I am working on problem 3 from the exercises following the section on Annihilators in the text "Finite Dimensional Vector Spaces".
Problem: Prove that if $y$ is a linear functional on an $n$-dimensional vector space $V$, then the set of all those vectors for which $[x,y]=0$ is a subspace of $V$; what is the dimension of that subspace.
Showing that it is a subspace was not terribly difficult, given $\chi_{1},\chi_{2}\in V$, such that $[\chi_{j},y]=0$, $j=1,2$ it follows from the linearity of the functional that $$[\beta_{1}\chi_{1}+\beta_{2}\chi_{2},y]=\beta_{1}[\chi_{1},y]+\beta_{2}[\chi_{2},y]=0$$
For all $(\beta_{1},\beta_{2})\in\mathbb{C}^{2}$
Now showing the dimension of the subspace has been more difficult. A few sections previous to this question it was introduced the concept of a dual space for which the set of linear functionals over an $n$-dimensional vector space has an $n$-dimensional basis $\lbrace y_{j}\rbrace $. Where $\left[x_{i},y_{j}\right]=\delta_{ij}$, $\lbrace x_{i}\rbrace$ being an $n$-dimensional basis for $V$. My intuition tells me that if a subspace of $V$ vanishes over a subspace of the dual space $V'$ then the dimension of the subspace of $V$ will be equal to that of the dimension of the subspace of the dual. If anyone has any hints on how to proceed I would greatly appreciate it.