How do I show that if $f(v)=0$ for all $f\in U^0$ then $v\in U$?
I know there are other answers on this but they fall short or I don’t understand them, in particular because most don’t prove linearity of the functional they create.
I tried extending a basis for $U$ but to no avail. I’m stuck!
Let $V$ be a finite dimensional vector space over $K$ and $U$ a subspace of $V$.
Let $\{u_1,\ldots,u_r\}$ be a basis of $U$ if $v\notin U$ then $\{v,u_1,\ldots,u_r\}$ is a linearly independent set and thus we can extend it to a basis of $V$, let this new basis be $\{v,u_1,\ldots,u_r,w_1,\ldots,w_m\}$ now we define $T\colon V\to K$ as follows \begin{align*} T\left(av+\sum_{i=1}^ra_iu_i+\sum_{i=1}^mb_iw_i\right)=a+\sum_{i=1}^mb_i \end{align*}
$T$ is well defined because $\{v,u_1,\ldots,u_r,w_1,\ldots,w_m\}$ is a basis of $V$, now we notice that $T$ is linear since \begin{align*} &T\left(\lambda(av+\sum_{i=1}^ra_iu_i+\sum_{i=1}^mb_iw_i)+(bv+\sum_{i=1}^r\alpha_iu_i+\sum_{i=1}^m\beta_iw_i)\right) \\ =&T\left((\lambda a+b)v+\sum_{i=1}^r(\lambda a_i+\alpha_i)u_i+\sum_{i=1}^m(\lambda b_i+\beta_i)w_i\right) \\ =& \lambda a+b+\sum_{i=1}^m(\lambda b_i+\beta_i) \\ =&\left(\lambda(a+\sum_{i=1}^mb_i)\right)+\left(b+\sum_{i=1}^m\beta_i\right) \\ =& \lambda T\left(av+\sum_{i=1}^ra_iu_i+\sum_{i=0}^mb_iw_i\right)+T\left(bv+\sum_{i=1}^r\alpha_iu_i+\sum_{i=0}^m\beta_iw_i\right). \end{align*}
We also notice that $T\in U^0$ since given $u=\sum_{i=1}^ra_iu_i\in U$ we have $$ T(u)=T\left(0v+\sum_{i=1}^ra_iu_i+\sum_{i=1}^m0w_i=0\right)=0+\sum_{i=1}^m0=0. $$ Finally, it's clear that $T(v)=1$ and thus there's $f\in U^0$ such that $f(v)\not=0$.