Results Concerning Annihilator of Vector Subspace

Theorem

Let $G$ be an $n$-dimensional vector space over a field.

Let $G^*$ be the algebraic dual of $G$.

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

Let $J: G \to G^{**}$ be the evaluation isomorphism.

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 $M$.

Let $J^\gets: \powerset {G^{**} } \to \powerset G$ be the inverse image mapping of $J$.

Then the following results hold:

Dimension of Annihilator on Algebraic Dual

$M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$.

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

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

where $J \sqbrk M$ denotes the image of $M$ under $J$.

Dimension of Preimage under Evaluation Isomorphism of Annihilator on Algebraic Dual

$\map {J^\gets} {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$

where

Mapping to Annihilator on Algebraic Dual is Bijection

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.

Inverse of Mapping to Annihilator on Algebraic Dual is Bijection

The inverse of $\phi$ is the bijection:

$N \to \map {J^\gets} {N^\circ}$

Sources

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