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Let $M$ be a symmetric positive-definite matrix and $$A = (I+M)^{-1}(I-M)$$ we know that eigenvalues of matrices $I+M$ and $I-M$ are as $1+\mu_i$ where $\mu_i$ is eigenvalue of $M$. Who we can determine eigenvalues of matrix $A$?

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Write $M=P^tDP$ with $P$ orthogonal and $D$ diagonal. Then $I+M=P^t(I+D)P$ hence $(I+M)^{-1}=P^t(I+D)^{-1}P$ and $A=P^t(I+D)^{-1}PP^t(I-D)P$, that is, $A$ is similar to $(I+D)^{-1}(I-D)$. We conclude that the eigenvalues of $A$ are $(1+\mu_j)^{-1}(1-\mu_j)$, where the $\mu_j$'s are the eigenvalues of $M$.

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  • $\begingroup$ Why eigenvalues of $(I+D)^{-1}(I-D)$ are as form $(1+\mu_j)^{-1}(1-\mu_j)$? Is it a general fact about product of diagonal matrices? $\endgroup$ – SKMohammadi Apr 28 '14 at 11:27
  • $\begingroup$ Yes: $I+D$ is diagonal and invertible, hence its inverse is diagonal. $\endgroup$ – Davide Giraudo Apr 28 '14 at 11:28

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