Let $a\in (0,1)$. Let $N,T,J$ be positive integers with $N>J, N>T$, $J>T$. Consider the following matrix $$ B\equiv \begin{pmatrix} I_N & A_1\\ I_N & A_2\\ ... & ...\\ I_N & A_T\\ \end{pmatrix} $$ where $I_N$ is the $N\times N$ identity matrix and, for $t=1,...,T$, $A_t$ is an $N\times J$ matrix such that:
- Each row of $A_t$ contains one and only one element equal to $a$, while the other elements are $0$. (This implies that the sum of all the elements of $A_t$ is $N*a$).
My questions:
(Q1) Show that the columns of $B$ are not linearly independent.
(Q2) Determine what is the column-rank of $B$.
(Q3) My suspect is that the column-rank of $B$ is $N+J-1$. Consider the system of equations $$ Y=(B'B)X $$ where $Y$ and $X$ are $(N+J)\times 1$ vectors. Given that $B$ is not full column-rank, this system will have infinitely many solutions (assuming that it is consistent). I want to show that we can freely set anyone element of $X$ equal to a known value (for example, zero) and, in turn, the system of equations will have a unique solution with respect the other $N+J-1$ unrestricted elements of $X$.
My thoughts:
[Q1] I think that Q1 can be answered as follows (your advise would neverthless be appreciated).
As suggested here, $B$ is full column rank if and only if $$ C\equiv \begin{pmatrix} A_2-A_1\\ A_3-A_1\\ \cdots\\ A_T-A_1\\ \end{pmatrix} $$ is full column rank.
Consider any $t\in \{2,...,T\}$. Given the structure of each $A_1$ and $A_t$, we have that each row of $A_t-A_1$ is:
either $0_{1\times J}$;
or, one of its elements is $a$, one of its elements is $-a$, all the other elements are $0$.
Therefore, $C$ is not full column rank if and only if there exists scalars $p_1,...,p_J$ not all equal to zero solving a system of at most $N(T-1)$ equations, where each equation looks like $$ p_k-p_j=0 \hspace{1cm} \text{ for some $k,j\in \{1,...,J\}$ with $k\neq j$} $$
This system admits as solution $p_1=p_2=...=p_J\neq 0$. Hence, $C$ is not full column rank. Therefore, $B$ is not full column rank.
[Q2] I don't know how to answer this. But, from doing some simulations, I believe that the column-rank should be $N+J-1$.
[Q3] I don't know how to answer this. It is true that, by correctly setting one of the elements of $X$ equal to zero, the system will have a unique solution with respect to the other $N+J-1$ elements of $X$. However, in general, we are not completely free in deciding which element of $X$ can be normalised. Nevertheless, I believe that in my specific example we are free. How can I show it?
