How to solve the quadratic matrix equation $A = BC B^H$?

Question

Given $6 \times 6$ matrix $A$ and $2 \times 2$ matrix $C$, I would like to solve the following quadratic matrix equation in $6 \times 2$ matrix $B$

$$A = BC B^H$$

How to find the matrix $B$? Many thanks in advance.

Answer 1

You should be more specific about your matrices $A$ and $C$, otherwise the problem may not have a solution. For instance, consider the following matrices:

$$A = \begin{matrix} \begin{bmatrix} -1 & a_{12} & \cdots & a_{16}\\ a_{21} & a_{22} & \cdots & a_{26} \\ \vdots & \vdots & \ddots & \vdots \\ a_{61} & a_{62} & \cdots & a_{66}\\ \end{bmatrix}\\ \end{matrix}~\text{and}~~ C = \begin{matrix} \begin{bmatrix} 1 & 0\\ 0 & 0 \\ \end{bmatrix}\\ \end{matrix}.$$

Applying the $\text{vec}$ operator on $A = BCB^{H}$ we have,

$$\text{vec}(A) = \text{vec}(BCB^{H}) = (B^{*}\otimes B)\text{vec}(C).$$

$B^{*}\otimes B$ is a $36 \times 4$ matrix whose first element is $|b_{11}|^2$. Hence, we would have

$$|b_{11}|^2 = -1,$$ which does not provide any solution.