My Math Forum Eigenvalues of piecewise linear systems

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 January 31st, 2015, 06:44 PM #1 Newbie   Joined: Jan 2015 From: Mtl Posts: 1 Thanks: 0 Eigenvalues of piecewise linear systems Hi all, I am a theoretical ecology M.Sc student and I'm struggling with the calculation of the eigenvalues of a fairly simple system. I have a difference equations system where the following 3x3 state matrix: line 1:[0, 1-x, 1-x] line 2:[k1, 0, 0] line 3:[0, k2, k3] is valid over x element of [0,k4] and line 1:[0, 0, 0] line 2:[k1, 0, 0] line 3:[0, k2, k3] is valid over x element of ]k4,inf Would there be a simple way to do a hand calculation of the eigenvalues and eigenvectors of such a system? Could someone point me towards documentation that would indicate how to do such a calculation? Thanks in advance! Julien
 February 4th, 2015, 05:01 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Yes, it is reasonably easy to find the eigenvalues "by hand". To find eigenvalues for the first you need to solve the equation $\displaystyle \left|\begin{array}{ccc}-\lambda & 1- x & 1- x \\ k_1 & -\lambda & 0 \\ 0 & k_2 & k_3- \lambda \end{array}\right|= 0$. Expanding on the middle row, that is $\displaystyle -k_1\left|\begin{array}{cc}1- x & 1- x \\ k_2 & k_2- \lambda\end{array}\right|+ \lambda\left|\begin{array}{cc}-\lambda & 1- x \\ 0 & k_3- \lambda \end{array}\right|$$\displaystyle = -k_1((1- x)(k_2- \lambda)- k_2(1- x)+ \lambda(-\lambda(k_3- \lambda)= 0$ a cubic equation for $\displaystyle \lambda$. The second is far easier. Since it is a triangular matrix, its eigenvalues are just the numbers on the "main diagonal", 0 and $\displaystyle k_3$ with 0 as a double eigenvalue.

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