Is the following Proof Correct? I am confident of the logic of the argument i am merely unsure whether the level of detail is sufficient.
I had asked for a similar critique here Existence of Non-Zero Annihilator but now i have re-written the argument slightly since i am coming back to it after some time and could not retrace my previous line of thought.
Theorem. Given that $V$ is finite dimensional and $U$ is a subspace of $V$ such that $U\neq V$. Then there exists a linear function $\phi:V\to \mathbf{F}$ such that $\forall u\in U(\phi u = 0)$ but $\phi\neq 0$.
Proof. Since $U\neq V$ it follows by theorem $\textbf{3.106}$ that $\dim V - \dim U = \dim U^0 >0$, equivalently the basis of $U^0$ will consist of at least one linear functional consequently choosing any one of the linear functionals constituting the basis of $U^0$ can serve as the required linear functional.
$\blacksquare$
NOTE: $U^0$ denotes the annihilator of the subspace $U$.
You are complicating things. why use annihilators? Take a basis for U and extend it to a basis for V. You can define a linear functional by by giving arbitrary values to basis elements and extending by linearity. Simply give the value 0 to the basis elements in U and, say, 1 to the remaining.