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.


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 $A$.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.