QUESTION
Consider the positive integers, $N,T,J$, where $N>T$, $N>J$. $1_{a}$ 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 $NT\times 1$ vector; $D$ is an $NT\times N$ matrix; $F$ is an $NT\times J$ 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. I'm asking your help, because perhaps you have already seen linear systems with the same structure and hence you may be able to advise.
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 $T$ 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 $T\times J$.
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 $N=4$, $T=2$, $J=3$, we could have $$ F\equiv \Big\{\underbrace{\begin{pmatrix} 0.3 & 0 & 0\\ 0 & 0.1 & 0\\ \end{pmatrix}}_{F_1}, \underbrace{\begin{pmatrix} 0.7 & 0 & 0\\ 0.2 & 0 & 0\\ \end{pmatrix}}_{F_2}, \underbrace{\begin{pmatrix} 0 & 0 & 0.5\\ 0 & 0.3 & 0\\ \end{pmatrix}}_{F_3}, \underbrace{\begin{pmatrix} 0 & 0 & 0.6\\ 0 & 0 & 0.4\\ \end{pmatrix}}_{F_4}\Big\} $$ 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 $T\times N$.
Property 4: For each $i\in \{1,...,N\}$, the submatrix $D_i$ has its $i$-th column equal to $1_{T}$ and all the other elements equal to zero.
For example, for $N=4$ and $T=2$, we have $$ D\equiv \Big\{\underbrace{\begin{pmatrix} 1 & 0 & 0 & 0\\ 1 & 0 & 0& 0\\ \end{pmatrix}}_{D_1}, \underbrace{\begin{pmatrix} 0 & 1 & 0& 0\\ 0 & 1 & 0& 0\\ \end{pmatrix}}_{D_2}, \underbrace{\begin{pmatrix} 0 & 0 & 1& 0\\ 0 & 0 & 1& 0\\ \end{pmatrix}}_{D_3}, \underbrace{\begin{pmatrix} 0 & 0 & 0&1\\ 0 & 0 & 0&1\\ \end{pmatrix}}_{D_4}\Big\} $$
Given, the properties above, $D'D$ and $F'F$ are diagonal matrices with strictly positive diagonal elements, where
$$ \underbrace{D'D}_{N\times N}= \begin{pmatrix} T & & \\ & \ddots & \\ & & T \end{pmatrix}, \underbrace{F'F}_{J\times J}=\begin{pmatrix} \sum_{k=1}^{NT}F(k,1)^2 & & \\ & \ddots & \\ & & \sum_{k=1}^{NT}F(k,J)^2 \end{pmatrix} $$
Further, $$ F'D=\begin{pmatrix} \sum_{t=1}^T F(1,t) & \sum_{t=T+1}^{2T} F(1,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(1,t)\\ \vdots & \vdots & \vdots & \vdots\\ \sum_{t=1}^T F(J,t) & \sum_{t=T+1}^{2T} F(J,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(J,t)\\ \end{pmatrix} $$
As I note in my comment, we can write $$ W = \pmatrix{D & F}' \pmatrix{D & F}. $$ As such, $W$ is invertible if and only if the columns of $\pmatrix{D & F}$ are linearly independent.
Now, rearrange the rows of the matrix $\pmatrix{D & F}$ such that we end up with a matrix of the form $$ M_1 = \pmatrix{I_N & G_1\\ I_N & G_2\\ \vdots & \vdots \\ I_N & G_T}, $$ where $I_N$ denotes the size $N$ identity matrix. For the example that you gave, we could end up with $$ M_1 = \left[\begin{array}{cccc|ccc}1&0&0&0& 0.3 & 0 & 0\\ 0 & 1 &0 & 0 & 0.7 & 0 & 0\\ 0 & 0 & 1& 0 & 0 & 0 & 0.5\\ 0 & 0 & 0 & 1 & 0 & 0 & 0.6\\ \hline 1 & 0 & 0 & 0 & 0 & 0.1 & 0\\ 0 & 1 & 0 & 0 & 0.2 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0.3 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0.4 \end{array}\right]. $$ Now, apply row operations on the matrix $\pmatrix{D&F}$. In particular, subtract the first block-row from each of the others to get the matrix $$ M_2 = \pmatrix{I_N & G_1\\ 0 & G_2 - G_1\\ \vdots & \vdots \\ 0 & G_T - G_1}. $$ We see that $M_2$ has linearly independent columns if and only if the smaller matrix $$ M_3 = \pmatrix{G_2 - G_1\\ \vdots\\ G_T - G_1} $$ has linearly independent columns.
In terms of the original problem, we can reach the following conclusion. For each of the blocks of $F$, obtain the matrix $H_i$ by removing the frist row and subtracting the first row of $F_i$ from the other rows. For your example, we would have $$ H_1 = \pmatrix{-0.3 & 0.1 & 0}, \dots, H_4 = \pmatrix{0 & 0 & -0.2}. $$ $W$ will be invertible if and only if the block-matrix $H = \{H_1,\dots,H_4\}$ has linearly independent columns. We can see this by noting that $H$ can be attained by rearranging the rows of the matrix $M_3$ obtained via the the process described above.