# Is there a more efficient way of obtaining Eigenvectors for this matrix?

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.

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