Kernel of nonzero linear function in vector space Let $l$ be a nonzero linear function on a vector space $V$, i.e. $l:V\to \mathbb{k}$. Show that $\text{ker}(l)$ is a maximal nontrivial linear subspace in $V$. Also prove that $V/\text{ker}(l)$ has dimension 1.
Remark: It is not necessary that $V$ is finite-dimensional space.
This is quite a famous problem and I was able to show that $\text{ker}(l)$ is maximal linear subspace of $V$. In other words, if $\text{ker}(l)\subseteq W\subseteq V$ then $W=\text{ker}(l)$ or $W=V$. 
Also since $l$ is a nonzero it means that $\exists v\in V$ such that $l(v)\neq 0$. Then one can show that $V/\text{ker}(l)$ has a basis $v+\ker (l)$ which means that $V/ \ker(l)$ has dimension.
But I have spent some time and was not able to show that $\ker(l)$ is nontrivial subspace of $V$, which means that there is some element $x\neq 0\in V$ such that $l(x)=0$.
I think that this is wrong.
Anyway I would be very grateful if anyone can show how to prove or disprove it.

In a vector space of dimension $1$, a maximal subspace is trivial. To show it is nontrivial in dimension greater than one, simply take a one dimensional subspace as a counterexample to the containment condition. 
