TITLE: Dimension of Annihilator: $\text{dim} \, U^0 + \text{dim} \, U = \text{dim} \, V$
QUESTION [8 upvotes]: First there is a vector space $V$ and $U$ is vector subspace of $V$.
Furthermore, $U^{0}=\{\varphi \in V^{*} |\space\forall u \in U: \varphi(u) = 0\}$ is the annihilator of $U$.
I need to show that:
$$\text{dim} \, U^0 + \text{dim} \, U = \text{dim} \, V$$ 
Do I need the rank and nullity theorem?

REPLY [3 votes]: The idea behind this proof is motivated by category theory.
$\mathbf {Proof }:$ Now suppose $i:S\rightarrow V$ be the natural inclusion map i.e $i(s)=s,\; \forall s \in S$. Consider its dual  map $i':V'\rightarrow S'$ defined as follows $i'(\phi)=\phi\,\circ i$. Thus $i'$ is a linear map from $V'$ to $S'$. Then $dim\,range(i')+dim\,null(i')=dim\,V'=dim\,V$. Now my first claim is,
$\it{Claim \;1}:$ $null(i')=S^0.$
$\it{Pf}:$ If $\phi \in null(i')\Leftrightarrow i'(\phi)=0\Leftrightarrow \phi\,\circ i=0\Leftrightarrow(\phi\,\circ i)(s)=0,\,\forall s\in S\Leftrightarrow\phi(i(s))=0,\,\forall s\in S$ $\Leftrightarrow\phi(s)=0,\, \forall s\in S\Leftrightarrow \phi \in S^0\,\bullet$
Now we have $dim\,range(i')+dim\,S^0=dim\,V$. This means that  $dim\,range(i')$ must be equal to $dim\,S=dim\,S'$. But $range\,(i')\subseteq S'$. So it means that $i'$ must be surjective i.e $range(i')=S'$. Indeed,
$\it{Claim\;2}:$ $range(i')=S'$.
$\it{Pf}:$ Let $\phi \in S'$. So $\phi:S\rightarrow \Bbb F$. We extend $\phi $ to a functional $\psi :V\rightarrow \Bbb F$ such that $\psi(s)=\phi(s),\,\forall s\in S$. So $i'(\psi)=\psi\,\circ i$. But $(\psi\, \circ i)(s)=\psi(i(s))=\psi(s)=\phi(s),\,\forall s \in S$. Hence $\psi\, \circ i = \phi \Leftrightarrow i'(\psi)=\phi$. So $S'=range(i')\,\bullet$.
Finally we have $dim\,S'+dim\,S^0=dim\,V\Rightarrow dim\,S+dim\,S^0=dim\,V.\,\lhd$