$$ \newcommand{\bsth}{{\boldsymbol\theta}} \newcommand{\nptime}{\textsf{NP}} \newcommand{\ptime}{\textsf{P}} \newcommand{\disteq}{\overset{d}{=}} %linalg \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\detmat}[1]{\begin{vmatrix} #1 \end{vmatrix}} \newcommand{\spanb}[1]{\text{span}\{ #1 \}} \DeclareMathOperator{\conv}{conv} % convex hull \DeclareMathOperator{\cone}{cone} \DeclareMathOperator{\vectorize}{vec} \DeclareMathOperator{\matricize}{mat} \DeclareMathOperator{\adj}{adj} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator*{\proj}{proj} % brackets, norms, cardinalities \newcommand{\pa}[1]{ \left({#1}\right) } \newcommand{\ha}[1]{ \left[{#1}\right] } \newcommand{\ca}[1]{ \left\{{#1}\right\} } \newcommand{\inner}[1]{\left\langle #1 \right\rangle} \newcommand{\innercpy}[1]{\inner{ #1, #1 }} \newcommand{\norm}[1]{\left\| #1 \right\|} \newcommand{\abs}[1]{\left|{#1}\right|} \newcommand{\card}[1]{\left\vert{#1}\right\vert} % math vectors \newcommand{\va}{\textbf{a}} \newcommand{\vb}{\textbf{b}} \newcommand{\vc}{\textbf{c}} \newcommand{\vd}{\textbf{d}} \newcommand{\ve}{\textbf{e}} \newcommand{\vf}{\textbf{f}} \newcommand{\vg}{\textbf{g}} \newcommand{\vh}{\textbf{h}} \newcommand{\vi}{\textbf{i}} \newcommand{\vj}{\textbf{j}} \newcommand{\vk}{\textbf{k}} \newcommand{\vl}{\textbf{l}} \newcommand{\vm}{\textbf{m}} \newcommand{\vn}{\textbf{n}} \newcommand{\vo}{\textbf{o}} \newcommand{\vp}{\textbf{p}} \newcommand{\vq}{\textbf{q}} \newcommand{\vr}{\textbf{r}} \newcommand{\vs}{\textbf{s}} \newcommand{\vt}{\textbf{t}} \newcommand{\vu}{\textbf{u}} \newcommand{\vv}{\textbf{v}} \newcommand{\vw}{\textbf{w}} \newcommand{\vx}{\textbf{x}} \newcommand{\vy}{\textbf{y}} \newcommand{\vz}{\textbf{z}} \newcommand{\vzero}{\textbf{0}} \newcommand{\vone}{\textbf{1}} \newcommand{\valpha}{{\boldsymbol\alpha}} \newcommand{\vepsilon}{{\boldsymbol\epsilon}} \newcommand{\vnu}{{\boldsymbol\nu}} \newcommand{\vpi}{{\boldsymbol\pi}} \newcommand{\veta}{{\boldsymbol\eta}} \newcommand{\vsigma}{ {\boldsymbol\sigma}} \newcommand{\vbeta}{ {\boldsymbol\beta}} \newcommand{\vtheta}{ {\boldsymbol\theta}} \newcommand{\vdelta}{ {\boldsymbol\delta}} \newcommand{\vlambda}{ {\boldsymbol\lambda}} \newcommand{\vmu}{ {\boldsymbol\mu}} % common math sets \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\T}{\mathbb{T}} % limits \def\sumn{\sum_{n=0}^\infty} \def\limn{\lim_{n\rightarrow\infty}} \def\prodn{\prod_{n=0}^\infty} % mathcal \newcommand{\mcA}{\mathcal{A}} \newcommand{\mcB}{\mathcal{B}} \newcommand{\mcC}{\mathcal{C}} \newcommand{\mcD}{\mathcal{D}} \newcommand{\mcE}{\mathcal{E}} \newcommand{\mcF}{\mathcal{F}} \newcommand{\mcG}{\mathcal{G}} \newcommand{\mcH}{\mathcal{H}} \newcommand{\mcI}{\mathcal{I}} \newcommand{\mcJ}{\mathcal{J}} \newcommand{\mcK}{\mathcal{K}} \newcommand{\mcL}{\mathcal{L}} \newcommand{\mcM}{\mathcal{M}} \newcommand{\mcN}{\mathcal{N}} \newcommand{\mcO}{\mathcal{O}} \newcommand{\mcP}{\mathcal{P}} \newcommand{\mcQ}{\mathcal{Q}} \newcommand{\mcR}{\mathcal{R}} \newcommand{\mcS}{\mathcal{S}} \newcommand{\mcT}{\mathcal{T}} \newcommand{\mcU}{\mathcal{U}} \newcommand{\mcV}{\mathcal{V}} \newcommand{\mcW}{\mathcal{W}} \newcommand{\mcX}{\mathcal{X}} \newcommand{\mcY}{\mathcal{Y}} \newcommand{\mcZ}{\mathcal{Z}} % distribs, probability \newcommand{\disteq}{\overset{d}{=}} \newcommand\independent{\perp \!\!\! \perp} \DeclareMathOperator{\Laplace}{Laplace} \DeclareMathOperator{\Poisson}{Poisson} \DeclareMathOperator{\Exponential}{Exponential} \DeclareMathOperator{\Multinomial}{Multinomial} \DeclareMathOperator{\Bernoulli}{Bernoulli} \DeclareMathOperator{\Categorical}{Categorical} \DeclareMathOperator{\Uniform}{Uniform} \DeclareMathOperator{\Binomial}{Binomial} \DeclareMathOperator{\Hypergeometric}{Hypergeometric} \DeclareMathOperator{\GammaDist}{Gamma} \DeclareMathOperator{\NegativeBinomial}{NegativeBinomial} \DeclareMathOperator\sub{sub} \renewcommand{\d}[1]{\mathop{\mathrm{d} #1 }} \newcommand{\dkl}[2]{\mathop{D_\mathrm{KL}}\left({#1}\;\middle\|\;{#2}\right)} \newcommand{\sg}{\mathop{\mathrm{SG}}} \newcommand{\se}{\mathop{\mathrm{SE}}} %operators \DeclareMathOperator{\power}{{\mathcal{P}}} \DeclareMathOperator{\var}{var} \DeclareMathOperator{\cov}{cov} \DeclareMathOperator\mathProb{\mathbb{P}} \DeclareMathOperator\mathExp{\mathbb{E}} \DeclareMathOperator*\mathExpUnder{\mathbb{E}} \DeclareMathOperator*\fat{fat} \renewcommand{\P}{\mathProb} % need to overwrite stupid paragraph symbol \newcommand{\E}{\mathExp} % need to overwrite stupid paragraph symbol \newcommand{\set}[2]{ \left\{ #1 \,\middle|\, #2 \right\} } \newcommand{\CE}[2]{ \mathExp\left[ #1 \,\middle|\, #2 \right] } \renewcommand{\CP}[2]{ \mathProb\left\{ #1 \,\middle|\, #2 \right\} } $$
A First Attempt
Let $\mathbf{x}^\top=\begin{pmatrix}\mathbf{x}_0&\cdots&\mathbf{x}_n\end{pmatrix}$ for conformal $\mathbf{x}_i$ of sizes $s_0,s,\cdots,s$. Inspecting the $i$-th block of $\mathbf{x}=\mathbf{1}+ P\mathbf{x}$ yields for $1<i<n$ that \(\mathbf{x}_{i}=\mathbf{1}_s+U\mathbf{x}_{i-1}+C\mathbf{x}_{i}+V^\top \mathbf{x}_{i+1}\,\,,\) which can be rewritten as \(\mathbf{x}_{i}=(I-C)^{-1}(\mathbf{1}_s+U\mathbf{x}_{i-1}+V^\top \mathbf{x}_{i+1})\,\,,\) since $\rho(C)<1$ is implied by $\rho(P)<1$, as $C$ is the restriction of $P$ to the subspace for the $i$-th block, so $I-C$ is also nonsingular.
By itself, expressing the $i$-th block in terms of the $(i-1)$-st and $(i+1)$-st isn’t too helpful: such a recurrence doesn’t bottom out (at least not obviously). Naively, if $U$ was somehow nonsingular then we could express $\mathbf{x}{i-1}$ in terms of $\mathbf{x}_i,\mathbf{x}{i+1}$, which does clearly terminate.
