dimension of Proper Subspace Less Than its Superspace.


 * $$(4)$$    Its inverse is the bijection $$N \to J^{-1} \left({N^\circ}\right)$$.

The final assertion follows by the definition of the inverse of a bijection.