Lemma (Zero vector) Let $X$ be a finite dimensional vector space. If $x_0\in X$ has the property that $f(x_0)=0$ for all $f\in X^{*}$ (algebric dual), then $x_0=0.$
Proof. Let $\{e_1,\dots, e_n\}$ be a basis for $X$ and $x_0=\sum\eta_je_j$. Then $$f(x_0)=\sum_{j=1}^n\eta_jf(e_j)=\sum_{j=1}^n\eta_j\alpha_j,\quad\text{where}\;f(e_j):=\alpha_j$$
By assuntion this is zero for every $f\in X^*$, that is $\color{RED}{\text{for every choice of $\alpha_1,\dots, \alpha_n$}}.$
Why it is true the red phrase?
I think that, whenever choose $\alpha_1,\dots,\alpha_n$ this determines a linear functional on $X$, it suffies define that $\alpha_1=f(e_1),\dots, \alpha_n=f(e_n).$
It is true?
Yes, it is exactly that. Every $f(e_j)$ corresponds to an $a_j$. Thus, since $f(x_0) = 0$ for any such $f$, the yielded sum should also be equal to zero.