# Help to spot my mistake in my attempt to show a result on linear forms

Considering the following linear forms $$f_1,f_2,...,f_n$$ of $$E$$ a real vector space of dimension $$m$$ and $$f \in \mathcal{L}(E,\mathbb{R})$$ such that $$f\neq 0$$, I would like to show (without using duality or quotient spaces) that :

$$\displaystyle \bigcap_{i=1}^{n} \mathrm{Ker}(f_i)\subset\mathrm{Ker}(f) \Longrightarrow f \in \mathrm{span}(f_1,...,f_n)$$

As $$f \in E^*$$ and $$f \neq 0$$, then $$\dim(\mathrm{Ker}(f))=m-1$$. So let $$(e_1,...,e_{m-1})$$ be a basis of $$\mathrm{Ker}(f)$$, I can find $$e_m$$ such that $$(e_1,\dots,e_m)$$ is a basis of $$E$$.

If $$\forall i\in \{1,\dots,n\}, f_i(e_m) = 0$$ then $$e_m \in \displaystyle \bigcap_{i=1}^{n} \mathrm{Ker}(f_i)$$ but $$e_m \notin \mathrm{Ker}(f)$$ thus $$\displaystyle \bigcap_{i=1}^{n} \mathrm{Ker}(f_i) \not \subset \mathrm{Ker}(f)$$.

Consequently by contraposition if $$\displaystyle \bigcap_{i=1}^{n} \mathrm{Ker}(f_i)\subset\mathrm{Ker}(f)$$ then we can find $$i_0 \in\{1,\dots,n\}$$ such as $$f_{i_0}(e_m)\neq 0$$.

Thus $$\mathrm{Im}(f_{i_0}) = \mathbb{R}f_{i_0}(e_m)$$ and as $$e_m \not \in \mathrm{Ker}(f_{i_0})$$ we have $$\mathrm{Ker}(f_{i_0}) \subset \mathrm{span}(e_1,\dots,e_{m-1})=\mathrm{Ker}(f)$$ and $$\dim(\mathrm{Ker}(f)) = \dim(\mathrm{Ker}(f_{i_0}))$$ thus $$\mathrm{Ker}(f_{i_0}) = \mathrm{Ker}(f)$$.

Finally if I take $$\lambda_{i_0} = \dfrac{f(e_m)}{f_{i_0}(e_m)}$$ then $$f = \lambda_{i_0} f_{i_0}$$, thus $$f \in \mathrm{span}(f_1,...,f_n)$$.

I have the feeling I have made a horrible mistake. Can you help me spot it?

'As $$e_m \notin Ker (f_{i_0})$$ we have $$ker (f_{i_0}) \subset span (e_1,e_2,..,e_{m-1})$$' is not correct. We can have a linear combination of all the e_i's in $$ker (f_{i_0})$$ (with non-zero coefficient of $$e_m)$$. (For example $$ae_1+be_m$$ could be in $$ker (f_{i_0})$$ with $$b \neq 0$$)