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Consider the following square matrix \begin{align} A = \left(\matrix{d & 0 & -\frac12 & 0 & 0 & 0 & 0 & 0 \\ 0 & d & -d+1 & -\frac12 & 0 & 0 & 0 & 0 \\ -\frac12 & -d+1 & d & 0 & -\frac12 & 0 & 0 & 0 \\ 0 & -\frac12 & 0 & d & -d+1 & -\frac12 & 0 & 0 \\ 0 & 0 & -\frac12 & -d+1 & d & 0 & -\frac12 & 0 \\ 0 & 0 & 0 & -\frac12 & 0 & d & -d+1 & -\frac12 \\ 0 & 0 & 0 & 0 & -\frac12 & -d+1 & d & 0 \\ 0 & 0 & 0 & 0 & 0 & -\frac12 & 0 & d }\right), \end{align} and the following rectangular matrix \begin{align} B = \left(\matrix{1 & \frac12 & \frac12 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac12 & \frac12 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & \frac12 & \frac12 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac12 & \frac12 & 1 }\right), \end{align} then \begin{align} B A B^\intercal = \left(\matrix{d & 0 & -\frac14 & 0 \\ 0 & d & -d+\frac12 & -\frac14 \\ -\frac14 & -d+\frac12 & d & 0 \\ 0 & -\frac14 & 0 & d}\right). \end{align}

I am looking for a matrix $C$, the same size as $B$ such that \begin{align} C A C^\intercal = \left(\matrix{\frac{d}2 & 0 & -\frac14 & 0 \\ 0 & \frac{d}2 & -\frac{d}2+\frac12 & -\frac14 \\ -\frac14 & -\frac{d}2+\frac12 & \frac{d}2 & 0 \\ 0 & -\frac14 & 0 & \frac{d}2}\right). \end{align}

I get into a very complex system of equations in order to find $C$ that I am having a very hard time solving.

Does somebody see a way to profit from the sparsity and symmetries of $A$ to obtain $C$?

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    $\begingroup$ crosspost math.stackexchange.com/questions/2960731/… $\endgroup$ – Will Jagy Oct 18 '18 at 17:56
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    $\begingroup$ The matrix $B$ is constant. Do you insist that $C$ be constant too, or may it depend on $d$? $\endgroup$ – Noam D. Elkies Oct 18 '18 at 20:58
  • $\begingroup$ @NoamD.Elkies Thanks for asking! It may depend on $d$. $\endgroup$ – Astor Oct 18 '18 at 21:21
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Answered in the crosspost.

Credit to @amd.

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