# Matrix equation of the form $C A C^\intercal = D$

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 $$C$$?

• Commented Oct 18, 2018 at 17:56
• The matrix $B$ is constant. Do you insist that $C$ be constant too, or may it depend on $d$? Commented Oct 18, 2018 at 20:58
• @NoamD.Elkies Thanks for asking! It may depend on $d$. Commented Oct 18, 2018 at 21:21