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In this question, matrix are real-valued. Let $\alpha \in \mathcal{S}$ be a symmetric matrix and $\beta \in \mathcal{M}_n$ be a matrix. I want to find $\lambda \succeq 0$ a symmetric semi-definite positive matrix such that $$- \|\lambda\|^2 + \langle 2\alpha + 4(2I+\lambda)^{-1} \beta^* \beta (2I+\lambda)^{-1} \ , \ \lambda \rangle = 0 $$ where $\langle \cdot , \cdot \rangle$ denotes the standard matrix Frobenius scalar product and $\| \cdot \|$ is the associated scalar product. So, $0$ is an obvious solution but I believe there is at least one more. How to find them ?

Background :

This comes from an optimization problem (the one in this question, for which I realized the answer was wrong). I wrote the Lagrangian of the optimization problem and used this question to formulate the lagrangian with a matrix $\lambda$, the same way that I did for the scalar case.

thanks a lot.

Tripo
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  • In case $\alpha = \beta = 0$, the solution $\lambda = 0$ is unique. – gerw Apr 25 '25 at 11:17
  • Consider the following relationship $$\eqalign{ \def\a{\alpha} \def\b{\beta} \def\l{\lambda} \l &= 2\a + 4,(\l+2I)^{-1}\cdot\operatorname{Sym}(\b^*!\b)\cdot(\l+2I)^{-1} }$$ It should be easy to construct iterative algorithms based on this constraint. – greg Jun 07 '25 at 17:37

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