Dimension of Annihilator: $\text{dim} \, U^0 + \text{dim} \, U = \text{dim} \, V$
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?
Solution 1:
Notation and Definitions
First, we need to know the following notations
- $V$ and $W$ are vector spaces over the field $\Bbb{F}$.
- $U$ is a subspace of the vector field $V$.
- $\mathcal{L}(V,W)$ denotes the vector space of all linear maps $T:V \to W$.
- $V'=\mathcal{L}(V,\Bbb{F})$ denotes the dual space of V.
- $T':W' \to V'$ is the dual map of $T$ defined by $T'(w')=w' \circ T$ where $\circ$ is the composition.
- $U^0=\{v' \in V': \forall u \in U, \, v'(u)=0\}$ denotes the annihilator of $U$.
Proof
Here is a proof by the fundamental theorem of linear maps (rank-nullity theorem). First define the following linear map
$$\begin{align} T : &U \to V \\ & u \to u \end{align}$$
then its dual map $T'$ will be
$$\begin{align} T' : &V' \to U' \\ & v' \to v' \circ T \end{align}$$
next, using definitions for $T$ and the dual space, we can write
$$\begin{align} T' : &\mathcal{L}(V,\Bbb{F}) \to \mathcal{L}(U,\Bbb{F}) \\ & v' \to v' \end{align}$$
so you can see that the only thing that $T'$ is doing is to change the domain of $v'$ from $V$ to $U$. This fact can help you to prove that $T'$ is surjective. More specifically, for proving the surjectivity, you should show that any linear map can be extended to a larger domain. Next, you can show the first result below by definition of null space and the second one by the fact I mentioned
$$\begin{align} \text{null} \, T' &= U^0 \\ \text{range} \, T' &= U' \end{align}$$
then apply the fundamental theorem of linear maps to $T'$ to get
$$\begin{align} \text{dim null} \, T' + \text{dim range} \, T'&= \text{dim} \, V' \\ \text{dim} \, U^0 + \text{dim} \, U'&= \text{dim} \, V' \\ \text{dim} \, U^0 + \text{dim} \, U &= \text{dim} \, V \end{align}$$
where the last step is achieved by knowing that $\text{dim} \, V'=\text{dim} \, V$ and $\text{dim} \, U'=\text{dim} \, U$.
Solution 2:
Assuming you are dealing with finite dimensional spaces, you can just use a dual basis argument : Suppose $\{v_1, v_2, \ldots, v_m\}$ is a basis for $U$, which can be extended to a basis $\{v_1, v_2, \ldots, v_m, v_{m+1}, \ldots, v_n\}$ for $V$.
Let $\{\varphi_1, \varphi_2, \ldots, \varphi_n\}$ be a dual basis for $V^{\ast}$, then $$ \{\varphi_{m+1}, \varphi_{m+2}, \ldots, \varphi_n\} \subset U^{\circ} $$ Now check that this set forms a basis for $U^{\circ}$
Solution 3:
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 $dim\,S'+dim\,S^0=dim\,V\Rightarrow dim\,S+dim\,S^0=dim\,V.\,\lhd$