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 $G$

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

$(4) \quad$ 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 $\displaystyle t' = \sum_{k=1}^n \lambda_k a'_k \in M^\circ$.

Then:

	\(\displaystyle \)	\(\displaystyle \forall j \in \left[{1 \, . \, . \, m}\right]:\)	\(\displaystyle \)	\(\displaystyle \lambda_j\)	\(=\)	\(\displaystyle \sum_{k=1}^n \lambda_k a'_k \left({a_j}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\sum_{k=1}^n \lambda_k a'_k}\right) \left({a_j}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle t' \left({a_j}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $n - m$.

$\Box$

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

$\Box$

$(2) \quad J^{-1} \left({N^\circ}\right)$ 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.

$\Box$

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.

$(4) \quad$ 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.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 28$: Theorem $28.10$