Why do extreme points lie on linearly independent binding hyperplanes? I know the algebraic notion of an extreme point. I am confused about the geometrical aspect of an extreme point in terms of the hyperplanes as mentioned below.

The point $x'$ lies on $n$ of the hyperplanes defining the polyhedron
$P$, such that the rows of A associated with those hyperplanes are
linearly independent. Thus, $x'$ satisfies $n$ linearly independent
tight constraints.

I would be very grateful if someone would please explain the text above.
In particular, why the corresponding rows of the coefficient matrix have to be linearly independent.
 A: Suppose ${\bf x}'$ does not lie on $n$ linearly independent binding hyperplanes.  (Remember that a hyperplane in $\mathbb{R}^n$ is a linear equation in $n$ variables.)  Then the maximum number of linearly independent hyperplanes binding at ${\bf x}'$ is $r < n$.  Let $G$ be the $r \times n$ matrix of constraint coefficients, and let $G{\bf x}' = {\bf g}$.  Since the rank of $G$ is $r$, the nullspace of $G$ has dimension $n-r > 0$.  Thus there exists some ${\bf d} \neq {\bf 0}$ such that $G{\bf d} = {\bf 0}$.  Therefore $G({\bf x}' + {\bf d}) = G({\bf x}' - {\bf d}) = {\bf g}$, which means that both ${\bf d}$ and $-{\bf d}$ are feasible directions at ${\bf x}'$ (i.e., for a sufficiently small distance from ${\mathbf x}'$ in these directions we're still in $P$, as the only constraint hyperplanes to worry about sufficiently close to ${\bf x}'$ are those that are binding at ${\bf x}'$).  Then there exists $\epsilon > 0$ such that ${\bf x}_1 = {\bf x}' + \epsilon {\bf d}$ and ${\bf x}_2 = {\bf x}' - \epsilon {\bf d}$ are both in $P$.  Since ${\bf x}' = \frac{1}{2} {\bf x}_1 + \frac{1}{2} {\bf x}_2$, ${\bf x}'$ cannot be an extreme point.  
Thus if ${\bf x}'$ is an extreme point, it must lie on $n$ linearly independent binding hyperplanes.
