I am working with the following $2N\times 2N$ matrix;

$\pmatrix{\bf0&\bf T^{-1}\\\bf -T^{\dagger}&\bf UT^{-1}}$

where $\bf 0, T, U$ are $N\times N$ matrices. $\bf T$ is a matrix of the form $\bf T = \pmatrix{ t_{11}& 0\\t_{21}& t_{11}}$ where $t_{11},0,t_{21}$ are general complex and of dimension $N/2\times N/2$. Every element of $0$ is zero. $\bf U = \epsilon I - V$ where $\bf V$ is a Hermitian matrix, $\bf I$ is the identity matrix and $\epsilon$ is a small imaginary component $\approx1\times 10^{-6}$. $\bf 0$ is a matrix where every element is $0$.

My question is given the nature of the matrix, is there a more efficient way of obtaining the Eigenvectors than using a Schur decomposition?

Secondly, once the Eigenvectors are calculated, I organise them into the matrix $\bf M$ which is $2N\times 2N$ and each column is an Eigenvector organised as $|\lambda_1|\leq|\lambda_2|\leq...\leq|\lambda_{2N}|$ where $\lambda_x$ are the corresponding Eigenvalues.

Then assuming $\bf M$ is in the form $\pmatrix{\bf a&\bf b\\\bf c&\bf d}$, I require the inverse of $\bf d$ and was wondering if there was a cheap way of obtaining this given the above.

Many thanks in advance

  • $\begingroup$ What is $T^\dagger$ here? $\endgroup$
    – A.Γ.
    May 30, 2019 at 17:21
  • $\begingroup$ $\bf T^{\dagger}$ is the Hermitian conjugate of $\bf T$ $\endgroup$
    – AlexD
    May 30, 2019 at 18:00
  • $\begingroup$ You say that $T$ is Hermitian, so it is just $T$, right? $\endgroup$
    – A.Γ.
    May 30, 2019 at 18:01
  • $\begingroup$ Whoops.. I should have proof read my question again.. $\bf T$ is not Hermitian. Question edited to indicate this $\endgroup$
    – AlexD
    May 30, 2019 at 19:35


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