Is the solution to a specific system of linear equations bounded in the unit hypercube? Let $B \in \mathbb{R}^{m \times n}$, where $m \leq n$, be such that every entry is in $[0, 1]$ and each row sums to $1$. Let $A = BB' \in \mathbb{R}^{m \times m}$, except replace each number on the diagonal with $1$. For example, if $$B = \begin{pmatrix} 0.4 & 0.6 & 0 \\ 0 & 0.5 & 0.5 \end{pmatrix}$$ then $$BB' = \begin{pmatrix} 0.52 & 0.3 \\ 0.3 & 0.5 \end{pmatrix}, \qquad A = \begin{pmatrix} 1 & 0.3 \\ 0.3 & 1 \end{pmatrix}$$ Let $v = [1, \dots, 1]' \in \mathbb{R}^m$. Assuming that $A$ is invertible, then is it true that $A^{-1}v$ (the solution to $Ax = v$) is in $[0, 1] \times \dots \times [0, 1] \in \mathbb{R}^m$? Are there any results or theorems that I should look at that could help clarify my thinking?
I've performed some simulations and haven't been able to find a counterexample, but I also haven't been able to see exactly why this is true. Please do note that

*

*The definitions seem so contrived because I've created them for a real-world application in social science, and


*I'm pretty rusty in linear algebra.
 A: WORK IN PROGRESS.
As a starting point, we note that the claim is true for $m=2$. Indeed, in this case $BB^T$ is a symmetric $2\times 2$ matrix and so $A$ is of the form
$$
\begin{array}{cc} 
A=\begin{bmatrix}1 & b \\ b &  1 \end{bmatrix}, & b\in (0, 1).\end{array}$$
The fact that $b\in(0,1)$ follows from Cauchy-Schwarz. Indeed, $b$ is determined by the matrix $B\in\mathbb R^{2\times m}$; precisely,$b=B^{(1)}\cdot B^{(2)}$, where $B^{(j)}$ denotes the $j$-th row. By assumption, the entries of each $B^{(j)}$ are nonnegative and sum to $1$, so in particular $$\lvert B^{(j)}\rvert^2=\sum_k (b^j_k)^2\le (\sum_k b^j_k)^2=1.$$
The fact that $b<1$ immediately follows.
We conclude the $m=2$ case by direct computation:
$$
A^{-1}\begin{bmatrix} 1 \\ 1 \end{bmatrix} =\begin{bmatrix} \frac{1}{1+b} \\ \frac{1}{1+b}\end{bmatrix}$$
and indeed $\frac{1}{1+b}\in (0,1).$
PARTIAL PROGRESS IN THE GENERAL CASE.
We are working on the general case $m>2$ in comments. So far we reached the following point. The matrix $A$ reads
$$
A=\begin{bmatrix} 1 & a_{12} & a_{13}& \ldots \\ 
a_{12} & 1 & a_{23} & \ldots \\ 
\ldots & \ldots & \ldots&\ldots \end{bmatrix}$$
where $a_{ij}=B^{(i)}\cdot B^{(j)}$ as before. Considering a solution $x$ to $Ax=v$ we have, by definition of $v=(1,\ldots, 1)^T$,
$$\tag{1}
\begin{cases} 
x_1+a_{12}x_2+\ldots+a_{1m}x_m=1, \\ 
a_{12}x_1+x_2+\ldots+a_{2m}x_m=1, \\ 
\vdots \\ 
a_{1m}x_1+a_{2m}x_2+\ldots+x_m=1.
\end{cases}
$$
The goal is to prove that $x_j\in[0,1]$ for all $j\in\{1, \ldots, m\}$. It is actually enough to prove that $x_j\ge 0$, because then the bound $x_j\le 1$ would be immediate from (1), since all $a_{ij}$ are nonnegative.
