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

Question

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.

Answer 1

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.