How to solve for $A$ in $A - BAB^T = CC^T$?

Question

Considering an unknown real symmetric matrix $A$, and two known matrix $B$ and $C$.

If we have the equation: $$ A - BAB^T = CC^T $$

Can we get an analytical solution of A?

Answer 1

Let $vec(A)$ denote the vectorization operator, and let $\otimes$ denote the Kronecker product. We can then rewrite both sides of the equation to get $$ (I - B \otimes B) \,vec(A) = vec(CC^T) $$ Assuming invertibility, we have $$ vec(A) = (I - B \otimes B)^{-1}vec(CC^T) $$

Answer 2

If $B$ is small, say $\|B\| < 1$ where $\|\cdot\|$ stands for any matrix norm, then $A$ has a solution in the form of a converging series $$A = \sum\limits_{n=0}^\infty B^n CC^T (B^T)^n$$ Depends on the form of $B$, you may simplify this further.

If $B$ is not small, the system need not have any solution. e.g, when $B$ is an orthogonal matrix, $$\text{Tr}(A - BAB^T) = \text{Tr}(A- B^TBA) = 0$$ If $\text{Tr}(CC^T) \ne 0$, there are no solution for your equation.