# Finding the kernel of a map

It's a simple problem, but I can't complete the argument. Let $$F$$ be a n-dimensional vector space and let $$S$$ be its dual space. Considering $$\delta^1, \dots, \delta^n$$ vectors from a basis of $$S$$, let $$S_0$$ be the subspace of $$S$$ given by the elements $$a = \sum_{i=1}^n a_i\delta^i \in S$$ such that $$\sum_{i=1}^n a_i = 0.$$

Then, let \begin{aligned} \rho: F \longrightarrow S_0^* ~~~~~~~f \longmapsto \varphi, \end{aligned} be the map where $$\varphi$$ is the functional \begin{aligned} \varphi: S_0 \longrightarrow \mathbb{R} ~~~~~~~a \longmapsto a(f). \end{aligned}

I'm trying to show that $$ker(\rho)$$ is the set of constants of $$F$$.Obviously, constant vectors are in the kernel. By "constant vectors" understand the vectors $$(f_1,\dots,f_n) \in F$$ such that $$f_1 = f_2 = \dots = f_n$$. My problem is to prove that only they are there. Someone could help me?

• $\rho$ is onto. $S_0$ has dimension $n-1$, and hence so does $S_0^*$. So the kernel is one dimensional. Mar 19 at 3:08
• I don't understand what you mean by 'constant vector'. Could you please clarify? Mar 19 at 4:46
• My apologies. For me, a vector $(f_1,\dots,f_n)$ is constant if $f_1= f_2 = \dots = f_n$ Mar 19 at 13:39
• @ArturoMagidin thanks. I understood your argument and, for me, it makes sense Mar 19 at 13:51