Dual vector space properties ($\forall \phi\in E^*,\langle\phi,x\rangle=0\implies x=0$) $E$ is a finite-dimensional vector space over a field K and $E^∗$ its dual.
$$\boxed{x\in E, x=0\iff \forall \phi\in E^*,\langle\phi,x\rangle=0}$$
The proof for ($\implies$) seems to be obvious since $x=0\implies\langle\phi,x\rangle=\phi(x)=0$
But for ($\Longleftarrow$)that is a little more difficult, below the proof:
$\bigg [\forall \phi\in E^∗,\langle \phi,x\rangle=0\implies x=0\bigg ]\iff \bigg[x\in E, x\neq0\implies \exists \phi \in E^*, \; \langle\phi,x\rangle\neq 0\bigg]$
Let $x\neq 0$ and $\mathcal{B}=\big\{e_1,e_2,...e_n\big\} \text{a basis of $E$} \implies x=\lambda_1e_1+\lambda_2e_2+...+\lambda_ie_i+...+\lambda_ne_n\quad i\in[\![0,n]\!]$ so there exists a basis of $E$ such that $\mathcal{B}'=\big\{x,u_2,u_3,..u_n\big\}$
Let $f$ such that:
$\begin{array}lf:&E&\to K\\&\lambda_iu_i&\to \lambda_i\end{array}\qquad\implies f\in E^* \text{ and } f(x)=1$
If the proof is correct, I understand the logical path, but I don't really understand it?
I know there is a similar question but it doesn't help me: Can a non-zero vector have zero image under every linear functional?
 A: Let $dim(E)=n$ and $A=\{v_1,v_2...v_n\}$ a basis of $E$.
Now let $x \in E$,then  $x=a_1v_n+a_2v_2+...a_nv_n$ for some $a_1...a_n \in F$
Take the set of linear functionals $B=\{f_1,f_2...f_n\}$ where $f_i(v_j)=0$ if $i \neq j$ and $f_i(v_j)=1$ if $i=j$.
We have that  $$0=f_i(x)=a_1f_i(v_1)+a_2f_i(v_2)+...+a_if_i(v_i)+...+a_nf_i(v_n)=a_i,\forall i \in \{1,2....n\},$$
Thus $a_i=0,\forall i \in \{1,2....n\} \Rightarrow x= \bar{0}$
A: Hint:
Choose a basis $(e_1,\dots, e_n)$ in $E$, and consider the dual basis $(e_1^*,\dots e_n^*)$ in $E^*$. Each $e_i^*$ is just  the $i$-th coordinate map relative to the chosen basis in $E$.
A: $\bigg [\forall \phi\in E^∗,\langle \phi,x\rangle=0\implies x=0\bigg ]\iff \bigg[x\in E, x\neq0\implies \exists \phi \in E^*, \; \langle\phi,x\rangle\neq 0\bigg]$
Let $x\neq 0$ and $\mathcal{B}=\big\{e_1,e_2,...e_n\big\} \text{a basis of $E$} \implies x=\lambda_1e_1+\lambda_2e_2+...+\lambda_ie_i+...+\lambda_ne_n\quad i\in[\![0,n]\!]$ so there exists a basis of $E$ such that $\mathcal{B}'=\big\{x,u_2,u_3,..u_n\big\}$
Let $f$ such that:
$\begin{array}lf:&E&\to K\\&\lambda_iu_i&\to \lambda_i\end{array}\implies f(x)=1$
Is it correct?
