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 the following results hold:

Dimension of Preimage under Evaluation Isomorphism of Annihilator on Algebraic Dual

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

Mapping to Annihilator on Algebraic Dual is Bijection

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

Inverse of Mapping to Annihilator on Algebraic Dual is Bijection

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

Proof

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

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

By definition: $\paren {\map {J^{-1} } {N^\circ} }^\circ = \set {z' \in G: \forall x \in G: \forall t' \in N: \map {t'} x = 0: \map {z'} x = 0}$

Thus $N \subseteq \paren {\map {J^{-1} } {N^\circ} }^\circ$.

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


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

The final assertion follows by the definition of an inverse mapping.