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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?

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    $\begingroup$ If $A_1$ is invertible then the first equation involving $K$ determines it by $K = A_1^{-1}K_1A_1^{-1}$. So there will be no solutions to the system unless the other two equations for $K$ happen to give the same result. If there's a kernel then you have some freedom on the kernel, but it's still almost never going to be the case that there's a solution to the system. $\endgroup$
    – Nik Weaver
    Aug 20, 2018 at 15:52
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    $\begingroup$ $A_{1}$, being symmetric and positive definite, is invertible, isn't it ? $\endgroup$ Aug 20, 2018 at 15:55
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    $\begingroup$ Depends what you mean by "positive definite". If you mean all eigenvalues are strictly positive, then yes. $\endgroup$
    – Nik Weaver
    Aug 20, 2018 at 17:53
  • $\begingroup$ If I recall correctly, positive definite (in the sense that $x^{T}Mx$ for every nonzero column vector $x$) implies strictly positive eigenvalues. $\endgroup$ Aug 20, 2018 at 19:56
  • $\begingroup$ Yes, I meant strictly positive eigenvalues. And yes, we can assume that $K_1$ is generated by a kernel function. Sorry, I didn't mention that $A_1, A_2, A_3$ are unknowns. $\endgroup$
    – Gab
    Aug 21, 2018 at 17:40

2 Answers 2

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

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  • $\begingroup$ Good idea, but it is not a complete solution: you have to choose the $U_i$ appropriately so that $A_i$ and $K$ are positive definite, and that does not seem immediate to do. $\endgroup$ Aug 23, 2018 at 7:24
  • $\begingroup$ @ Frederico: I agree. $\endgroup$ Aug 23, 2018 at 21:23
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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 $\sum_{i}\lambda_{\min}(A_i) - \lambda_{\max}(K_{inv})$.

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