Let $\mathbf{1}$ denote an $n\times 1$ vector with all entries equal to 1. Given an $n\times n$ matrix A with *strictly positive entries*, and non-negative *diagonal* matrices $D_1$ and $D_2,$ evaluate vectors $\mathbf{v}$ and $\mathbf{u}$ given by:

\begin{align} D_1AD_2\mathbf{1} & =:\mathbf{v}\ ;\\ \mathbf{1}^{\mathrm{T}}D_1AD_2 & =:\mathbf{u}^{\mathrm{T}}\ . \end{align}

Given the positive entried matrix $A$ and the vectors $\mathbf{v},\mathbf{u},$ is it true that the pair of non-negative diagonal matrices $D_1,D_2$ for which the above two equalities hold is unique up to scaling?

Clearly, we can choose for any $\alpha>0,$ $D_1^\prime = \alpha D_1$ and $D_2^\prime = \frac{1}{\alpha}D_2.$ Are these all the solutions for $(D_1^\prime,D_2^\prime)$?

More precisely, we have a specific system of $(2n-1)$ polynomials (one for each entry of $\mathbf{v}$ and $\mathbf{u},$ however one of them is redundant, since sum of row sums = sum of column sums). Each polynomial has degree $2$ and there are $2n$ variables (the diagonal entries of $D_1$ and $D_2$). I am asking if the dimension of the affine algebraic set that is the zero-locus of this system of polynomials is exactly 1. (Please note that the fact that all entries of $A$ are strictly positive is crucial. As a bad example, take A as the zero matrix.)