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I am working with a class of matrices $A$ which are non-negative-definite, not symmetric, and have maximum eigenvalue less than 1. I am interested in the spectral properties of the matrix $H = A(I - A)^{-1}$.

Is there a spectral theorem for this context, which gives sufficient conditions on $A$ for the matrix $H$ to be diagonalizable? Are there sufficient conditions on $A$ to guarantee that the eigenvalues of $H$ decay rapidly (e.g., exponentially)?

I am sure that such questions have been analyzed in the past, perhaps in the economics literature. However, I have been unable to find a reference. While I phrased these question in terms of matrices, I of course would also be interested in the more general context of linear operators on Hilbert spaces.

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If you take a Schur form $A=QTQ^T$, then $H=QT(I-T)^{-1}Q^T$, and you can ignore the orthogonal factors $Q$. You might also want to set $N=I-T$, so that $Q^THQ=N^{-1}-I$. Now the problem looks much simpler.

  • $H$ diagonalizable $\Leftrightarrow$ $N^{-1}$ diagonalizable $\Leftrightarrow$ $N$ diagonalizable $\Leftrightarrow$ $T$ diagonalizable $\Leftrightarrow$ $A$ diagonalizable
  • the eigenvalues of $H$ are $\frac{\lambda_i}{1-\lambda_i}$, where $\lambda_i$ are the eigenvalues of $A$ (you can read it off $H=QT(I-T)^{-1}Q^T$). So they decay to zero if the eigenvalues of $A$ do so.

Some pointers: the first chapter of Higham's book Functions of matrices treats with much general cases, and what you are doing is applying a special Cayley transform to the matrix $A$.

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  • $\begingroup$ @Federico Poloni: Tom LaGatta says $A$ is not symmetric. How does the orthogonal matrix $Q$ diagonalize $A$? If it can not, what is the relevance of tranforming $A$ to $T$? $\endgroup$ – Hans May 6 '13 at 19:24
  • $\begingroup$ T is triangular. en.wikipedia.org/wiki/Schur_form $\endgroup$ – Federico Poloni May 6 '13 at 21:12
  • $\begingroup$ @Fededrico Poloni: I see. Thank you, Federico. $\endgroup$ – Hans May 6 '13 at 22:24

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