How to solve this system of Matrix equations? (Coupled riccati equations?)

Question

I am trying to solve for K in the following problem:

$ 3I = A_1 + A_2 + A_3$

$ A_1 K A_1 = K_1 $

$ A_2 K A_2 = K_2 $

$ A_3 K A_3 = K_3 $

Where $I$ is the identity, $K, K_1, K_2, K_3, A_1, A_2, A_3$ are known to be symmetric and positive definite. $K, A_1, A_2, A_3$ are unknown. $K_1, K_2, K_3$ are know.

• I have attempted to shape the problem into a large $A K A = K_{1,2,3}$ problem where I just have to solve for $K$ but have been unsuccessful.

• I have also tried to solve by substitution, but I always get multiplications among the $A$ matrices that I am unable to decouple.

• The formulation almost resembles that of the Algebraic Riccati equation in Optimal Control (if we could make it look something like $CK_1C = K_2$ and solve using Riccati).

Any ideas?

Answer 1

Not sure if I misunderstood something, but the following seems to work. Let us change the unknown positive-definite variables from the tuple $(A_{1},A_{2},A_{3},K) \mapsto (X_{1},X_{2},X_{3},K)$ where $X_{i} := A_{i}K^{1/2}$ for $i=1,2,3$ (all positive definite matrices have unique pos. def. square root).

Then the last three equations become $X_{i}X_{i}^{\top} = K_{i}$, which have solutions $X_{i} = K_{i}^{1/2}U_{i}$, $i=1,2,3$, for arbitrary square orthogonal matrices $U_{1},U_{2},U_{3}$. Therefore,

$$\displaystyle\sum_{i}X_{i} = 3 K^{1/2} = \displaystyle\sum_{i} K_{i}^{1/2}U_{i} \quad \Rightarrow \quad K = \left(\frac{1}{3}\displaystyle\sum_{i} K_{i}^{1/2}U_{i}\right)^{2}.$$

Back to original unknowns: $A_{i} = X_{i}K^{-1/2} = K_{i}^{1/2}U_{i} \left(\frac{1}{3}\displaystyle\sum_{i} K_{i}^{1/2}U_{i}\right)^{-1}$.

Answer 2

Too long to comment.

Not a complete answer, but certainly can yield a certificate of non-existence of solutions. Consider the following opt problem. $$ \max t\\ \mbox{subject to}\\ \begin{bmatrix} K_i & A_i\\ A_i & K_{inv} \end{bmatrix} \succeq 0, \forall i\\ A_1 + A_2 + A_3 = 3I, A_i\succeq 0 \forall i\\ \lambda_{\min}(A_i) \geq t, \forall i,~~ \lambda_{\min}(K_{inv}) \geq t $$ If there is no a solution for the above problem (can be found using CVXPY) or the opt value is 0, then there is no a solution to the original problem.

I'm wondering if we can think of a concave cost function to minimize the gap in the Schur complement of the LMIs. Maybe something like $\sum_{i}\lambda_{\min}(A_i) - \lambda_{\max}(K_{inv})$.