Writing $U=L\Sigma R^\top$ as the full SVD for unitary $s\times s$ matrices $L, R$, and $\Sigma=\begin{pmatrix}
I_r & 0
0 & 0{s-r}
\end{pmatrix}$ for the rank $r$ of $U$, let’s multiply each block of the equation $(I-P)\mathbf{x}=\mathbf{1}$ by $L^\top$ on the left and $R$ on the right. Note we don’t need to treat the edge case $r=0$ differently. Letting $\mathbf{y}_i= R^\top \mathbf{x}_i$ for every block $n\ge i >1$ we are left with for every $i$,
\(\begin{gather*}
&-L^\top L\Sigma R^\top R\mathbf{y}_{i-1}+
L^\top(I-C)R\mathbf{y}_i-
L^\top V^\top R\mathbf{y}_{i+1}\\
=&\begin{pmatrix}
-I_r & 0\\
0 & 0
\end{pmatrix}\mathbf{y}_{i-1} + A\mathbf{y}_i+B\mathbf{y}_{i+1}\\
=&L^\top\mathbf{1}\,\,.
\end{gather*}\)
Above, $A,B$ are suitably defined, where we note that since $L,R$ are unitary, $A$ is nonsingular, and we let $\mathbf{y}{n+1}=\mathbf{0}$ for the last equation.
Now, to consider fully determined systems, we need to think about re-indexing. Our current block indexing is by $s_0,s,\cdots,s$:
\(\mathbf{y}^\top=\begin{pmatrix}\mathbf{x}_0&\mathbf{y}_1&\cdots&\mathbf{y}_{n-1}&\mathbf{x}_n & \mathbf{0}_s\end{pmatrix}\,\,,\)
but consider instead blocking by $s_0 + r, s, \cdots s$ by extending with more zeros so that
\((\mathbf{y}')^\top=\begin{pmatrix}\mathbf{y}_0'&\mathbf{y}_1'&\cdots&\mathbf{y}_{n-1}'&\mathbf{y}_n' & \mathbf{0}_s\end{pmatrix}\,\,,\)
where $(\mathbf{y}0’)^\top =\mathbf{x}_0^\top|(\mathbf{y}_1){1:r}^\top$, $(\mathbf{y}{n-1}’)^\top =(\mathbf{y}{n-1}){r+1:s}^\top|(\mathbf{x}_n){1:r}^\top$, $(\mathbf{y}{n}’)^\top =(\mathbf{x}{n}){r+1:s}^\top|\mathbf{0}_r^\top$. All this hard work yields the shifted equations
\(\begin{pmatrix}
A_{s-r,r} & A_{s-r}\\
-I_r & 0
\end{pmatrix}\mathbf{y}_{i-1}' + \begin{pmatrix}
B_{s-r,r} & B_{s-r}\\
A_{r} & A_{r,s-r}
\end{pmatrix}\mathbf{y}_i'+\begin{pmatrix}
0 & 0\\
B_{r} & A_{r,s-r}
\end{pmatrix}\mathbf{y}_{i+1}'
=\begin{pmatrix}
(L^\top)_{s-r,s}\\(L^\top)_{r, s}
\end{pmatrix}\mathbf{1}\,\,,\)
where we conformably break the $s\times s$ matrix $A=\begin{pmatrix}
A_r & A{r,s-r}
A_{s-r,r} & A_{s-r}
\end{pmatrix}$, and similarly for others, e.g. $L^\top=\begin{pmatrix}
(L^\top){r, s}
(L^\top){s-r,s}
\end{pmatrix}$.
For all the marbles, we ask whether $\begin{pmatrix}
A_{s-r,r} & A_{s-r}
-I_r & 0
\end{pmatrix}$ is invertible. I haven’t figured this out yet (TODO). But, assuming it is, then we can solve the linear system above for $\mathbf{y}{i-1}’=T(\mathbf{y}_i,\mathbf{y}{i+1})$.
TODO make it square
\[(\mathbf{y}_0,\mathbf{x}_1,\mathbf{x}_2)=T_0(\mathbf{x}_2,\mathbf{x}_3, \mathbf{x}_4,\mathbf{1}_s)=T_0T^{n-1}(\mathbf{x}_{n-1},\mathbf{x}_n,\mathbf{1}_s)\,\,,\]TODO exponentiate as above
Backsolve x