if $\operatorname{Ker}(\phi)$ can be generated with a basis of $E$ then $\phi = 0$ While thinking about linear form and doing some test, I was wondering if the following is true : 
let's say I have a linear form $\phi : \operatorname{End}(E) \to \mathbb{K}$ where $E$ is a finite-dimensional  vector space. 
Moreover, we have the following property : 
$\exists x \in E, Span\{f_1(x),...,f_p(x)\} =E$  where the $f_i$ are a basis of : $\ker \phi$. 
Then do we have necessarily $\phi = 0$ ? 
I don't know if this is true but I am quite sure it is. i thought about using Riesz theorem, so that we have :
$$\forall f, \phi(f) = \langle a,f\rangle , a \in \operatorname{End}(E)$$
But I can't relate this to the value of $f(x)$.
 A: Let $n$ be a positive integer greater than $1$.  Set $E:=\mathbb{K}^n$ and write $e_1,e_2,\ldots,e_n$ for the standard basis vectors of $\mathbb{K}^n$.  Define $\tau_{i,j}\in\text{End}_\mathbb{K}(E)$ to be the linear map sending $$\sum\limits_{k=1}^n\,\lambda_k\,e_k\mapsto \lambda_i\,e_j\,,$$ where $\lambda_1,\lambda_2,\ldots,\lambda_n\in \mathbb{K}$.  The maps $\tau_{i,j}$ for $i,j\in\{1,2,\ldots,n\}$ form a basis of $\text{End}_\mathbb{K}(E)$.  
Now, suppose that $\phi:\text{End}_{\mathbb{K}}(E)\to \mathbb{K}$ be the map sending $$\sum_{i=1}^n\,\sum_{j=1}^n\,\kappa_{i,j}\,\tau_{{i,j}}\mapsto \kappa_{1,1}\,,$$
where $\kappa_{i,j}\in\mathbb{K}$ for all $i,j=1,2,\ldots,n$.  Then,  the maps $\tau_{i,j}$ for $(i,j)\in\{1,2,\ldots,n\}^2\setminus\big\{(1,1)\big\}$ form a basis of $\ker(\phi)$.  Clearly, $\phi\not\equiv 0$, since $\phi\big(\tau_{1,1}\big)=1\neq 0$.
Pick $x:=e_1+e_2+\ldots+e_n$.  Ergo, $e_1=\tau_{n,1}(x)$, $e_2=\tau_{1,2}(x)$, $e_3=\tau_{2,3}(x)$, $\ldots$, $e_n=\tau_{n-1,n}(x)$ form a basis of $E$.  Hence, the answer to the question is no.
