Consider an $m\times m$ positive definite and Hermitian matrix $\mathbf{M}$ and an arbitrary $m\times n (m>n)$ para-unitary matrix $\mathbf{R}$, i.e., $\mathbf{R}^H\mathbf{R}=\mathbf{I}_n$.
then， is there any strict proof for: Let $\mu_1,...,\mu_{n}$ and $\lambda_1,...,\lambda_{n}$ denote the eigenvalues in descending order of $(\mathbf{R}^H\mathbf{M}\mathbf{R})^{-1}$ and $\mathbf{R}^H\mathbf{M}^{-1}\mathbf{R}$, respectively, we have $\mu_k \le \lambda_k, \forall k$. ?
I use matlab and generate many random matrices to check the result. I think it must be true. However, how to prove it?
By a change of orthonormal basis, we may assume that $R^H=\pmatrix{I_n&0}$. Let us write $$ M=\pmatrix{A&B^H\\ B&C}, \text{ so that } M^{-1}=\pmatrix{(A-B^HC^{-1}B)^{-1}&\ast\\ \ast&\ast} $$ where $A-B^HC^{-1}B$ is the Schur complement of $C$ in $M$. Then $$ (R^HMR)^{-1}=A^{-1}\prec(A-B^HC^{-1}B)^{-1}=R^HM^{-1}R $$ and the result follows.