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?
Best Answer
Q1:
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.
Q2:
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.