Bounding spectral radius of special matrix (extension of the extension)

Question

This is an extension of Bounding spectral radius of special matrix (extension), which has been already solved.

Let $A$ be an $n \times n$ matrix with all nonnegative entries and row sums strictly less than one, let $V$ be an $n \times n$ nonnegative diagonal matrix satisfying $V \leq I$ (entrywise), let $D_0$ be an $n \times n$ nonnegative diagonal matrix such that $I \leq D_0 < I + \mathrm{diag} \left\{ \iota - A \iota \right\} \left( \mathrm{diag} \left\{ A \iota \right\} \right)^{-1}$, let $X$ be a vector in the $n$-dimensional simplex (i.e., $x_j \geq 0,\sum_j^n x_j=1$), let $D_1$ and $D_2$ be two strictly positive diagonal $n \times n$ matrices. Define $B \equiv \left(I - AV\right)^{-1}$, $B_{1} \equiv \left(I - D_0 AV\right)^{-1}$ and $B^{*}_{1} \equiv\left(I - D_0 A\right)^{-1}$. Finally, let $$M \equiv \left(\mathrm{diag}\left\{ B^{T}X\right\} \right)^{-1}B^{T}\left[V\mathrm{diag}\left\{ X\right\} +\left(I-V\right)\mathrm{diag}\left\{ B^{T}X\right\} \right]D_{1} B_{1} D_{2}.$$ I want to show that the spectral radius of $M$ is less than or equal to one, $\rho(M)\leq 1$, provided that the following condition holds $$\tag{1} D_{1} B^{*}_{1} D_{2} \iota\leq\iota.$$

Observe that if $D_0 = I$, then we get the same problem as in Bounding spectral radius of special matrix (extension).

Also, I used subscript $1$ for matrix $B_{1}^{*}$ to distinguish it from matrix $B^{*} \equiv \left(I - A\right)^{-1}$ that might be useful for a proof.

Update on May 25, 2022: Changed notation and, importantly, dropped dependence of matrix $D_{2}$ on $D_{0}$. We don't need this dependence: numerically $\rho \left(M \right) < 1$ with independent $D_{2}$ and $D_{0}$.

Update on Sep 22, 2022: I thought that there might be monotonicity of the spectral radius with respect to $D_{0}$. In particular, I thought that once the dependence of $D_{2}$ on $D_{0}$ is dropped, then $\rho \left(M \right)$ is falling as we increase diagonal elements of $D_{0}$. But this is not true.

Answer 1

This is not an answer, but some thoughts.

Consider the case when $D_{2}$ has only one positive diagonal entry, and all other diagonal entries equal to $0$ -- similar to what we did here.

First, as in here, it is enough to consider the unit vectors $X = e_{j}$ for $j = 1,\dots,n$. We have $\mathrm{diag} \left\{ B^{T} e_{j} \right\} = \mathrm{diag} \left\{ b_{j1}, \dots, b_{jn} \right\}$.

Consider any $e_{j}$. Assume that $d_{2,i} > 0$ for some $i$ and $d_{2,k} = 0$ for $k \neq i$. Then $\rho(M)$ is given by \begin{align*} \rho(M) = \dfrac{b_{ji}v_{j}d_{1,j}b_{1,ji}d_{2,i}+\sum_{k}b_{ki}\left(1-v_{k}\right)b_{jk}d_{1,k}b_{1,ki}d_{2,i}}{b_{ji}}, \end{align*} and condition (1) is given by $d_{1,k} b^{*}_{1, ki} d_{2,i} \leq 1$ for $k = 1,\dots, n$. This gives $d_{1,k} \leq \left( b^{*}_{1, ki} d_{2,i} \right)^{-1}$ and \begin{align*} \rho(M) \leq \dfrac{b_{ji}v_{j} \left[ b^{*}_{1,ji} \right]^{-1} b_{1,ji} + \sum_{k}b_{ki}\left(1-v_{k}\right)b_{jk} \left[ b^{*}_{1,ki} \right]^{-1} b_{1,ki} }{b_{ji}}. \end{align*} If we can show that the right-hand side of the above is not larger than $1$, then we would be done. So, we want to show that \begin{align*} v_{j} b_{ji} \dfrac{ b_{1,ji} }{ b^{*}_{1,ji} } + \sum_{k} \left(1-v_{k}\right) b_{ki} b_{jk} \dfrac{ b_{1,ki} }{ b^{*}_{1,ki} } \leq b_{ji} . \qquad (2) \end{align*}

I thought that the following is true: $b_{1,ki} \big/ b^{*}_{1,ki} \leq b_{ki} \big/ b^{*}_{ki}$ for all $k$ and $i$. However, it does not generally hold.