I am revising linear algebra and am stuck on a question. I want to show that if $v \notin U$ then there is some $f \in U^0$ such that $f(v) \neq 0$.
I am confused about how to go about proving this, and what constraints on the annihilator it follows from.
I was thinking it could be proven by contradiction, if you begin with the assumption that $v \notin U$, but then take the statement that for all $f \in U^0$ $f(v)= 0$. But this doesn't seem to lead me anywhere, as I don't think it follows from here therefore that $v \in U$.
Any help appreciated, thank you.
I was just doing this question and I came up with this solution. Note that I'm assuming this is the same question as the one I saw. Along with the statement from the OP, I was also given that $U$ is a subspace of a finite dimensional vector space $V$.
Consider the following:
Let $B = \{e_1, e_2, ...,e_k\}$ be that basis of $U$. Now extend this basis to the basis $B'$ of $V$ as $B' = \{e_1, e_2, ..., e_k, e_{k+1}, ..., e_n\}$.
Now define a functional say $f$ s.t $f(e_i)=0$ if $i \neq k+1$ and $f(e_i) = 1$ if $i = k+1$.
Now $\forall u \in U, f(u) = 0$ because $u \in span(B)$. Thus, $f \in U^0$ but $f(e_{k+1}) = 1$ and since $e_{k+1} \not\in U$ therefore the statement follows.