In section 2.A.2 of Quantum Ising Phases and Transition in Transverse Ising Models by Suzuki, et. al. the authors give the following in their derivation of the Bogoliubov transformation for a Hamiltonian
$H = \sum_{ij} c_i^\dagger A_{ij} c_j +\frac{1}{2} \sum_{ij}c_i^\dagger B_{ij}c_j^\dagger$
with $A$ Hermitian, and $B$ antisymmetric, and $c_i$ Fermionic operators.
One makes a linear transformation of the form
$$\eta_q = \sum_i \left( g_{qi}c_i+h_{qi}c_i^\dagger \right)$$
$$ \eta_q^\dagger = \sum_i \left( g_{qi}c_i^\dagger + h_{qi}c_i \right)$$
where $g_{qi}$ and $h_{qi}$ can be chosen to be real. For $\eta_q$ to satisfy fermionic anti-commutation relations we require
$$ \sum_i \left( g_{qi}g_{q'i} + h_{qi}h_{q'i} \right) = \delta_{qq'}$$ $$\sum_i \left( g_{qi}h_{q'i} - g_{q'i}h_{qi} \right) = \delta_{qq'}$$
My questions are the following:
I see why, if $B =0$, $A$ being a Hermitian matrix ensures that $H$ is a Hermitian operator. But, I don't see how this Hamiltonian is Hermitian for nonzero $B$.
I do not see how the second equation $$\sum_i \left( g_{qi}h_{q'i} - g_{q'i}h_{qi} \right) = \delta_{qq'}$$ follows. I can see that if comes from $[ \eta_q, \eta_{q'}]_+ = 0$
But, when I make the computation of $[ \eta_q, \eta_{q'}]_+$ I get:
$$\begin{align} [ \eta_q, \eta_{q'}]+ &= [\sum_i \left( g_{qi}c_i+h_{qi}c_i^\dagger \right), \sum_j \left( g_{q'j}c_j+h_{q'j}c_j^\dagger \right) ]_+ \\ &= \sum_{ij} g_{qi}g_{q'j} [c_i,c_j]_+ + g_{qi}h_{q'j} [c_i,c_j^\dagger]_+ h_{qi}g_{q'j} [c_i^\dagger, c_j]_+ + h_{qi}h_{q'j} [c_i^\dagger,c_j^\dagger]_+ \end{align}$$
Then, using, that $[c_i,c_j]_+ = [c_i^\dagger,c_j^\dagger,]_+ = 0$ and $[c_i,c_j^\dagger]_+ = [c_i^\dagger, c_j]_+ = \delta_{ij}I$, we get
$$\sum_i \left( g_{qi}h_{q'i} + g_{q'i}h_{qi} \right) = \delta_{qq'}$$
which is in conflict with the equation above. I could see the equation above following if we had commutation relations, but we're specifically talking about Fermionic operators. Where am I going wrong? Is the book accidentally doing the Bosonic case?