Let $X$ be a vector space. Assume that $f_1,...,f_n$ are linear functionals (may not be bounded) on $X.$

It can be shown that if $$\bigcap_{k=1}^n ker(f_k)=\{0\},$$ then $X$ is finite dimensional.

I would like to know whether the converse holds, that is,

If $X$ is finite dimensional, is it true that $$\bigcap_{k=1}^n ker(f_k) =\{0\}$$ for any $n\in\mathbb{N}?$

It seems true to me as linear functionals on finite dimensional vector space, say $\mathbb{R}^m,$ have trivial kernel. But I do not know how to prove it.

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    $\begingroup$ I am pretty sure you want to avoid examples like $f_k =0$ for all $k=1,2,\cdots$. What kind of functionals do you want to look at? $\endgroup$ – Sangchul Lee Feb 16 '18 at 4:36
  • $\begingroup$ @SangchulLee: Maybe exclude zero functional? That is, all linear functionals exclude zero functional. $\endgroup$ – Idonknow Feb 16 '18 at 4:44
  • $\begingroup$ There are three possibilities for $\rm{ker ~f}$: $$1) \{0\} , ~~~~~~~~~~~~~~~~~2) \rm{n-1 ~ dimension ~ subspacce} ~~~~~~~~~~~~~3) X$$ $\endgroup$ – Red shoes Feb 16 '18 at 4:46
  • $\begingroup$ @Redshoes: I see. So the converse does not hold for any linear functional, right? Because codimension of kernel of linear functional is always $1.$ $\endgroup$ – Idonknow Feb 16 '18 at 4:48
  • $\begingroup$ Now that I think about it carefully, it seems that converse fails even in $X = \mathbb{R}^2.$ Take $f_1$ and $f_2$ to be planes passing through origin. Then their intersection may not be trivial. $\endgroup$ – Idonknow Feb 16 '18 at 4:49

This is obviously false as a theorem reads as follows

If $f_1,\cdots,f_n$ are linear functionals over a linear space $V$, then each linear functional vanishing on $\bigcap_{k=1}^n\ker f_k$ is a linear combination of $f_1,\cdots,f_n$.

The converse also holds apparently and thus numerous counterexamples for your question can be constructed.


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