Prove that for every subspace we can find a finite number of linear functionals such that $W=\ker l_{1}\cap\cdots\cap \ker l_{k}$

Question

In need of some assistance regarding this questions from a University textbook (I'm learning by myself). Its about Dual Spaces:

Let there be $V$ a finite vector space (Has a basis) over $\mathbb{F}$. Prove that for every subspace $W$ of $V$ we can find a finite number of linear functionals $l_{1}, \dots,l_{k}$ such that $W=\ker l_{1}\cap\cdots\cap \ker l_{k}$

Alright, so I don't quite understand what they want me to prove. All the linear functionals are functions from the subspace $W$ to the field $F$, and as I understand it, the Kernel for each functional are all the vectors that the functional sends to $0_{\mathbb{F}}$

What I don't understand is what I'm supposed to do with that information, in order to prove the question :(

Anyways, any help is appreciated

Thank you!

Answer 1

I'm not entirely sure if this is what you wanted but here is a sketch. Let me know of any confusions:

We have $V$ a finite dimensional vector space, say dimension $n$, and so the subspace $W$ is also finite dimensional, say of dimension $n-k$. Let's choose a basis for $W$, say $(w_1, \ldots, w_{n-k})$, and then extend this to a basis of $V$, say $(w_1, \ldots, w_{n-k}, v_1, \ldots, v_k)$.

Once you have this basis, you can explicitly write down linear functionals $l_i: V\to \mathbb{F}$ such that $\ker l_i = \operatorname{span}(w_1, \ldots, w_{n-k}, v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_k)$ for $i=1,\ldots, k$. So taking $\ker l_1 \cap \cdots \cap \ker l_k$ gives you $W$.

Remark (if you know differential manifolds): this problem may seem somewhat random, but it parallels an important fact in differential manifolds: if $X$ is a $n$-dimensional manifold and $Y\subset X$ is a $n-k$ dimensional submanifold, then $Y$ can be locally "carved out" by $k$ smooth functions.