Dimension of Annihilator on Algebraic Dual
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Theorem
Let $G$ be an $n$-dimensional vector space over a field.
Let $G^*$ be the algebraic dual of $G$.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $M^\circ$ be the annihilator of $M$.
Then:
- $M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$.
Proof
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:
\(\ds \forall j \in \closedint 1 m: \, \) | \(\ds \lambda_j\) | \(=\) | \(\ds \map {\sum_{k \mathop = 1}^n \lambda_k {a_k}'} {a_j}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\sum_{k \mathop = 1}^n \lambda_k {a_k}'} } {a_j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map t {a_j}\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
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 $n - m$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$