Dimension of a null-space (Halmos)

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.

Show that if $y\not=0$ then the null space of $y$ has co-dimension $1$. That is, fix $z$ where $y(z)\not=0$. Observe that any $x$ is $az+w$ for some scalar $a$ and some $w$ satisfying $y(w)=0$. Let $a=y(x)/y(z)$ and $w=x-az$. Hence if $B$ is a vector-space basis for the null space of y, then $Bu\{z\}$ is a basis for $V$, so B has $n-1$ members.
• Let me reiterate. So if $y$ is a non-zero functional then there exists a vector $z\in V$ such that $[z,y]\neq 0$. That being the case you decompose take a $w,x$ which are not linear combinations of $z$ and show that those are in the kernel of the linear functional $y$ and are $n-1$ dimensional. – Tucker Aug 10 '15 at 0:18
The square bracket notation is rather unusual. Most mathematicians tend to use the triangular brackets for this pairing: $\langle x,y\rangle$, if $x\in V$ and $y\in V^*$. (The star notation is also the usual way to denote the dual space.)
Let me answer in the most general form; that is, when the base field is an arbitrary field $F$. To answer your question, the dimension of $\text{ker}(y)=\big\{x\in V\,\big|\,\langle x,y\rangle=0\big\}$ depends on $y$. If $y$ is the zero map, then the dimension of $\text{ker}(y)=V$ is clearly $\dim_F(V)=n$. If $y$ is nonzero, then $\text{im}(y)=\left\{\langle x,y\rangle\,\big|\,x\in V\right\}$ is $F$ itself. Now, we know that $$\dim_F\big(\text{ker}(y)\big)+\dim_F\big(\text{im}(y)\big)=\dim_F(V)=n\,.$$ As $\dim_F(F)=1$, we conclude that $\dim_F\big(\text{ker}(y)\big)=n-1$.
A next question you might answer yourself is: if $y_1,y_2\in V^*$, what is the dimension of $\text{ker}\left(y_1\right)\cap\text{ker}\left(y_2\right)$? There are three cases to consider.