Calculating a dimension What is the dimension of $$Z(F):=\{v\in V:\forall\alpha\in F:\alpha(v)=0\},$$ where $V$ is a finite-dimensional vector space and $F\subset V^\ast$ a subspace.
Maybe it is $\dim V-\dim F,$ but I am not sure and can't finde a proof.
 A: Yes that's true. Let's denote the space of linear functionals on $V^{\ast}$ that send $F$ to 0 with $F^o$
With the isomorphism $i_V\colon V\to V^{\ast\ast}, i_V(v)(\phi)=\phi(v)$ one sees $i_V(Z(F)=F^o$ because $F^o=\{\psi\in V^{\ast\ast} \colon \psi(\phi)=0 \forall \phi\in F\}=\{i_V(v)\colon v\in V, i_V(v)(\phi)=0\forall \phi\in F\}$
Now use $dim(V)=dim(F)+dim(F^o)$ and you should get the desired result.
For the latter equality: that follows from the kernel-range-dimension formula when looking at the surjective(!) dual map $\alpha^\ast$ induced by the inclusion map $\alpha\colon F \to V$ and identify the kernel of $\alpha^\ast$ with $F^o$.
A: Let $\{b_1^*, \ldots, b_k^*\}$ be the basis for the subspace $F$ and let $\{b_1^*, \ldots, b_n^*\}$ be its extension to a basis for $V^*$.
Denote $\{b_1, \ldots, b_n\}$ the basis for $V$ such that $\{b_1^*, \ldots, b_n^*\}$ is its dual basis.
We claim that $Z(F) = \operatorname{span}\{b_{k+1}, \ldots, b_n\}$.
Clearly $\operatorname{span}\{b_{k+1}, \ldots, b_n\} \subseteq Z(F)$ because $b_1^*, \ldots, b_k^*$ annihilate $b_{k+1}, \ldots, b_n$ by definition, so they also annihilate $\operatorname{span}\{b_{k+1}, \ldots, b_n\}$ by linearity.
Conversely, let $x \in Z(F)$ and let $x = \sum_{i=1}^n \alpha_ib_i$.
For any $j = 1, \ldots, k$ we have
$$0 = b_j^*(x) = b_j^*\left(\sum_{i=1}^n \alpha_ib_i\right) = \alpha_j$$
Hence $x = \sum_{i=k+1}^n \alpha_ib_i \in  \operatorname{span}\{b_{k+1}, \ldots, b_n\}$. 
We conclude $Z(F) = \operatorname{span}\{b_{k+1}, \ldots, b_n\}$ so $$\dim Z(F) = n - k$$
since $\{b_{k+1}, \ldots, b_n\}$ is linearly independent.
