Prove there is $a_0\in L$ s.t. $\phi(a_0)=1$

Question

Let $L$ be a linear space over $F$ and $\phi$ a linear functional s.t. $\phi \neq0$.

Prove there is $a_0\in L$ s.t. $\phi(a_0)=1$.

My try:

Suppose $dimL=n+1$ and let $a_0\in L$. Let's complete $a_0$ to a basis of $L$: $B_L=\{a_0,a_1,\dots,a_n\}$ and $\phi:=\phi_0$ to a basis of $L^*$: $B_{L^*}=\{\phi_0,\phi_1,\dots,\phi_n\}$ such that $B_{L^*}$ is the dual basis of $B_L$. (I'm not sure I can do that, do I need to prove it? If so, how?).

That is $\phi_i(a_j)=\delta_{ij}$, hence $\phi_0(a_0)=1$ and $\phi(a_0)=1 $.

Is my proof ok?

Answer 1

There is $a \in L$ such that $\phi(a) \ne 0$. Put $a_0=\frac{1}{\phi(a)}a$.

Then:

$$\phi(a_0)=1.$$