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?

  • $\begingroup$ Think about the value of $f_2$ on $\ker(f_1)$. $\endgroup$
    – Kapil
    Oct 17, 2020 at 18:21

1 Answer 1


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$$f(v)=-\frac{\lambda_1}\lambda f_1(v)-\frac{\lambda_2}\lambda f_2(v).$$

  • $\begingroup$ Thanks for the answer! But why does such $\varphi$ exist? $\endgroup$
    – Ilya
    Oct 17, 2020 at 19:06
  • 1
    $\begingroup$ 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 $\varphi$. $\endgroup$ Oct 17, 2020 at 19:43

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