I want to determine the form of following $\theta_1\times \theta_2$ matrix $$[I_{\theta_1} I_{\theta_1} \cdots][I_{\theta_2} I_{\theta_2},\cdots]':=AB$$ where $A$ is a $\theta_1\times T$ matrix, $B$ a $T\times \theta_2$ and $I_d$ the $d\times d$ identity matrix.
My attempt
The first row of $A$ is of form $1,0,0,\dotsc,1,0,0,\dotsc$ where $1$ repeats $\theta_1$-periodically. Analogously, the first column of $B$ has the same form, but $1$ occurs $\theta_2$-periodically. Denote the $1$st row of A by $a:=(a_{1},\dotsc,a_{T})$ and the $1$st column of $B$ by $b:=(b_{1},\dotsc,b_{T})$. Both entries of $a$ and $b$ are $1$ at indices satisfying $1+k_1\theta_1=1+k_2\theta_2\iff k_1\theta_1=k_2\theta_2$ (i.e., everytime the multiples coincides). This suggests that the intersection $\{n:a_n=1,b_n=1\}=\{1+k\theta_x\}_{k=0}^K$, where $K\in\mathbb{N}$ satisfies $1+K\theta_x\leq T$ and $\theta_x$ is the least common multiple of $\theta_1$ and $\theta_2$. The both $a_n,b_n$ coincide with one, $\theta_x$-periodically. Hence the $(AB)_{1,1}=K+1$.
So the form of $AB$ is a matter of counting periodic points. The problem is to determine the other entries. For example, consider the entry $(AB)_{1,2}$. In this case, the second column of $B$ is $b$ shifted forward one time. When the first intersection may occur? Suppose, it happens at $n=x$. Then we know that $\#(AB)_{1,2}=\# \{x+k\theta_x\}_{k=0}^{K'}$ for some $K'$ which is easy to determine.
This was my approach. Can you suggest anything to me?
Just for visualization, take the particular case when $\theta_1=2, \theta_2=3$,
When $\theta_1=\theta_2$, is easy. $AB=diag(K_1,\dotsc,K_{\theta_1})$. S0, I mainly interested in cases when $\theta_1\neq \theta_2$.