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?

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 $\text{dim} \, U'=\text{dim} \, U$.
