Matrices with $M\binom ab\not<\binom 11$ Let $Q:=\{(x,y)\colon\max\{x,y\}<1\}$ and
$Q_0:=\{(x,y)\colon\max\{x,y\}\le1\}$. Also, let $\Gamma:=\mathbb N^2$.
Is there any comprehensible description of the set of all real square matrices $M$ of order $2$ such that $M\binom{1}{0}\in Q_0$, $M\binom{0}{1}\in Q_0$, and $M\Gamma$ is disjoint from $Q$, with the possible exception of the vector $M\binom{1}{1}$ which can be in $Q$?
As an example, $M$ has this property given that each of the two column sums of $M$ is $2/3$ at least. Another example:
 $$ M=\begin{pmatrix} 1 & -1 \\ -1/2 & 1 \end{pmatrix}. $$
In contrast, if all elements of $M$ are non-positive, then $M$ does not have the property in question.
 A: I think something analogous to your conjecture in your other question is correct.
For simplicity let:

*

*$\mathbb{N} = \{1, 2, 3, \dots\}$ and $\mathbb{N}_0 = \{0, 1, 2, 3, \dots\}$


*$A = \{ [0, 1]^T, [1, 0]^T \}, B = \{[1, 2]^T, [2 ,1]^T, [2, 2]^T\}, C = \{[p, q]^T \in \mathbb{N}^2 : p + q \ge 3\}$

Claim: Consider $M \in \mathbb{R}_{2 \times 2}$.  We have:
$$ (\forall v \in A: M v \in Q_0) \cap (\forall v \in B: M v \notin Q) \implies \forall v \in C: Mv \notin Q$$

In other words, to check if $M$ has the desired property all you need to check are the five points in $A$ and $B$.  IMHO this qualifies as a "comprehensible description" but that's for you to judge.  :)
Proof:
Suppose $M$ is invertible, i.e. its two columns are linearly independent.  We will interpret a column vector as a point in $(x,y)$ plane.  The points $\{ Mv : v \in  \mathbb{N}_0^2\}$ form a grid on the plane; to be more specific, it forms a quadrant of an infinite grid.  We are interested in $M$ s.t. the points $MA$ all $ \in Q_0$ and the points $MC$ all $\notin Q$.
The problem becomes much easier to visualize if we transform the plane and use the columns of $M$ as basis.  In this transformed space, the grid points are the integral lattice points $\mathbb{N}_0^2$ and now $Q$ and $Q_0$ are transformed to some other shape.  But what are these shapes?  The constraint $x \le 1$ transforms into a half-plane not through the origin, and same for the constraint $y \le 1$.  Thus $Q$ and $Q_0$ (the transformed versions) can be characterized by drawing two arbitrary intersecting lines, neither through the origin, and then picking one of the four quadrants.  $Q$ is the interior of the chosen quadrant and $Q_0$ includes the boundary.  The important thing (indeed the only important thing) is that the transformed $Q, Q_0$ are convex.
The claim is now geometrically intuitive.  The points are as follows:
y-axis
7 . C C C C C C C
6 . C C C C C C C
5 . C C C C C C C
4 . C C C C C C C
3 . C C C C C C C
2 . B B C C C C C
1 A . B C C C C C
0 . A . . . . . .
  0 1 2 3 4 5 6 7 x-axis

Since the two points $A \in Q_0$, if any of the points  $C \in Q$ then the triangle formed by these three points must include one of the $B$ points, and since $C$ is in the interior $Q$, the included $B$ is also in the interior $Q$, which is a contradiction.
(To be more clear: for any $C$ along the main diagonal $(x=y)$ the triangle would include $[2, 2]^T$, and for any $C$ above the main diagonal $(y > x)$ the triangle would include $[1, 2]$.)
Finally we are left with the case where the two columns of $M$ are linearly dependent.  I actually have a bit of trouble with this case algebraically, but geometrically the transformed space is basically collapsed into one dimension and $Q$ is just a semi-infinite interval, so it all "should" work...  Sorry, will think more about this degenerate case when I have more time.
A: Here's what I have so far; it's not complete and I don't think I will complete it.
Denote by $A$ the set of matrices satisfying your conditions, and $Z:=\{(a,b)\mid a\geq 1,b\geq1,(a,b)\neq(1,1)\}$. For $M$ a matrix, we will denote the coefficients by
$$\begin{pmatrix}
m_{11} & m_{12} \\ m_{21} & m_{22}
\end{pmatrix}$$
Let $J$ be the matrix of all $1$'s.
If $M\in A$ and $M<M'$ ($m'_ij>m_ij$ for all $i,j$), then $M'\in A$. This is because we test $M'$ on positive vectors. Moreover, $A$ is stable by permutation of rows and/or columns.
It suffices to find $M\in A$ such that there exists $(a_0,b_0)\in Z$ such that $m_{11}a_0+m_{12}b_0=1$. Indeed, denoting this subset by $A'$, we have $A=A'+\Bbb R_+J$ (the main argument is that $A$ is stable by $\Bbb R_+J$).
Let $M\in A'$. Then there exists $(a_0,b_0)\in Z$ such that $m_{11}a_0+m_{12}b_0=1$, so there exists $t\in\Bbb R$ such that $m_{11}=t\frac1{a_0}$ and $m_{12}=(1-t)\frac1{b_0}$. Reciprocally, if $M\in A$ such that $m_{11}$ and $m_{12}$ have this form then $M\in A'$.
Let $M$ be a matrix with $m_{11}$ and $m_{12}$ of the previous form. Then $M\in A$ iff $M\in A'$ iff
$$\forall (a,b)\in Z,\quad t\frac a{a_0}+(1-t)\frac{b}{b_0}<1\Rightarrow m_{21}a+m_{22}b\geq 1.$$
Repeating the previous reasoning, it suffices to find the matrices $M\in A$ of the form
$$\begin{bmatrix}
t\frac1{a_0} & (1-t)\frac1{b_0} \\
u\frac1{a_1} & (1-u)\frac1{b_1}
\end{bmatrix}$$
where $t\frac{a_1}{a_0}+(1-t)\frac{b_1}{b_0}<1$. The "only" thing left to check is that $u$ satisfies
$$\forall (a,b)\in Z,\quad t\frac a{a_0}+(1-t)\frac{b}{b_0}<1\Rightarrow u\frac{a}{a_1}+(1-u)\frac{b}{b_1}\geq1.\tag{$*$}$$
The reason I put quotes around "only" is because this is where I got stuck. If we can get some necessary and sufficient condition on $u$ for it to satisfy $(*)$, say $\mathcal P_{a_0,b_0,t,a_1,b_1}(u)$, then we can write
$$A=\left\{\begin{bmatrix}
t\frac1{a_0} & (1-t)\frac1{b_0} \\
u\frac1{a_1} & (1-u)\frac1{b_1}
\end{bmatrix}\left|\begin{array}{l}(a_0,b_0)\in Z,t\in\Bbb R, \\
(a_1,b_1)\in Z,t\frac{a_1}{a_0}+(1-t)\frac{b_1}{b_0}<1\\
u\in\Bbb R,\mathcal P_{a_0,b_0,t,a_1,b_1}(u)\end{array}\right.\right\}+\Bbb R_+J.$$
My conclusion: either we can calculate this property, and we get a messy characterization of $A$, or there's a simpler strategy I didn't see.
