Let $T(x)=Ax + b$, suppose that $T$ is injective and have strictly positive values (ie values in the first orthant). Then, must $T$ be constant?
I was thinking of somehow using the unboundedness and the analyticity of $T$ that, up to a shift, $T$ has to be identical $0$ on a set connected open subset of its domain. However, this is as far as I've gotten, and I'm not sure the claim is true.
Unless the original space is zero-dimension, $T$ can't be injective if it has strictly positive values.
Assume $A \neq 0$. Then for some $i$, $j$, $a_{ij} \neq 0$. Then $T(\lambda e_i)_j = a_{ij}\lambda + b_j$ is a non-constant linear function of $\lambda$. Taing $\lambda = {-b_j - 1}{a_{ij}}$ we get $T(\lambda e_i)_j = -1$ - thus $T$ doesn't have strictly posiitve values.
And if $A = 0$ then of course $T$ is constant.