Mapping to Annihilator on Algebraic Dual is Bijection

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

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

Let $G_m$ denote the set of all $m$-dimensional subspaces of $G$.

Let ${G^*}_{n - m}$ denote the set of all $n - m$-dimensional subspaces of $G^*$.

Let $\phi: G_m \to {G^*}_{n - m}$ be the mapping from $G_m$ to the power set of ${G^*}_{n - m}$ defined as:

$\forall M \in \powerset G: \map \phi M = M^\circ$

Then $\phi$ is a bijection.

Proof

From Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism, we have that:

$M^{\circ \circ} = J \sqbrk M$

From Evaluation Isomorphism is Isomorphism, $J: G \to G^{**}$ is a bijection.

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Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$