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:

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

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

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

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

Let $\sequence {a_n}$ be an ordered basis of $G$ such that $\sequence {a_m}$ is an ordered basis of $M$.

Let $\sequence {a'_n}$ be the ordered dual basis of $G^*$.

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

Then:

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

But $a'_k$ clearly belongs to $M^\circ$ for each $k \in \closedint {m + 1} n$.

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 $n - \paren {n - m} = m$.

But clearly $\map J M \subseteq M^{\circ \circ}$.

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

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

As a consequence, $\map {J^{-1} } {M^{\circ \circ} } = M$.

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


 * $(2) \quad \map {J^{-1} } {N^\circ}$ 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.