Calculating the boundary modes in Kitaev Chain In section 2 of the paper, 'Unpaired Majorana Fermions in Quantum Wires', equation (14), the following transformation:
\begin{equation}
 b^{'} = \sum_{j} (\alpha_+ ^{'} x_+ ^{j} + \alpha_- ^{'} x_- ^{j})c_{2j-1} \qquad
 b^{"} = \sum_{j} (\alpha_+ ^{"} x_+ ^{-j} + \alpha_- ^{"} x_- ^{-j})c_{2j}
\end{equation}
is supposed to bring the original Hamiltonian into the form :
\begin{equation}
H_{canonical}= \frac{i}{2} \sum_{m=1}^{N} \epsilon_m b^{'}_m b^{"}_m
\end{equation}
I am really confused whether the transformation of the original Majorana operators is limited only to two specific $ b^{'}_m$ and $b^{"}_m $, so that the matrix given in equation(12) in the paper has one of it's $\epsilon_m $ as zero and the corresponding Hamiltonian does not include $ b^{'}_m$ and $b^{"}_m $, giving rise to uncoupled Majorana modes. 
Or is the transformation related to the diagonalization of the A matrix, which is weird because $H_{canonical}$ is certainly not diagonal.(I saw other posts here on stackexchange where some people have said we need to diagonalize A to get the conditions on $x_{\pm}$.)
Could anyone help me with how to proceed on which front to get the conditions on 
$x_{\pm}$ ?
I am calculating on my first understanding but I am not sure if this is the correct way. So if this gets cleared up,it will be of great help and save time.
The paper can be found here.
 A: For $\epsilon=0$, $W$ may contain true eigenvectors of $A$, which can be arranged to be real (and only these matter for boundary modes). But in general for $A=A^* = -A^T$ the rows of $W$ contain real and imaginary parts of the complex eigenvectors of $A$ (or degenerate eigenvectors of $A^2 = -A^\dagger A \le 0$). 
Say 
$$
Au_j = i\epsilon_j u_j\\ 
Au^*_j = -i \epsilon_j u^*_j\\
u^\dagger_j u_k = u^\dagger_j u^*_k = 0\;\;\text{for}\;\; j\neq k, \;\; u^\dagger_j u^*_j = 0, \;\;u^\dagger_j u_j = 1
$$
and let 
$$
\xi_j = (u_j + u^*_j)/2, \;\; \eta_j = i( u_j^* - u_j)/2
$$ 
be their real and imaginary parts, satisfying 
$$
A \xi_j = -\epsilon_j \eta_j\\
A \eta_j = +\epsilon_j \xi_j
$$ 
From the orthogonality of the eigenvectors it can be checked that the real and imaginary parts are also orthogonal,
$$
\xi^T_j\xi_k = \eta^T_j\eta_k = \frac{1}{2}\delta_{jk}, \;\;\; \xi^T_j\xi_j + \eta^T_j \eta_j = 1\\
\xi^T_j\eta_k = \xi^T_j\eta_k = 0
$$
Then the rows of $W$ are ($j \ge 1$)
$$
W_{2j - 1} = \eta^T_j\\
W_{2j} = -\xi^T_j
$$
This gives indeed
$$
W_{2j-1}AW^T_{2k-1} = \eta^T_j A \eta_k = \eta^T_j\left(\epsilon_k \xi_k \right) = 0\\
W_{2j}AW^T_{2k} = \xi^T_j A \xi_k = \xi^T_j \left(-\epsilon_k \eta_k\right) = 0\\
W_{2j-1}AW^T_{2k} = - \eta^T_j A \xi_k = -\eta^T_j \left(-\epsilon_k \eta_k \right) = \epsilon_k \delta_{jk}  \\
W_{2j}AW^T_{2k-1} = - \xi^T_j A \eta_k = -\xi^T_j\left(\epsilon_k \xi_k \right) = \epsilon_k \delta_{jk}
$$ 
The $b'$ and $b^"$ operators read then
$$
b'_j = \sum_{k=1}^N{\eta_{j,k}c_{k}} \\
b^"_j = - \sum_{k=1}^N{\xi_{j,k}c_k}
$$ 
which can be re-expressed in terms of the complex eigenvectors $u_j$. There is still the issue of eqs.(14) for $b'$, $b^"$ of the zero mode, which use only one of the sets $\{c_{2j}\}$ or $\{c_{2j-1}\}$. I think this is a matter of convenience (symmetry?) since in this case the 2 eigenvectors are degenerate and can be redefined as needed.
