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Let $T$ denote a generic row vector $1 \times n$. By generic I mean I'm gonna construct a matrix with lots of $T$'s and each one can represent a different row vector. In the same way, let $D$ be a generic matrix $m \times mn$ given by $$D = \left[ \begin{array}{cccc} T & 0 & \ldots & 0\\ 0 & T & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & T \end{array} \right].$$

Finally, let $C_i$, for $i=1\ldots m$, be a generic matrix $n \times mn$ given by $$C_i = \left[ \begin{array}{ccccc} 0 & \ldots & T & \ldots & 0\\ 0 & \ldots & T & \ldots & 0\\ \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 & \ldots & T & \ldots & 0 \end{array} \right].$$

All $T$'s are at the $i$th "column" (it is in fact a block column) of $C_i$.

Write $-\textbf{1}$ for the $m \times m$ matrix composed by $-1$ in each entry. I want to know under what conditions the following matrix is invertible. $$\left[ \begin{array}{cccc} -\textbf{1} & D & D & D\\ \hline\\ 0 & 0 & C_1 & C_1\\ 0 & 0 & C_2 & C_2\\ \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & C_m & C_m\\ \hline\\ 0 & C_1 & 0 & C_1\\ 0 & C_2 & 0 & C_2\\ \vdots & \vdots & \vdots & \vdots\\ 0 & C_m & 0 & C_m\\ \hline\\ 0 & C_1 & C_1 & 0\\ 0 & C_2 & C_2 & 0\\ \vdots & \vdots & \vdots & \vdots\\ 0 & C_m & C_m & 0\\ \end{array} \right].$$

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It is never invertible: the first $m$-columns are identical, the matrices $C_i$ do not have full rank. No chance.

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  • $\begingroup$ All matrices $C_i, D$ and row vectors $T$ are supposed to be distinct. I used the same notation just to keep it simple. I can't see how there are lot of repeated rows. Anyway, you are right about the first $m$ columns. I failed to see this simple fact. Thanks. $\endgroup$ – Integral Mar 13 '18 at 14:39

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