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?

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


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