Let $\textsf{V}$ be a finite dimensional vector space and $\textsf{W}$ a subspace of $\textsf{V}$. Then $\operatorname{Ann}(\textsf{W})$, the annihilator of $\textsf{W}$) is a subspace of the dual space $\operatorname{Hom}(\textsf{V},F)$ and further
$$\dim \operatorname{Ann}(\textsf{W}) = \dim \textsf{V}-\dim \textsf{W}$$
My Proof : Let $\{w_1,\dots,w_m\}$ be a basis for $\textsf{W}$ such that $m<\dim \textsf{W}$.
Extend this set to a basis $\{w_1,w_2,\dots,w_m,w_{m+1},\dots,w_n\}$ for $\textsf{V}$ and let $\{w_1^*,\dots,w_n^*\}$ be the corresponding basis for $\operatorname{Hom}(\textsf{V},F)$.
Observe that $\{w_{m+1},\dots,w_n\}$ is a subset of linearly independent vectors in $\operatorname{Ann}(\textsf{W})$. Let $f\in\operatorname{Ann}(\textsf{W})$ be arbitrary. Then observe that
$$f=\sum^n_{i=1}\lambda_iw_i^*$$
Then for each $j=1,2,3,\dots,m$ we have $f(w_j)=\lambda_j=0$.
Consequently,
$$f= \sum^{n}_{i=m+1}\lambda_iw^*_i$$
so they do indeed span the annihilator.
Remark : $(1)$ I have shown that the annihilator is indeed a subspace.
$(2)$ I have removed the case where $m=\dim \textsf V$ intentionally.
Is my proof correct?
Best Answer
Sounds alright. You could note that $\hom(V,F) \simeq V$ implies $\dim V = \dim \hom(V,F)$, since you are using it to conclude.
Here's another proof: take $W \leq V$. We can always find $S$ such that $V = W \oplus S$. Now, consider the linear mapping
$$ \Gamma : \varphi \in \operatorname{Ann}(W) \mapsto \varphi|_S \in \hom(S,F). $$
This is surjective, as given $\psi : S \to F$ we can define $\widetilde{\psi}(w + s) = \psi(s)$ which is an element of $\operatorname{Ann}(W)$ such that $\Gamma(\widetilde{\psi}) = \psi$. It is also injective: if $\varphi|_S = 0$ for some $\varphi$ in the annihilator of $W$, then $\varphi = 0$, as we already know that $\varphi|_W = 0$ and $V = W \oplus S$.
The former shows that $\Gamma$ is an isomorpshism. Finally, since $\hom(S,F) \simeq S$, we obtain that
$$ \operatorname{Ann}(W) \simeq \hom(S,F) \simeq S $$
and thus
$$ \dim \operatorname{Ann}(W) = \dim S = \dim V - \dim W. $$