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If we have $n$ square matrices $M_{1},...,M_{n}$, call $M'$ an extension if it is a square matrix of the form $$M'=\left(\begin{array}{ccccc} M_{1} & A\\ B & M_{2}\\ & & \ddots\\ & & & M_{n} & X\\ & & & Y & Z \end{array}\right)$$ Is there always a non-singular extension for $n$ square matrices regardless of whether they are themselves non-singular? For $n=1$ this is indeed true, since we have $$M'=\left(\begin{array}{cc} M & I\\ I & 0 \end{array}\right) $$ with inverse $$M'^{-1}=\left(\begin{array}{cc} 0 & I\\ I & -M \end{array}\right) $$

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    $\begingroup$ Are the unwritten entries necessarily zero? $\endgroup$ – Omnomnomnom Jun 12 '18 at 23:48
  • $\begingroup$ In other words: is $M'$ supposed to be block-tridiagonal? $\endgroup$ – Omnomnomnom Jun 12 '18 at 23:50
  • $\begingroup$ No, the unwritten entries can be anything, not tridiagonal. $\endgroup$ – Joshua Tilley Jun 12 '18 at 23:56
  • $\begingroup$ Do you care about the size of the blocks? $\endgroup$ – Omnomnomnom Jun 12 '18 at 23:57
  • $\begingroup$ Only requirement is that the matrices $M_{1},...,M_{n}$ appear as blocks on the diagonal, the off diagonal blocks can be anything. $\endgroup$ – Joshua Tilley Jun 12 '18 at 23:59
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Let $M$ be any extension of $M_1,M_2,\dots,M_n$. If we choose an invertible extension $M'$ of $M$, then $M'$ is also an invertible extension of $M_1,M_2,\dots,M_n$.

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