How do I find the vertices of a polygon? Exercise: Denote $[n] = \{1,2,\ldots,n\}$. Let $P\subseteq \mathbb{R}^{m\times n}$ be the feasible region of the system $$\begin{split}x\geq &\, 0\\x_{ij}\leq&\, 1 \text{ for all $i\in[m]$ and $j\in[n]$}\\x_{ij} + x_{i'j} + x_{ij'} + x_{i'j'} \leq&\,3 \text{ for all $i,i'\in[m]$ and $j,j'\in[n]$ with $i\neq i'$ and $j\neq j'$}\end{split}$$
Now consider the case $m=n=3$. Show that $$x = \begin{bmatrix}\frac{3}{4}&\frac{3}{4}&\frac{3}{4}\\\frac{3}{4}&\frac{3}{4}&\frac{3}{4}\\\frac{3}{4}&\frac{3}{4}&\frac{3}{4}\end{bmatrix}$$ is a vertex of $P$.
What I've tried: It makes a lot of sense to me for $x$ to be a vertex of $P$, since increasing any element of $x$ would mean that $x$ was no longer a member of $P$. I have no clue whatsoever how to show that my intuition is correct though.
Question: How do I show that $x$ is a vertex of $P$?
Thanks in advance!
EDIT: Apparently since $x\in\mathbb{R}^{3\times 3}$, I have to show that $x$ satisfies $9$ equalities that are allowed in $P$, because this would mean that $x$ is a vertex of $P$. I don't know which inequalities/equalities I have to look for.
 A: An equivalent notion to a point being a vertex is that the point is a basic feasible solution. The definition (from Wikipedia):

For a polyhedron $P$ and a vector $\mathbf{x}^*\in\mathbb{R}^n$, $\mathbf{x}^*$ is a basic solution if:
  
  
*
  
*All the equality constraints defining $P$ are active at $\mathbf{x}^*$
  
*Of all the constraints that are active at that vector, at least $n$ of them must be linearly independent. Note that this also means that at least $n$ constraints must be active at that vector.
  
  
  A constraint is active for a particular solution $\mathbf{x}^*$ if it is satisfied at equality for that solution.
A basic solution that satisfies all the constraints defining $P$, or in other words, one that lies within $P$ is called a basic feasible solution.

So it suffices to show that proposed point satisfies $9$ linearly independent constraints with equality. Clearly, it doesn't satisfy any of the constraints $x_{ij}\geqslant0$ or $x_{ij}\leqslant1$ with equality, leaving only the third class of equalities. In fact, there are only 9 constraints in this class:
$$
\underbrace{\begin{bmatrix}
1&1&&1&1&&&&\\
1&&1&1&&1&&&\\
&1&1&&1&1&&&\\
1&1&&&&&1&1&\\
1&&1&&&&1&&1\\
&1&1&&&&&1&1\\
&&&1&1&&1&1&\\
&&&1&&1&1&&1\\
&&&&1&1&&1&1\\
\end{bmatrix}}_{\displaystyle=A}
\begin{bmatrix}
x_{11}\\x_{12}\\x_{13}\\x_{21}\\x_{22}\\x_{23}\\x_{31}\\x_{32}\\x_{33}
\end{bmatrix}
\leqslant\begin{bmatrix}3\\3\\3\\3\\3\\3\\3\\3\\3\end{bmatrix}
$$
It is easy to verify that the proposed point satisfies each of these with equality, since
$$\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=3.$$
Furthermore, the matrix $A$ above is full-rank, so all of the constraints are linearly independent. Since the proposed point satisfies $n=9$ linearly independent constraints with equality and is feasible, it is a basic feasible solution, and thus a vertex.
