This is how I understand the annihilator now, but I feel like it might be incorrect.
So for some $U \subset V$, the annihilator of $U$ is all of the linear functionals $t(v)$ in $V'$, such that $t(u)=0$ for $u \in U$.
So in other words, I understand it as the following: You pick the subspace U that you want to annihilate, then the annihilator of U is simply the set of linear functions that map each vector in U to zero?
Notation: $V'$ is the set of all linear maps that map vectors in $V$ from $V$ to $F$
Best Answer
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=\langle t,u\rangle$, the annihilator of $U$ consists of all $t \in V'$ such that $\langle t,u \rangle =0$ for all $u \in U$.