Congruence Property of Monotone Operators A map $T$ is called strictly monotone if for $x\ne y$, $\langle u-v,x-y\rangle>0$ for all $u\in T(x),v\in T(y)$. Let $A$ be an $m\times n$ matrix and $b\in\mathbb R^m$. I want to prove that if $T:\mathbb R^n\rightrightarrows\mathbb R^m$ is strictly monotone and $\text{rank}\;A=n$, then $S:=A^TT(Ax+b)$ is also strictly monotone. Let $u\in S(x)$ and $v\in S(y)$, then we have $$u=A^Tu',v=A^Tv',u'\in T(Ax+b),v'\in T(Ay+b)$$ and $$\langle u-v,x-y\rangle=\langle A^Tu'-A^Tv',x-y\rangle=\langle u'-v',Ax-Ay\rangle=\langle u'-v',Ax+b-(Ay+b)\rangle>0$$ since $T$ was strictly monotone and if $x\ne y$ then $Ax+b\ne Ay+b$, but I don't know why the condition that $\text{rank}\;A=n$ is necessary here?

If $rank(A) < n$, the strictness of $S$ may be lost, since it is now possible that $x \
e y$ but $S(x) \cap S(y) \
e \emptyset$. That is because there is $w \
e 0$ such that $Aw = 0$. You should be able to construct a counterexample to strict monotonicity using that information.