Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k) $

Question

Let $V$ a $\mathbb{K}$-vector space of finite dimension $n$, with $\{f_1,...,f_n\}$ a set linearly independent of $V^*$ and $f\in V^*$.

Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k).$

($\Leftarrow$) Let $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$.

Let $f\in\bigcap_{i=1}^k\ker(f_i) $ then $f\in \ker(f_1),...,f\in \ker(f_k)$.

By hypothesis we have if $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$

Here I'm a little stuck. Can someone help me?

($\Rightarrow$) I do'nt know how to prove this part. Help me, if you can. I will be very grateful.

Answer 1

You start well, but soon make a mistake: the statement $f\in\bigcap_{i=1}^k\ker(f_i)$ is wrong, because the kernels are subspaces of $V$ and $f\in V^*$.

What you have to prove is

If $f\in\operatorname{Span}(f_1,\dots,f_k)$ then $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$.

Suppose $f\in\operatorname{Span}(f_1,\dots,f_k)$; then $f=\alpha_1f_1+\dots+\alpha_kf_k$ for some scalars $\alpha_1,\dots,\alpha_k$. If $x\in\bigcap_{i=1}^k\ker(f_i)$, then $f_i(x)=0$, for $i=1,\dots,k$ and therefore $$ f(x)=\alpha_1f_1(x)+\dots+\alpha_kf_k(x)=0 $$ proving that $x\in\ker(f)$.

Now let's try the converse:

If $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$ then $f\in\operatorname{Span}(f_1,\dots,f_k)$.

Suppose $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$. Write $$ f=\alpha_1f_1+\dots+\alpha_kf_k+\alpha_{k+1}f_{k+1}+\dots+\alpha_nf_n $$ which is possible because $\{f_1,\dots,f_n\}$ is a basis of $V^*$.

Lemma. Every basis of $V^*$ is the dual of a basis $\{e_1,\dots,e_n\}$ of $V$.

Once accepted this lemma (for the proof, see https://math.stackexchange.com/a/1772676/62967), we have $\{e_1,\dots,e_n\}$ such that $$ f_i(e_j)=\begin{cases} 1 & i=j \\ 0 & i\ne j \end{cases} $$ In particular, $e_{k+1},\dots,e_n\in\bigcap_{i=1}^k\ker(f_i)$, so $$ f(e_j)=0,\quad j=k+1,\dots,n $$ and therefore, for $j=k+1,\dots,n$, \begin{align} 0=f(e_j)&=\alpha_1f_1(e_j)+\dots+\alpha_kf_k(e_j)+ \alpha_{k+1}f_{k+1}(e_j)+\dots+\alpha_jf_j(e_j)+\dots+\alpha_nf_n(e_j) \\ &=0+\dots+0+0+\dots+\alpha_j\cdot1+\dots+0 \\ &=\alpha_j \end{align} Hence $\alpha_j=0$ for $j=k+1,\dots,n$ and finally $$ f=\alpha_1f_1+\dots+\alpha_kf_k\in\operatorname{Span}(f_1,\dots,f_k) $$

Answer 2

$(\Leftarrow)$ Let $x\in V$ such that $x\in \bigcap ker(f_i)$ i.e. $x\in ker(f_i)$ for all $i=1,....,n$.

Hence $f_i(x)=0$ for all $i=1,....,n$.

Now as $f=a_1f_1+.....+a_kf_k$ which gives $f(x)=0$.

$(\Rightarrow)$ for good proof of this part refer to,

J.B.Conway - A Course in Functinal Analysis, 2nd Edition

Proposition 1.4(page no. 371)

Answer 3

Probably not the best proof but it was suggested by a colleague :

We will use duality...

Suppose that $\bigcap \limits_{i=1}^{k} \ker{f_i} \subset \ker{f}$.

Using properties of duality we have that $(\ker{f})^{\bot} \subset \left( \bigcap \limits_{i=1}^{k}\ker{f_i}\right)^{\bot} = \displaystyle \sum \limits_{i=1}^{k} (\ker f_i)^{\bot}$.

(Remember that for a set $A\subseteq V$, we have $A^{\bot} :=\{\phi\in V^* \ / \ \forall x \in A, \phi(x)=0 \}$).

If we suppose that all the linear forms are $\not \equiv 0$ then for all $i\in\{1,...,k\}$, all $(\ker{f_i})$ are hyperplanes of $V$ hence all the $(\ker{f_i})^{\bot}$ are of the form $\mathbb{K}a_i$ with $a_1,...,a_k\in \mathbb{K}^{\times}$ and the $(a_i)_i$ can be taken pairwise different.

Then $(\ker{f})^{\bot} \subset \sum \limits_{i=1}^{k} \mathbb{K}a_i$.

Answer 4

$(\Rightarrow)$ Suppose $1\leq k\leq n \bigcap_{j\neq k}ker(f_{j})\neq\bigcap_{j=1}^{n}ker(f_{j})$.

Then exists$y_{k}\in\bigcap_{j\neq k}ker(f_{j})$ tal que $y_{k}\notin\bigcap_{j=1}^{n}ker(f_{j})$. then $f_{j}(y_{k})=0$ if $k\neq j$ and $f_{j}(y_{k})\neq0$ if $j=k$.

Let $x_{k}=\frac{yk}{f_{k}(y_{k})}$(Note $f_{k}(y_{k})\neq0)$ Then, $f_{k}(x_{k})=f_{k}(\frac{y_{k}}{f_{k}(y_{k})})=\frac{f_{k}(y_{k})}{f_{k}(y_{k})}=1$.For other way, $f_{j}(x_{k})=f_{j}(\frac{y_{k}}{f_{k}(y_{k})})=\frac{f_{j}(y_{k})}{f_{k}(y_{k})}=0$

Let $f\in V^{*}y x\in\mathbb{R}$ .

Define $y=x-\sum_{k=1}^{n}f_{k}(x)x_{k}$ Then $f_{j}(y)=f_{j}(x-\sum_{k=1}^{n}f_{k}(x)x_{k})=f_{j}(x)-\sum f_{k}(x)f_{j}(x_{k})=f_{j}(x_{k})-f_{j}(x)f_{j}(x_{j})=0$.

Then, $f_{j}(x_{k})=0\,y\,f_{j}(x_{j})=1$.

By hyphotesis $f(y)=0$,then:

$0=f(x)-\sum_{k=1}^{n}f_{k}(x)f(x_{k})=f(x)-\sum_{k=1}^{n}\alpha_{k}f_{k}(x)\Rightarrow f(x)=\sum_{k=1}^{n}\alpha_{k}f_{k}\Rightarrow f=\sum_{k=1}^{n}\alpha_{k}f_{k}$

In consequence, $f\in span(f_{1},...,f_{n})$

$\smash{}$