Column rank of a matrix

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:

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?

• where did this problem come from and why do you think column rank should always be $N+J-1$? Suppose you consider the case that each $A_k$ is identical and in particular each has its first column equal to $a\cdot \mathbf 1_N$ and all other columns are zero-- in such a case $\text{rank}\big(B\Big) = N$. Commented Feb 13, 2021 at 19:09

If you sum the first $$N$$ columns of $$B$$ you get the vector of all ones. If you sum the last $$J$$ columns you get the vector of all $$a$$'s. These two vectors are proportional, hence the columns of $$B$$ are linearly dependent.
The rank is not uniquely determined by the conditions you have on $$B$$. For example, if $$T=1$$ then the rank of $$B$$ is $$N$$, the row rank. For $$T>1$$ it's easy to make examples with e.g. rank $$N+J-1$$ as suggested.