Is this matrix of maximal rank? Let $(A_0,A_1)\in (M_{p\times r}(\mathbb{R}))^2$, $(B_0,B_1)\in (M_{p\times q}(\mathbb{R}))^2$, and suppose that the matrix
$$\begin{pmatrix}A_1&B_0\\A_0&B_1\end{pmatrix}$$
has rank $2p$. Has the matrix
$$\begin{pmatrix}A_1&B_0&&&\\
&B_1&A_0&&\\
&&A_1&B_0&\\
&&&B_1&A_0\end{pmatrix},$$
where blank spaces are all zero, rank $4p$? Some numerical experiments seems to say so, but trying to make appear the matrix from the hypothesis and doing Gaussian elimination like in
$$\begin{pmatrix}A_1&B_0&&&\\
&B_1&A_0&&\\
&&A_1&B_0&\\
&&&B_1&A_0\end{pmatrix}\sim\begin{pmatrix}A_1&B_0&&&\\
A_0&B_1&A_0&&\\
A_1&&A_1&B_0&\\
&&A_0&B_1&A_0\end{pmatrix}$$
won't directly work, due to terms like the $A_1$ in "$(3,1)$" position here, which prevent me to put the matrix in canonical form. I would also be interested if there was a similar argument for matrices of the form
$$\begin{pmatrix}A_1&B_0&&&&\\
&B_1&A_0&&&\\
&&A_1&B_0&&\\
&&&B_1&A_0&\\
&&&&A_1&\dots\end{pmatrix},$$
where eventually you stop at some $A_0$ or $B_0$. Thanks in advance for any ideas!

For simplicity, set $$M = \begin{pmatrix} A_{1} & B_{0}\\ A_{0} & B_{1} \end{pmatrix} \in M_{2 p, q +r}(\mathbb{R}) \quad \text{and} \quad N = \begin{pmatrix} A_{1} & B_{0} & & & \\ & B_{1} & A_{0} & & \\ & & A_{1} & B_{0} & \\ & & & B_{1} & A_{0} \end{pmatrix} \in M_{4 p, 2 q +3 r}(\mathbb{R}) \, \text{.}$$ To prove that $N$ has rank $4 p$, it suffices to show that, if $X \in M_{1, 4 p}(\mathbb{R})$ satisfies $X N = 0$, then $X = 0$. Suppose that $X \in M_{1, 4 p}(\mathbb{R})$ is such that $X N = 0$. Writing $$X = \left( X_{1} \vert X_{2} \vert X_{3} \vert X_{4} \right) \quad \text{with} \quad X_{1}, X_{2}, X_{3}, X_{4} \in M_{1, p}(\mathbb{R}) \, \text{,}$$ we obtain the equations $$\left\lbrace \begin{array}{lr} X_{1} A_{1} = 0 & (1)\\ X_{1} B_{0} +X_{2} B_{1} = 0 & (2)\\ X_{2} A_{0} +X_{3} A_{1} = 0 & (3)\\ X_{3} B_{0} +X_{4} B_{1} = 0 & (4)\\ X_{4} A_{0} = 0 & (5) \end{array} \right. \, \text{.}$$ The equalities $(1) +(3) +(5)$ and $(2) +(4)$ can be rewritten as $$\left( X_{1} +X_{3} \vert X_{2} +X_{4} \right) M = 0 \, \text{.}$$ Since $M$ has rank $2 p$ by hypothesis, it follows that $X_{3} = -X_{1}$ and $X_{4} = -X_{2}$. In particular, we have $$\left\lbrace \begin{array}{lr} X_{1} A_{1} = 0 & (i)\\ X_{1} B_{0} +X_{2} B_{1} = 0 & (ii)\\ X_{2} A_{0} = 0 & (iii) \end{array} \right. $$ by the equations $(1)$, $(2)$ and $(5)$. The equalities $(i) +(iii)$ and $(ii)$ can be rewritten as $$\left( X_{1} \vert X_{2} \right) M = 0 \, \text{.}$$ Therefore, we have $X_{1} = 0$ and $X_{2} = 0$ since $M$ has rank $2 p$. Thus, we have $X = 0$, and the desired result is proved.
