Let $\Omega$ and $\Sigma$ be symmetric, PSD matrices, $\lambda$ be a positive scalar, and $I$ be the identity matrix. Further, let the $\text{diag}$ operator set all non-diagonal elements of a square matrix to zero. I want to solve the following equation (Eq. 1)
Eq. 1: $\Sigma - \Omega^{-1} + \lambda[\Omega - \text{diag}(\Omega)] = 0$
but I am not sure how to proceed. As a first step, I solved a simpler version of the equation that does not include the diag operator (Eq. 2).
Eq. 2: $\Sigma - \Omega^{-1} + \lambda\Omega = 0$
Rewriting Eq. 2, I got $\lambda\Omega^2 + \Sigma\Omega - I = 0$. I can solve this equation using the quadratic rule for equations of square matrices, which states $AX^2 + BX + C = 0 \rightarrow X = \left[-B \pm (B^2-4AC)^{1/2} \right]2^{-1}A^{-1}$ (for $A$ invertible). The solution I get is $\hat{\Omega} = \frac{1}{2\lambda} \left[ (\Sigma^2 + 4 \lambda I)^{1/2} -\Sigma \right]$ (the negative solution is ignored since $\Omega$ is PSD).
Assuming this solution to Eq. 2 is correct, can I use it as a springboard in solving the more complicated problem (Eq. 1)? How would one go about it?