Consider the following $6\times 6$ matrix
$$\begin{pmatrix} \alpha &0&0&0&0&0\\ 0 &b_{11}&0&0&b_{12}&-m_{12}\\ 0 &0&m_{11}&0&0&0\\ 0 &0&0&b_{22}&0&0\\ 0 &b_{21}&0&0&\beta&0\\ 0 &0&m_{21}&0&0&m_{22}\\ \end{pmatrix},$$
where $\alpha,\,\beta,\,b_{ij}$ and $m_{ij}$ are complex numbers, for $i,j\in\{1,2\}$.
Is it possible to 'transform' this matrix into the a matrix of the form
$$\begin{pmatrix} \alpha &0&0&0&0&0\\ 0 &b_{11}&b_{12}&0&0&0\\ 0 &b_{21}&b_{22}&0&0&0\\ 0 &0&0&m_{22}&m_{12}&0\\ 0 &0&0&m_{21}&m_{11}&0\\ 0 &0&0&0&0&\beta\\ \end{pmatrix}\,?$$
I have a feeling that one can do this with elementary row operations, and I have tried to obtain the latter matrix from the former in this way. However, by interchanging rows and columns, I only able to obtain the sub-matrix of the form
$$\begin{pmatrix} b_{11}&b_{12}\\ b_{21}&0 \end{pmatrix},$$
with the remaining $b_{22}$ term always 'dangling' diagonally below. Can one achieve this by ERO's? Or is there something more at work here?
Am I able to get somewhere if I were to rather assume that the entries of the initial entry were all non-negative in the sense of positive-definiteness for a Hermitian matrix?
Starting from the 6x6 matrix we are working with: $$ \mathbf{M} = \begin{pmatrix} \alpha &0&0&0&0&0\\ 0 &b_{11}&0&0&b_{12}&-m_{12}\\ 0 &0&m_{11}&0&0&0\\ 0 &0&0&b_{22}&0&0\\ 0 &b_{21}&0&0&\beta&0\\ 0 &0&m_{21}&0&0&m_{22}\\ \end{pmatrix} $$ there are a number of operations we can do to try to make this matrix easier to work with. The first operation that we're going to try involves some elementary row (and column) swapping. The cleanest way to do this is by defining a Unitary Matrix; the simplest choice of which (for our specific example) might be: $$ \mathbf{U} = \begin{pmatrix} 1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&1&0&0&0&0\\ 0&0&0&0&1&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&1\\ \end{pmatrix} $$ which has the properties $\mathbf{U}\mathbf{U}^{\ast}=\mathbf{U}^{\ast}\mathbf{U}=\mathbf{I}$ (where $\mathbf{U}^{\ast}$ denotes the conjugate transpose of $\mathbf{U}$) and $\det(\mathbf{U})=1$. This means (thanks to the identity $\det(\mathbf{AB})=\det(\mathbf{A})\det(\mathbf{B})$ used repeatedly) that we can creat a transformed matrix $\mathbf{T}=\mathbf{U}\mathbf{M}\mathbf{U}^{\ast}$ which will have $\det(\mathbf{T})=\det(\mathbf{M})$ which should make computing eigenvalues easier if we made a good choice of $\mathbf{U}$. If we put all of the above together, we have: $$ \mathbf{T} = \begin{pmatrix} \alpha&0&0&0&0&0\\ 0 &b_{22}&0&0&0&0\\ 0 &0&b_{11}&b_{12}&0&-m_{12}\\ 0 &0&b_{21}&\beta&0&0\\ 0 &0&0&0&m_{11}&0\\ 0 &0&0&0&m_{21}&m_{22}\\ \end{pmatrix} $$ which is as simple as we can get things via basic transformations. Fortunately, the above is a (Block) Upper Triangular matrix so it shouldn't be too difficult to calculate determinants or eigenvalues from here.