solution for Matrix equation

Question

$$ (w*(R_1*P*R_1^{-1})^{-1}+w*(R_2*P*R_2^{-1})^{-1})^{-1}=P_{th} $$ $R_i$ is a rotation matrix 2*2: $$ R_i=\left[\begin{matrix} cos\theta_i & sin\theta_i \\ -sin\theta_i & cos\theta_i \\ \end{matrix}\right] $$ $w$ is constant number $${0<w<1}$$and $P_{th}$ is a constant threshold matrix. how can solve this equation for squared matrix $P$ according to matrix mathematics? also it's seem that $P$ have to be a symmetric, positive definite matrix.indeed left term of above equation fuse two ellipse and i want to find those ellipse matrices that their combination satisfy threshold matrix $P_{th}$. I would really appreciate if someone can solve this problem or have useful discussion about it.

Answer 1

The equality can be rewritten $R_1P^{-1}R_1^{-1}+R_2P^{-1}R_2^{-1}=w^{-1}P_{th}^{-1}$. Let $R_1P^{-1}R_1^{-1}=X,R_2R_1^{-1}=R,w^{-1}P_{th}^{-1}=S$. Since $R_1,R_2,w,P_{th}$ are known, it suffices to find $X$ s.t. $X+RXR^{-1}=S$, that is $XR+RX=SR$, where $R$ is a rotation with angle $\theta$. This is a Sylvester equation

cf. http://en.wikipedia.org/wiki/Sylvester_equation

The previous equation admits a sole solution in $X$ when $\cos(\theta)\not= 0$.