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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?

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    Not following your argument. Since $\phi\neq 0$ there must be some vector $\vec v\in L $ such that $\phi(\vec v)\neq 0$. But $\forall \lambda \in F$ we have $\phi(\lambda \vec v) =\lambda \phi (\vec v)$ so... – 2017-02-22
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    @lulu I'm not getting you. Could you please try to explain? – 2017-02-22
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    If $\phi(\vec v)\neq 0$ then we can speak of $\frac 1{\phi(\vec v)}$ as an element of the field $F$. But then $\phi( \frac 1{\phi(\vec v)}\vec v)=\frac 1{\phi(\vec v)}\times \phi(\vec v)=1$. – 2017-02-22
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    @lulu I see it now. Could you please explain what's wrong with my proof? – 2017-02-22
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    I don't understand your proof. Rather, it appears circular...how do I know that I can complete $\phi$ to a basis of the dual space? That's more or less exactly the same as what you are asked to prove. – 2017-02-22

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

Then:

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

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    How simple. Could you please explain what's wrong with my proof? – 2017-02-22
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    You have assumed that $L$ is finite- dimensional. This must not be the case. – 2017-02-22
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    In my course we're dealing only with finite dimensional spaces. – 2017-02-22