I want to show that the matrix $A$ defined below is NOT full column rank. I have a proof but it does not look complete to me. Could you help to fix it?
Consider a matrix $A$ structured as follows $$ A=\begin{pmatrix} A_1\\ A_2\\ \vdots\\ A_N \end{pmatrix} $$ where, for each $n\in \{1,...,N\}$, $A_n$ is a $T_n\times (NK+KJ)$ matrix.
Consider some numbers $\{k_{nt}\}_{n=1,...,N, t=1,...,T_n}$ where each $k_{nt}\in \{1,...,K\}$.
Consider some numbers $\{j_{nt}\}_{n=1,...,N, t=1,...,T_n}$ where each $j_{nt}\in \{1,...,J\}$.
For each $n\in \{1,...,N\}$, every $t$-th row of the matrix $A_n$ is structured as follows:
The $\Big((n-1)K+k_{nt}\Big)$-th element is equal to $1$.
The $\Big(KN+(j_{nt}-1)K+k_{nt}\Big)$-th element is equal to some scalar $\delta_{n}>0$. Note here that $\delta_n$ is equal across rows of $A_n$.
All the other elements equal to zero.
Further, the matrix $A$ contains no zero columns and $\sum_{t=1}^{N} T_n \geq (NK+KJ)$.
MY PROOF:
Given the structure of $A$, we have that $A$ has full column rank if and only if the following system admits as unique solution $x=0_{(NK+KJ)\times 1}$: $$ \begin{aligned} x_{a_{nt}} + \delta_{n} x_{b_{nt}}=0\hspace{1cm} \quad & \text{$a_{nt}\equiv (n-1)K+k_{nt}$, } \text{$b_{nt}\equiv NK+(j_{nt}-1)K+k_{nt}$,}\\ & \text{for each $t\in \{1,...,T_n\}$ and $n\in \{1,...,N\}$}. \end{aligned} $$
Observe that $x_{a_{nt}}$ appears twice in the system if and only if there is $\tau\neq t\in \{1,...,T_n\}$ such that $k_{n\tau}=k_{nt}$. In this case, the system contains the equations $$ \begin{aligned} &x_{a_{nt}} + \delta_{n} x_{b_{nt}}=0,\\ &x_{a_{nt}} + \delta_{n} x_{c_{n\tau}}=0,\\ \end{aligned} $$ which admits any solution if $b_{nt}=c_{n\tau}$, or admits any solution such that $x_{b_{nt}}=x_{c_{n\tau}}$ if $b_{nt}\neq c_{n\tau}$.
Similarly, $x_{b_{nt}}$ appears twice in the system either because we are in the case above, or because there exists $n'\neq n \in \{1,...,N\}$ and $\tau\in \{1,...,T_{n'}\}$ with $k_{n'\tau}=k_{nt}$ and $j_{n'\tau}=j_{nt}$. In the last case, the system contains the equations $$ \begin{aligned} &x_{a_{nt}} + \delta_{n} x_{b_{nt}}=0,\\ & x_{d_{n'\tau}} + \delta_{n'} x_{b_{nt}}=0,\\ \end{aligned} $$ which admits any solution such that $x_{a_{nt}}=x_{d_{n'\tau}}$ if $\delta_n=\delta_{n'}$, or admits any solution such that $\frac{x_{a_{nt}}}{\delta_{n}}=\frac{x_{d_{n'\tau}}}{\delta_{n'}}$ if $\delta_n\neq \delta_{n'}$.
Therefore, $x=0_{(NK+KJ)\times 1}$ cannot be the unique solution of the system. I'm not sure I can conclude that because I have only looked at local sub-systems.