Solutions to non-linear equation with positive semidefinite matrix

Question

Let $\mathbf{A}$ and $\mathbf{B}$ be real $m\times n$ matrices with $m>n$, $\rho$ be a scalar and $\Sigma$ be a positive definite $n\times n$ matrix. I would like to find conditions on $\mathbf{A}$ and $\mathbf{B}$ which guarantee that the solutions $(\rho_1,\Sigma_1)$ of the following equation all have $\rho_1=\rho$. The equation is

$$\mathbf{A}\Sigma_1\mathbf{A}^\top+\rho_1\mathbf{B}\Sigma_1\mathbf{B}^\top=\mathbf{A}\Sigma\mathbf{A}^\top+\rho\mathbf{B}\Sigma\mathbf{B}^\top$$

There are no restrictions on $\rho_1$, but $\Sigma_1$ must be positive definite. Clearly, $\Sigma_1=\Sigma,\rho_1=\rho$ is a solution, but I would like to make assumptions such that all solutions have $\rho_1=\rho$.

Answer 1

One gets infinitely many solutions if $A$ and $B$ are of the form

\begin{align} A &= C\,X\,\Omega, \\ B &= \gamma\,C\,Y\,\Omega, \end{align}

with

• $C \in \mathbb{R}^{m \times n}$,
• $\Omega$ such that $\Omega\,\Sigma\,\Omega^\top = I$,
• $X,Y \in \mathbb{R}^{n \times n}$ orthogonal matrices, i.e. $X\,X^\top = Y\,Y^\top = I$,
• $\gamma \in \mathbb{R}$.

The solutions in this case are

$$ \Sigma_1 = \frac{1 + \gamma\,\rho}{1 + \gamma\,\rho_1}\,\Sigma,\ \forall\,\rho_1 \in \left\{x\in\mathbb{R} \left|\frac{1 + \gamma\,\rho}{1 + \gamma\,x} > 0\right.\right\}. $$

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By using vectorization and the Kronecker product the equation can also be written as

$$ \left(A \otimes A + \rho_1\,B \otimes B\right) \text{vec}(\Sigma_1) = \text{vec}(A\,\Sigma\,A^\top + \rho\,B\,\Sigma\,B^\top), $$

which only has a solution if $\text{vec}(A\,\Sigma\,A^\top + \rho\,B\,\Sigma\,B^\top)$ lies in the column space of $A \otimes A + \rho_1\,B \otimes B$. Because of the linear contribution, due to $\rho_1$, of $B \otimes B$ to the column space, I believe there either can be only one solution ($\rho_1=\rho$) or infinitely many. However, I am not entirely sure if the cases from the previous section are the only ways to get infinitely many solutions.