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

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