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I know this question, but I would like to know the middle square matrix $B$.

Given positive definite matrix $A \in \mathbb R^{2 \times 2}$ and non-zero matrix $C \in \mathbb R^{2 \times 3}$, find $3 \times3$ matrix $B$

$$A = C B C^t$$

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JimSD
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1 Answers1

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Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).

Then $C^+=C^T(CC^T)^{-1}$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_{2,2},0_{2,1}],C^+=\begin{pmatrix}D^{-1}\\0_{1,2}\end{pmatrix}$.

(*) is equivalent to $C^+A(C^+)^T=SBS$ or

$\begin{pmatrix}D^{-1}A(D^{-1})^T&0\\0&0\end{pmatrix}=\begin{pmatrix}a&b&0\\d&e&0\\0&0&0\end{pmatrix}$ where the unknown $B=\begin{pmatrix}a&b&c\\d&e&f\\g&h&p\end{pmatrix}$.

Note that $U=[u_{i,j}]=D^{-1}A(D^{-1})^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied

$a=u_{1,1},e=u_{2,2},b=d=u_{1,2}$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.