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 $28.10$