If $T : V \to k$ is not the zero map, there is $v \in V$ such that $T(v) = 1$. If $V$ is a $k$-vector space and $T : V \to k$ is a homomorphism (linear map) that is not the zero map, is it true that some element of $V$ is mapped to $1$ of $k$? Of course $k$ is a 1-dimensional $k$-vector space generated by $1$ but how can I see that there exists a $v \in V$ such that $T(v) = 1$?

As $T$ is not the zero map, there is $v \in V$ such that $T(v) = a \neq 0$. Then $\frac{1}{a}T(v) = 1$, so by the linearity of $T$, $T(\frac{1}{a}v) = 1$.
