Proof: Each linear subspace of $\Bbb R^n$ (or $\Bbb Q^n$) is also the intersection of finitely many linear hyperplanes I am currently in the process of reading Theory of Linear and Integer Programming by Alexander Schrijver and encountered the following theorem and proof:

Theorem 3.1. Each linear subspace of $\Bbb R^n$ (or $\Bbb Q^n$) is generated by finitely many vectors, and is also the intersection of finitely many
  linear hyperplanes.
Proof. The proof is well-known from linear algebra. If $L$ is a linear subspace of $R^n$, then any maximal set of linearly independent
  vectors in $L$ is finite and generates $L$. Similarly, $L$ is the
  intersection of finitely many linear hyperplanes $\{x|a_ix = 0\}$,
  where the $a_i$ form a maximal set of linearly independent vectors
  generating $L^* $$:=$ $\{z|zx =0$ for all $x$ in $L\}$.

Although I understand the first part of the proof, I have trouble understanding why every linear subspace is the intersection of finitely many linear hyperplanes. 
 A: Hint: extend a basis $b_1, \ldots, b_l$ for $L$ to a basis $b_1, \ldots, b_l, \ldots b_n$ for $\Bbb{R}^n$. Consider the hyperplanes $H_k$ comprising the points $\sum_i x_ib_i$ where $x_k = 0$. $L$ can be expressed as the intersection of some of the $H_k$.
A: The book's proof can be expanded to be more explicit: it is, for every subspace $L$, defining a subspace usually known as $L^{\perp}$ perpendicular to the original in the sense that $z\in L^{\perp}$ if and only if for every $x\in L$ we have that the inner product $z\cdot x = 0$ - i.e. that $z$ is "perpendicular" to every $x\in L$.
The idea is that if you let $z_1,\ldots,z_k$ be a basis for $L^{\perp}$, then you can define hyperplanes $H_i$ to be the set of $x$ such that $z_i\cdot x = 0$. The claim is that the fact that $z_1,\ldots,z_k$ is a basis for $L^{\perp}$ implies that $H_1\cap H_2\cap \ldots \cap H_k$ must equal $L$ and therefore that $L$ is a finite intersection of hyperplanes.
For instance, if you had the subspace $L=\{(0,t,t) : t\in\mathbb R\}$ in $\mathbb R^3$, one would find that $L^{\perp}$ could be described as $\{(x,y,-y):x,y\in\mathbb R\}$ which has a basis $\{(1,0,0),(0,1,-1)\}$ which then says that $L$ is the intersection of the two hyperplanes defined by the tuples $(x,y,z)$ such that $x=0$ and $y-z=0$.
This does subtly rely on the nontrivial fact that $L^{\perp\perp} = L$.

Though it's not necessary to understand this proof, the idea behind it is encapsulated in the notion of duality which is basically a method in linear algebra in which one transforms lines (which can be represented by vectors!) into hyperplanes and spanning sets into intersections of hyperplanes. The motivation for the proof in the book is to apply this reasoning (i.e. to look at $L^{\perp}$) to try to get a new statement out of our preexisting knowledge of bases.
