so basically I've got two "proofs"; one proving the equality in the title holds and the other disproving it but I can't seem to figure out where I'm going wrong. Here is the question and my two proofs:
Q: Let $\phi \in V^{\prime}$ be a nonzero linear functional acting on the finite-dimensional vector space $V$. prove or disprove $(null\, \phi)^0 = span(\phi)$.
proof the equality holds: $$\begin{aligned} dim\,(null \,\phi)^0 &= dim\, V -dim\, (null \ \phi) \\ &=dim \, range(\phi) \\ &=1 \end{aligned}$$ since $\phi \in (null \, \phi)^0 $ and $(null\, \phi)^0 $ is one-dimensional we conclude that $(null\, \phi)^0 = span(\phi)$.
proof the equality fails to hold:
Let $u_1 \ldots u_n $ be a basis of $null \, \phi$. We can extend it to become a basis of $V$. Let $u_1 \ldots u_n v_1 \ldots v_m$ be a basis of $V$. Now we define a linear functional $\psi$ by letting $\psi(u_i)=0$ for $1 \leq i \leq n$ and choosing $\psi(v_j)$ for $1 \leq j \leq m$ such that there is no scalar $\gamma$ such that $\psi(v_j)=\gamma \phi (v_j)$ for all $1 \leq j \leq m$. Hence we have $\psi\in (null\, \phi)^0$ but $\psi \not \in span(\phi)$ disproving $(null\, \phi)^0 = span(\phi)$.
So what am I missing here? Thanks in advance.