QUESTION
Consider the positive integers, $N,\{T_n\}_{n=1}^N,J$, where $N>\max\{T_1,...,T_N\}$, $N>J$. $1_{a}$ denotes the $a\times 1$ vector of ones.
Consider the system of equations $$ \begin{pmatrix} D'Y\\ F'Y \end{pmatrix}= \underbrace{\begin{pmatrix} D'D & D'F\\ F'D & F'F \end{pmatrix}}_{\equiv W} \beta $$ where $Y$ is an $\sum_{n=1}^N T_n\times 1$ vector; $D$ is an $\sum_{n=1}^N T_n\times N$ matrix; $F$ is an $\sum_{n=1}^N T_n\times J$ matrix; $\beta$ is an $(N+J)\times 1$ vector.
Could you help me to find some necessary (and, if possible, also sufficient) conditions such that the system has a unique solution wrto $\beta$. That is, such that the matrix $W$ is invertible?
I report below some properties of the system that I analyse.
Note: a recent question of mine (here) seems very similar. However, there is a key difference that prevents me from implementing the same answer: it is the fact that here the matrices $D$ and $F$ can be decomposed in blocks that do not have necessarily the same dimension. If you believe that the same answer can be applied, please advise on how.
SOME PROPERTIES OF THE SYSTEM
$Y$ does not have elements equal to zero. The matrices $F$ and $D$ have specific structures.
With regards to the matrix $F$:
Let me decompose $F$ in $N$ submatrices, but cutting it every $T_1,...,T_N$ rows. Then, with some abuse of notation, let me write $F$ as the collection of such $N$ submatrices, $F\equiv \{F_1,F_2,...,F_N\}$, where each submatrix $F_i$ has size $T_i\times J$.
Property 1: For each $i\in \{1,...,N\}$, the submatrix $F_i$ is such that in each row, one and only one element takes value in $(0,1)$ and all the other elements takes value zero.
Property 2: If the submatrix $F_i$ has each of its nonzero elements positioned in the same columns as the submatrix $F_k$, then it should be that $F_i=F_k$.
Construct the bipartite graph such the nodes in one side (hereafter, side 1) represent the column indices $j\in \{1,...,J\}$ and the the nodes in the other side (hereafter, side 2) represent the submatrix indices $i\in \{1,...,N\}$. Draw an edge between nodes $j\in \{1,...,J\}$ and $i\in \{1,...,N\}$ if at least one element of the $j$-th column of $F_i$ has a nonzero element.
Property 3: The aforementioned bipartite graph should be such that each node in side 1 is indirectly connected to at least another node in side 1. Further, the bipartite graph should be such that each node in side 2 is indirectly connected to at least another node in side 2. This is called a connected bipartite graph.
For example, for $N=4$, $T_1=2$, $T_2=3$, $T_3=1$, $T_4=4$, and $J=3$, we could have $$ F\equiv \Big\{\underbrace{\begin{pmatrix} 0.3 & 0 & 0\\ 0 & 0.1 & 0\\ \end{pmatrix}}_{F_1}, \underbrace{\begin{pmatrix} 0.7 & 0 & 0\\ 0.2 & 0 & 0\\ 0 & 0 & 0.1 \end{pmatrix}}_{F_2}, \underbrace{\begin{pmatrix} 0 & 0 & 0.5\\ \end{pmatrix}}_{F_3}, \underbrace{\begin{pmatrix} 0 & 0 & 0.6\\ 0 & 0 & 0.4\\ 0.2 & 0 & 0\\ 0 & 0.8 & 0\\ \end{pmatrix}}_{F_4}\Big\} $$ with bipartite graph
With regards to the matrix $D$:
As done for $F$, let me write $D\equiv \{D_1,D_2,...,D_N\}$, where each submatrix $D_i$ has size $T_i\times N$.
Property 4: For each $i\in \{1,...,N\}$, the submatrix $D_i$ has its $i$-th column equal to $1_{T_i}$ and all the other elements equal to zero.
For example, for $N=4$, $T_1=2$, $T_2=3$, $T_3=1$, and $T_4=4$, we have $$ D\equiv \Big\{\underbrace{\begin{pmatrix} 1 & 0 & 0 & 0\\ 1 & 0 & 0& 0\\ \end{pmatrix}}_{D_1}, \underbrace{\begin{pmatrix} 0 & 1 & 0& 0\\ 0 & 1 & 0& 0\\ 0 & 1 & 0& 0\\ \end{pmatrix}}_{D_2}, \underbrace{\begin{pmatrix} 0 & 0 & 1& 0\\ \end{pmatrix}}_{D_3}, \underbrace{\begin{pmatrix} 0 & 0 & 0&1\\ 0 & 0 & 0&1\\ 0 & 0 & 0&1\\ 0 & 0 & 0&1\\ \end{pmatrix}}_{D_4}\Big\} $$
Given, the properties above, $D'D$ and $F'F$ are diagonal matrices with strictly positive diagonal elements, where
$$ \underbrace{D'D}_{N\times N}= \begin{pmatrix} T_1 & & \\ & \ddots & \\ & & T_N \end{pmatrix}, \underbrace{F'F}_{J\times J}=\begin{pmatrix} \sum_{k=1}^{\sum_{n=1}^N T_n}F(k,1)^2 & & \\ & \ddots & \\ & & \sum_{k=1}^{\sum_{n=1}^N T_n}F(k,J)^2 \end{pmatrix} $$
Further, $$ F'D=\begin{pmatrix} \sum_{t=1}^{T_1} F(1,t) & \sum_{t=T_1+1}^{T_1+T_2} F(1,t) & ... \\ \vdots & \vdots & \vdots \\ \sum_{t=1}^{T_1} F(J,t) & \sum_{t=T_1+1}^{T_1+T_2} F(J,t) & ... \\ \end{pmatrix} $$
As in the earlier post, $W$ is invertible if and only if $[D\ \ F]$ has a trivial nullspace.
Extend the columns of $D$ to form an orthogonal basis for $\Bbb R^k$ (where $k = \sum_n T_n$). Let $P$ denote the matrix whose columns are these additional vectors. Note that we have $$ D'P = 0, \quad P'D = 0, $$ and both $D'D$ and $P'P$ are diagonal. Now, note that $[D\ \ F]$ has a trivial nullspace if and only if $[D\ \ P]'[D\ \ F]$ has a trivial nullspace. Compute $$ \pmatrix{D & P}' \pmatrix{D & F} = \pmatrix{D'D & D'F\\P'D & P'F} = \pmatrix{D'D & D'F\\0 & P'F}. $$ We can conclude that $W$ is invertible if and only if $P'F$ has a trivial nullspace.