linearized Boltzmann equation $$\left({\cal L}+{\cal V}+{\cal C}\right)\psi_{\mathbf{k}\lambda}\left(\mathbf{q},\omega\right)=S_{\mathbf{k}\lambda}\left(\mathbf{q},\omega\right).\label{eq:Kin_Eq}$$ $\mathcal{L}$ is the Liouville operator and given by $${\cal L}=-i\omega+i\mathbf{q}\cdot\mathbf{v}_{\mathbf{k}\lambda}$$ The linearization of the collision operator can be expressed in the form $${\cal C}\psi_{\mathbf{k}\lambda}\approx\frac{1}{w_{k}}\sum_{\lambda'}\int_{\mathbf{k}'}\frac{\delta\left(C\psi\right)_{\mathbf{k}\lambda}}{\delta\psi_{\mathbf{k}'\lambda'}}\psi_{\mathbf{k}'\lambda'},$$ where the weight function $w_{k}$ was introduced above.

Let the $\psi_{\boldsymbol{k}}$ be element of a function space with inner product $$\left\langle \phi\mid\psi\right\rangle =\sum_{\lambda}\int_{\mathbf{k}}w_{k}\phi_{\mathbf{k}\lambda}^{*}\psi_{\mathbf{k}\lambda},\label{eq:Scalar_product}$$ such that $$\begin{aligned}
\left\langle \phi\left|{\cal C}\right|\psi\right\rangle  & = & \sum_{\lambda}\int_{\mathbf{k}}w_{k}\phi_{\mathbf{k}\lambda}^{*}{\cal C}\psi_{\mathbf{k}\lambda}\nonumber \\
 & = & \sum_{\lambda\lambda'}\int_{\mathbf{k}\mathbf{k'}}\phi_{\mathbf{k}\lambda}^{*}\frac{\delta\left(C\psi\right)_{\mathbf{k}\lambda}}{\delta\psi_{\mathbf{k}'\lambda'}}\psi_{\mathbf{k}'\lambda'}.\end{aligned}$$ One can show that the entropy production in the absence of external driving terms is $\frac{\partial S}{\partial t}=k_{B}\left\langle \psi\left|{\cal C}\right|\psi\right\rangle $ which ensures that the collision operator is positive definite. In fact, ${\cal C}$ is Hermitian under the above scalar product. Therefore its eigenvalues are real and its eigenfunction form an orthonormal basis of the function space.

The right hand side of Eq. (\[eq:Kin\_Eq\]) is determined by the forces acting on the system. The three force terms studied here are due to electric fields, thermal gradients and viscous forces. For an electric field oriented along the $x$-axis, $\mathbf{E}_{0}=E_{0}\hat{\mathbf{e}}_{x}$, the force term reads $$S_{E}=-eE_{0}\cos\theta\left(\lambda v\beta\right),\label{eq:Electric_force}$$ where $\theta$ is the polar angle of the momentum $\mathbf{k}$. The corresponding term for a thermal gradient $\nabla T$ is given by $$S_{T}=-k\left|\nabla T\right|\cos\theta k_{B}\left(v\beta\right)^{2}.\label{eq:Thermal_force}$$ A 