If $W$ is a $k$-dimensional subspace of an $n$-dimensional vector space $V$, then $W$ is the intersection of $(n- k)$ hyperspaces in $V$.
Rephrasing problem to my taste:
If $W$ is a $k$-dimensional subspace of an $n$-dimensional vector space $V$, then $W=\bigcap_{k\lt i\leq n}N_{f_i}$ $=\{x\in V|f_i(x)=0_F\text{, }\forall k\lt i\leq n\}$, where $\{f_1,…,f_n\}$ is dual basis of $B_V$ (extended from $B_W$).
My attempt: Let $B_W=\{\alpha_1,…,\alpha_k\}$ be basis of $W$. By theorem 5 section 2.3, $\exists B_V\subseteq V$ such that $B_V$ is finite basis of $V$ and $\{\alpha_1,…,\alpha_k\}\subseteq B_V$. Let $B_V=\{\alpha_1,..,\alpha_k,\alpha_{k+1},..,\alpha_n\}$ be basis of $V$. Let $B^*=\{f_1,…,f_n\}$ be dual basis of $B_V$. By definition of dual basis, $f_i(\alpha_j)=\delta_{ij}$, $\forall j\in J_n$. In particular, $f_i(\alpha_i)=\delta_{ii}=1_F$. So $f_i\neq 0_{L(V)}$, $\forall i\in J_n$. Which implies $N_{f_i}$ is hyperspace, $\forall i\in J_n$.
We show $f_i\in W^0$, $\forall k\lt i\leq n$. Let $i\in J_n\setminus J_k$. Let $\alpha \in W$. Since $\mathrm{span}(\{\alpha_1,…,\alpha_k\})=W$, we have $\alpha=\sum_{j=1}^ka_j\cdot \alpha_j$. So $f_i(\alpha)$ $=f_i(\sum_{j=1}^ka_j\cdot \alpha_j)$ $= \sum_{j=1}^ka_j\cdot f_i(\alpha_j)$ $= \sum_{j=1}^ka_j\cdot \delta_{ij}$. Since $j\leq k\lt i$, we have $f_i(\alpha)$ $= \sum_{j=1}^ka_j\cdot \delta_{ij}=0_F$. So $f_i(\alpha)=0_F$, $\forall \alpha \in W$. Since $i$ was arbitrary, $f_i\in W^0$, $\forall k\lt i\leq n$. Thus $W\subseteq N_{f_i}$, $\forall k\lt i\leq n$. Hence $W\subseteq \bigcap_{k\lt i\leq n}N_{f_i}$.
Conversely, suppose $x\in \bigcap_{k\lt i\leq n}N_{f_i}\subseteq V$. Then $f_i(x)=0_F$, $\forall k\lt i\leq n$. By theorem 15 section 3.5, $x=\sum_{i=1}^nf_i(x)\cdot \alpha_i$ $=\sum_{i=1}^kf_i(x)\cdot \alpha_i$. So $x= \sum_{i=1}^kf_i(x)\cdot \alpha_i$$\in \mathrm{span}(\{\alpha_1,…,\alpha_k\})$ $=W$. Thus $\bigcap_{k\lt i\leq n}N_{f_i}\subseteq W$. Hence $W= \bigcap_{k\lt i\leq n}N_{f_i}$. Is my proof correct?
Best Answer
Yes it's correct. Notice however that sometimes it's simpler to reason in coordinates!
Let's take the basis $B$ of your proof. Then $W$ can be represented in coordinates as follows: $$W:\begin{cases}x_{k+1}=0 \\ \ldots \\ x_n=0\end{cases}.$$ This means that $W$ is the intersection of the subspaces $N_{k+1},\ldots,N_{n}$ where for all $i=1,\ldots, n-k$: $$N_i:x_i=0.$$ Clearly each $N_i$ is an hyperspace.