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?


1 Answer 1


'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 $b \neq 0$)


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