Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In this paper the authors have the dynamical system

$$\begin{align} T_f \dot{y}_f & = -y_f + (1-\alpha(v))\varphi(z,d) &(1)\\ T_r \dot{y}_r & = -y_r + \alpha(v) \varphi(z,d) &(2)\\ \dot{z} & = -\varphi(z,d) + y_r + u &(3) \end{align}$$

and they state that the eigenvalues of the linearization(1-3) at the equilibrium points $(\overline{y}_f, \overline{y}_r, \overline{z})$ are

$$\begin{align} \lambda_1 & = -T_f^{-1} &(4)\\ \lambda_2 + \lambda_3 & = -\varphi_z(\overline{z},d) - T_r^{-1} &(5)\\ \lambda_2 \lambda_3 & = T_r^{-1} \phi_z(\overline{z},d)(1-\alpha(\overline{v})) &(6)\\ \end{align}$$

I try linearize it by myself

$$\left( \begin{array}{ccc} -T_f^{-1}-\lambda & 0 & T_f^{-1}(1-\alpha (v))\varphi _z(z,d) \\ 0 & -T_r^{-1}-\lambda & T_r^{-1}\alpha (v)\varphi _z(z,d) \\ 0 & 1 & -\varphi _z(z,d)-\lambda \end{array} \right)$$

Disclosure determinant:

$$\begin{align} (-T_f^{-1}-\lambda)(-T_r^{-1}-\lambda)(-\varphi _z(z,d)-\lambda)-T_r^{-1}\alpha (v)\varphi _z(z,d)(-T_f^{-1}-\lambda)=0 \end{align}$$

$$\begin{align} -(T_f^{-1}+\lambda)(T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d)(T_f^{-1}+\lambda)=0 \end{align}$$

$$\begin{align} -(T_f^{-1}+\lambda)((T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d))=0 \end{align}$$

$$\begin{align} T_f^{-1}+\lambda=0 \end{align}$$

$$\begin{align} \lambda _1=-T_f^{-1} &(4^{*}) \end{align}$$

$$\begin{align} (T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d)=0 \end{align}$$

remove brackets

$$\begin{align} T_r^{-1}\varphi _z(z,d)+T_r^{-1}\lambda+\varphi _z(z,d)\lambda+\lambda^{2}+T_r^{-1}\alpha (v)\varphi _z(z,d)=0 \end{align}$$

$$\begin{align} (1+\alpha (v))T_r^{-1}\varphi _z(z,d)+T_r^{-1}\lambda+\varphi _z(z,d)\lambda+\lambda^{2}=0 &(7) \end{align}$$

Can someone explain to me how they got (5) and (6) from (7)?

UPD Mathematica has solved like this

$$\begin{align} -\frac{1}{T_f} \\ \frac{-1-T_r \varphi _z-\sqrt{1+4 T_r \alpha \varphi _z-2 T_r \varphi _z+T_r^2 \varphi _z^2}}{2 T_r} \\ \frac{-1-T_r \varphi _z+\sqrt{1+4 T_r \alpha \varphi _z-2 T_r \varphi _z+T_r^2 \varphi _z^2}}{2 T_r} \end{align}$$

share|cite|improve this question
up vote 2 down vote accepted

You are doing the wrong thing to find the eigenvalues. To find an eigenvalue $\lambda$ of a matrix $A$, you need to subtract $\lambda$ times the identity matrix from $A$, and then compute the determinant. You are subtracting the matrix $\mathrm{diag}(\lambda_1, \lambda_2, \lambda_3)$ from $A$, whereas you want to be subtracting $\mathrm{diag}(\lambda, \lambda, \lambda)$.

This gives you a polynomial whose order is the dimension of $A$ (in your case $A$ has dimension three, so the polynomial will be a cubic). The roots of this polynomial are the eigenvalues.

In your case there will be one easily computed root of the polynomial, which is $\lambda=-T_f^{-1}$. You can then factor this out and be left with a quadratic in $\lambda$, which you can solve with the standard method.

share|cite|improve this answer
Thanks, I understand. Sometimes i feel myself like dumb) Is my linearizing correct without regard for lambda? – mike_price Jul 11 '11 at 14:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.