Eigenvalues of diff-system(can't understand) In this paper the authors have the dynamical system
$$\begin{align}
T_f \dot{y}_f & = -y_f + (1-\alpha(v))\varphi(z,d) \\
T_r \dot{y}_r & = -y_r + \alpha(v) \varphi(z,d) \\
\dot{z}       & = -\varphi(z,d) + y_r + u
\end{align}$$
and they state in eqns (8-10) that the eigenvalues of the linearization at the equilibrium points $(\overline{y}_f, \overline{y}_r, \overline{z})$ are
$$\begin{align}
\lambda_1 & = -T_f^{-1} \\
\lambda_2 + \lambda_3 & = -\varphi_z(\overline{z},d) - T_r^{-1} \\
\lambda_2 \lambda_3 & = T_r^{-1} \phi_z(\overline{z},d)(1-\alpha(\overline{v}))
\end{align}$$
Can someone explain to me how these are derived?
 A: Do you know how to linearize a dynamical system around an equilibrium?
The idea is that you have $x\in \mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ and you define the system $\dot{x}=f(x)$.
Now to find the linearization of the system you have to expand to a Taylor polynomial around the equilibrium and keep only the linear terms. Practically you find the Jacobian matrix of $f$. In most cases you move the equilibrium to $0$ and you end up with
$$\dot x=Jx$$
with $J$ the Jacobian of $f$. For example consider
$$\dot{x_1}=x_1+x_1 x_2$$
$$\dot{x_2}=2x_1+x_1^2-x_2$$
The equilibrium is $(0,0)$ already and the Jacobian of $f$ here is
$$\left[ \begin{array}{cc}
1+x_2 & x_1 \\
2+2x_1 & -1  \end{array}\right]$$
at $(0,0)$ this becomes
$$\left[ \begin{array}{cc}
1 & 0 \\
2 & -1  \end{array}\right]$$
and the linearization of the system is
$$\left[ \begin{array}{c}
\dot x_1 \\
\dot x_2  \end{array}\right]=\left[ \begin{array}{cc}
1 & 0 \\
2 & -1  \end{array}\right]\left[ \begin{array}{c}
x_1 \\
x_2  \end{array}\right].$$
The eigenvalues of the equilibrium are the eigenvalues of the Jacobian at the equilibrium.
I hope this helps.
