# If $\ker f\supset \ker f_1\cap \ker f_2$ then $f\in \operatorname{span}\{f_1,f_2\}$

Let $$f,f_1,f_2$$ be linear functionals on vector space V (infinite dimensional) and $$\ker f\supset \ker f_1\cap \ker f_2$$. I want to obtain that $$f\in \operatorname{span}\{f_1,f_2\}$$.

I tried to use factors, but i don't think that $$V/\ker f \subset V/(\ker rf_1 \cap \ker f_2)$$...

Another approach is to find vectors $$y, z$$ s.t. $$\forall x\ f(x-yf_1(x)-zf_2(x))=0$$, but it didn't work out for me.

Any hints?

• Think about the value of $f_2$ on $\ker(f_1)$. Oct 17 '20 at 18:21

Consider the linear map$$\begin{array}{rccc}\Psi\colon&V&\longrightarrow&\Bbb R^3\\&v&\mapsto&\bigl(f_1(v),f_2(v),f(v)\bigr).\end{array}$$Then $$(0,0,1)\notin\Psi(V)$$ and therefore there is a linear map $$\varphi\colon\Bbb R^3\longrightarrow\Bbb R$$ such that, for each $$w\in\Psi(V)$$, $$\varphi(w)=0$$ and that $$\varphi(0,0,1)\ne0$$. Then, for each $$(a,b,c)\in\Bbb R^3$$,$$\varphi(a,b,c)=\lambda_1 a+\lambda_2b +\lambda c,$$for some $$\lambda_1,\lambda_2,\lambda\in\Bbb R$$, with $$\lambda\ne 0$$ (otherwise, $$\varphi(0,0,1)=0$$). But then, for each $$v\in V$$,$$\lambda_1 f_1(x)+\lambda_2 f_2(v)+\lambda f(v)=0,$$and therefore$$f(v)=-\frac{\lambda_1}\lambda f_1(v)-\frac{\lambda_2}\lambda f_2(v).$$
• Thanks for the answer! But why does such $\varphi$ exist?
• Since $\Psi(V)$ is a subspace of $\Bbb R^3$ which is not the whole space, it is contained in a plane passing through the origin. And every such plane is the kernel of a linear functional $\varphi$. Oct 17 '20 at 19:43