Question regarding annihilators of subsets (not subspaces) of a vector space

Question

I have a question regarding annihilators. The following theorem is from "Finite Dimensional Vector Spaces" by Halmos:

If $\mathbb{M}$ is an m-dimensional subspace of an n-dimensional vector space $\mathbb{V}$, then $\mathbb{M}^{0}$ is an ($n-m$)-dimensional subspace of $\mathbb{V}'$.

This theorem is pretty easy to imagine. For example if I take $\mathbb{R}^2$ and a 1-dimensional subspace of $\mathbb{V}$ then I get that $\mathbb{M}^{0}$ is a 1-dimensional subspace of $\mathbb{V}'$. This subspace contains the linear functional which also contains the before mentioned subspace of $\mathbb{V}$ which is equivalent to the linear functional evaluated to zero.

Since annihilators are defined on subsets rather then subspaces my question is - given a subset can we obtain some information from it about its annihilator? For example if I have a line in $\mathbb{R}^2$ which doesn't go through the origin, how does its annihilator look like? From what I understand the annihilator is empty. Let's say I have a non-zero vector in $\mathbb{R}^2$ then there is obviously a non empty annihilator. When I have for example the vector \begin{pmatrix}3\\3\end{pmatrix} then I could define the linear functional that annihilates the vector to be $y(x)=1*x_{1}-1*x_{2}$. So basically my question is - what can we say about the annihilator knowing how the subset looks like which we are observing.

Thanks for any help!

Answer 1

Annihilators of subsets are annihilators of subspaces: a functional $f$ annihilates $X$ iff it annihilates the subspace comprising every vector of the form $a_1x_1 + \ldots a_nx_n$ where $a_i$ are scalars and $x_i \in X$. In the usual jargon, the annihilator of $X$ is the same as the annihilator of the subspace spanned by $X$. So for example, if you have a line in $\Bbb{R}^2$ that does not go through the origin, it spans all of $\Bbb{R}^2$ and its annihilator is the same as the annihilator of $\Bbb{R}^2$, i.e., the set $\{0\}$ comprising the $0$ functional (not the empty set).