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Here is the statement :

Let $E$ a finite $\mathbb{K}$-vectorial space and $\phi_1, \ldots, \phi_p, \psi$ a collection of linear forms on $E$ such that : $\bigcap \limits_{i=1}^p\ker(\phi_i)\subset \ker(\psi)$. Then $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$.

The suggestion is to make an induction on $p\ge 1$ (I already know that this method is not the fastest way) :

For $p=1$ then $\ker(\phi_1)\subset \ker(\psi)$.

If $\phi_1 \equiv 0$ then $\ker(\phi_1)=E$ so $\psi \equiv 0$ and the statement is true.

If not, we have the fact that $\phi_1$ and $\psi$ are linearly dependent so $\psi \in \operatorname{span} \{\phi_1\}$ and the statement is also true.

Now we want to check the statement for $p+1$.

We suppose $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \ker(\psi)$ and we want to prove that $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_{p+1}\}$.

If all the linear forms are null then $\psi \equiv 0$ and the statement is true.

If not, we have $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \bigcap \limits_{i=1}^p \ker(\phi_i) \subset \ker(\psi)$ so by the induction's hypothesis, $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$.

Moreover (not sure) $\ker(\phi_{p+1})\subset \ker(\psi)$ so $\psi \in \operatorname{span} \{\phi_{p+1}\}$.

Then I'm stuck...

Thanks in advance

Maman
  • 3,300

1 Answers1

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Take $p \in \mathbb Z^+$and suppose that for any $\mathbb K$-vector space $F$ and any $p$ linear forms $f_1,\dots,f_p$ on $F$ we have that $$\textstyle (\forall g \in F^*) \quad \bigcap_{i=1}^p \ker f_i \subseteq \ker g \ \ \Rightarrow \ \ g \in \operatorname{span}\{f_1,\dots,f_p\}.$$ Now, consider $p+1$ linear forms $\phi_1,\dots,\phi_{p+1}$ on $E$, and take $\psi \in E^*$ such that $\bigcap_{i=1}^{p+1} \ker \phi_i \subseteq \ker \psi$. Denote $\ker \phi_{p+1}$ by $K$ and observe that $\phi_1|_K,\dots,\phi_p|_K$ are $p$ linear forms on $K$ such that $\bigcap_{i=1}^p \ker \phi_i|_K \subseteq \ker \psi|_K$ because if $v \in \bigcap_{i=1}^p \ker \phi_i|_K \subseteq K$, then $v \in \bigcap_{i=1}^{p+1} \ker \phi_i \subseteq \ker \psi$ and so $0 = \psi(v) = \psi|_K(v)$.

Thus, by induction hypotheses there are $a_1,\dots,a_p \in \mathbb K$ such that $$\psi|_K = a_1\phi_1|_K + \cdots + a_p\phi_p|_K.$$ Finally, the linear form $j := \psi-(a_1\phi_1+\cdots+a_p\phi_p)$ on $E$ vanishes on $K$, and so $\ker \phi_{p+1} \subseteq \ker j$, which implies that $j$ is a scalar multiple of $\phi_{p+1}$, meaning that $\psi \in \operatorname{span}\{\phi_1,\dots,\phi_{p+1}\}$.

azif00
  • 20,792
  • Thank you ! Nice method ! Notice that the induction works because of the finite dimension of $E$ ! – Maman Dec 27 '20 at 12:49
  • Your welcome! Why is that? I don't see any problem if we do not assume finite dimensions. – azif00 Dec 27 '20 at 21:04
  • I meant for the linear forms. If we have an infinite collections of linear forms and $E$ is not a finite vector space, the statement is not verified. – Maman Dec 28 '20 at 00:04