So i'm trying to prove the following:
Let $V$ be a vector space, $\dim(V)=n, f_1,…,f_k:V\to \mathbb{R}$ are linear functionals on $V$.
then $\dim(\ker(f_1) ∩ \ …\ ∩ \ker(f_k)) ≥ n-k$.
I've tried a few different approaches, but they got me nowhere.
like:
let $B_i=\{u_{i1},…,u_{in-1}\}$ be a basis of $\ker(f_i)$ where $1 \le i \le k$
now $\dim(\ker(f_1) ∩ \ker(f_2)) = \dim([B_1]∩[B_2]) ≤ n-1$ and continue on like this, but that would lead me to $\dim(\ker(f_1) ∩ \ …\ ∩ \ker(f_k))≤ n-1$.
Another approach that I've tried is:
Define $F:V\to \mathbb{R}\times \cdots \times \mathbb{R}$ ($k$ products of $\mathbb{R}$) as $F(x)=(f_1(x),…,f_k(x))$. Then $\ker(F)= \ker(f_1) ∩ … ∩ \ker(f_k)$ but i don't see a way how to proceed from that or if it would even lead somewhere
Best Answer
Assume that $f_1, \ldots, f_k$ are linearly independent and extend them to a a basis $\{f_1, \ldots, f_n\}$ for the dual space $V^*$.
Let $\{b_1, \ldots, b_n\}$ be a basis for $V$ such that $\{f_1, \ldots, f_n\}$ is its dual basis.
Note that $\bigcap_{i=1}^k \ker f_i = \operatorname{span}\{b_{k+1}, \ldots, b_n\}$.
Indeed, obviously $b_j \in \ker f_i$ for $i = 1,\ldots, k$ and $j = k+1, \ldots, n$.
Conversely, let $x = \sum_{i=1}^n \alpha_ib_i \in \bigcap_{i=1}^k \ker f_i $. We have
$$0 = f_i(x) = \alpha_i, \quad\forall i = 1, \ldots, k$$
so $x = \sum_{i=k+1}^n \alpha_ib_i \in \operatorname{span}\{b_{k+1}, \ldots, b_n\}$.
Since $b_{k+1}, \ldots, b_n$ are linearly independent, we conclude $\dim \bigcap_{i=1}^k \ker f_i = n-k$.
If $f_1, \ldots, f_k$ are not linearly independent, we can assume that the first $f_1, \ldots, f_l$ are linearly independent, and that $f_{l+1}, \ldots, f_k$ are their linear combinations.
Then notice that $$\bigcap_{i=1}^l \ker f_i \subseteq \ker f_j , \quad\forall j = k+1,\ldots, k$$
so $\bigcap_{i=1}^l \ker f_i = \bigcap_{i=1}^k \ker f_i$.
From the above proof follows that
$$\dim \bigcap_{i=1}^k \ker f_i =\dim \bigcap_{i=1}^l \ker f_i = n-l \ge n-k$$