Linear independent functionals over finite vector space

Question

Let $V$ be a $n$-dimensional vector space. Let $f_1,\dots,f_m\in V^*$ be linear functionals. Show that $f_1,\dots,f_m$ are linearly independent if and only if $\cap _{i=1}^{m}\ker(f_i)$ has dimension $n-m$.

Remark: I know that this problem is quite popular but almost all the topics in MSE with problem have not detailed proof but I have troubles with details.

My approach:

$\Rightarrow$ Define the map $F:V\to \mathbb{k}^m$ by rule $F(x)=(f_1(x),\dots,f_m(x))$ for all $x\in V$. Then one can easily show that $F$ is linear and $\ker(F)=\cap _{i=1}^{m}\ker(f_i)$. Then by rank-nullity theorem we get that: $n=\dim \ker(F)+\dim \text{Im}(F)$.

So the result follows easily if we can show that $\text{Im}(F)=\mathbb{k}^m$. However I have troubles with this. Take any $\bar{\alpha}=(\alpha_1,\dots,\alpha_m)\in \mathbb{k}^m$ and I want to show that there is $x\in V$ such that $F(x)=\bar{\alpha}$. But I do not know how to use that $f_1,\dots, f_m$ are linearly independent.

$\Leftarrow$ Suppose that $f_1,f_2,\dots,f_m$ are linearly dependent and WLOG assume that $f_1$ is linear combination of $f_2,\dots,f_m$ then easy to see that $\cap _{i=2}^{m}\ker(f_i)\subseteq \ker (f_1)$. I do not know what to do after that.

Please do not duplicate this question since most of the topics does not have any details. Would be very thankful if you can show me how complete my reasoning?

Answer 1

If $f_1,f_2,\dots,f_m$ are linearly independent, then you can complete them to a basis $f_1,\dots,f_m,f_{m+1},\dots,f_n$ of $V^*$. Such a basis is the dual of a basis $e_1,\dots,e_n$ of $V$, see https://math.stackexchange.com/a/1772676/62967.

Now $F(e_1)=(1,0,\dots,0)$ and similarly for $F(e_i)$, $1\le i\le m$. Thus $F$ is surjective, because its range contains a basis.

Let's try the converse. If $f_1$ is a linear combination of $f_2,\dots,f_m$, then $$ \bigcap_{i=1}^m\ker f_i=\bigcap_{i=2}^m \ker f_i $$ One inclusion is obvious, namely $\subseteq$, but you proved the other. Now you can consider the linear map $$ G\colon V\to \mathbb{k}^{m-1},\qquad G(x)=(f_2(x),\dots,f_m(x)) $$ and the rank nullity theorem tells you that $$ \dim\ker G=n-\dim\operatorname{Im}G\ge n-(m-1)=(n-m)+1>n-m $$ a contradiction, because $$ \ker G=\bigcap_{i=2}^m\ker f_i=\bigcap_{i=1}^m \ker f_i $$ has dimension $n-m$ by assumption.

Answer 2

Well, I did it this way... Define, $F = (f_1,\cdots,f_m):V\to\mathbb{R}^{m}$, we can prove easily that $F$ is linear, by some theorem i don't recall the name, we know $\text{dim Im}F= n-\text{dim Ker}F$.

$(\Leftarrow)$ In this case, $\text{dim Im }F = m$

So, $F\vert_{\tilde{V}}:\tilde{V}\to\mathbb{R}^m$ is a linear isomorphism(both vector spaces has the same dimension and F is surjective), here $V=\tilde{V}+\text{Ker }F$ some decomposition of $V$(that can be achieved by choosing any basis for the kernel and complete it to get a basis for $V$, then the elements used to complete generate another subspace that we call $\tilde{V}$). Thus, $\exists v_i\in\tilde{V}\subseteq V \text{ s.t. }F(v_i)=e_i$ then $\sum a^if_i(v_j)=a^j$ thus $f_i$'s are linearly independent.

$(\Rightarrow)$ If $f_i$'s are linearly independent then the rank of any matrix that represents $F$ is equal to $m$ thus $\text{dim Im }F=m$ and then follows what we want.