Proving that dimension of intersection of kernels of $k$ linear functionals on $n$ dimensional vector space is $\ge n-k$

Question

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

Answer 1

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 $\dim \bigcap_{i=1}^k \ker f_i = n-k$.

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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

$$\dim \bigcap_{i=1}^k \ker f_i =\dim \bigcap_{i=1}^l \ker f_i = n-l \ge n-k$$

Answer 2

Well, using the second approach the statement follows easily...

$$\text{dim} ker F+ \text{dim} Im F = n$$

Since, at most $\text{dim} Im F = k$ then as minimum $\text{dim} Ker F\geq n-k$. Note, $F$ is a linear map and $\text{dim} Ker F=\text{dim}\left[ker f_1\cap ker f_2\cap \cdots\cap ker f_k\right]$.

Additionally, you can prove that $f_1,\cdots,f_k$ are linearly independent if and only if $\text{dim}Ker F = n-k$.