Uniqueness of the solution of a linear system of equations 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,
$$
F'D=\begin{pmatrix}
\sum_{t=1}^T F(1,t) & \sum_{t=T+1}^{2T} F(1,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(1,t)\\
\vdots & \vdots & \vdots & \vdots\\
\sum_{t=1}^T F(J,t) & \sum_{t=T+1}^{2T} F(J,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(J,t)\\
\end{pmatrix}
$$
 A: 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.
