A stretch of desert is populated by two species of animals, roadrunners and coyotes, who are engaged in an endless game of rivalry and mischief. The populations r ( t ) and c ( t ) of roadrunners and coyotes t years from now can be modelled by
r ( t + 1 ) c ( t + 1 ) = = 3 r ( t ) − 2 c ( t ) 5 r ( t ) − 3 c ( t ) − 1 0 0
If there are 120 roadrunners and 110 coyotes initially (at time t = 0 ), how many roadrunners will there be 1001 years later (at time t = 1 0 0 1 )?
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Exactly, the system is periodic with a period of 4. (+1) This is happening because the eigenvalues of the matrix A = [ 3 5 − 2 − 3 ] are ± i , so that A 4 = I 2 since i 4 = 1 .
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Sir...how do we find eigen values in general.
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They are the roots of the characteristic polynomial, det ( λ I n − A ) . In our case, this polynomial is det [ λ − 3 − 5 2 λ + 3 ] = λ 2 + 1 , with roots ± i .
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@Otto Bretscher – Wow...That's really interesting. I always wonder how this works. Thanks a lot sir. Is it possible to get periods other than 4?
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@Ashwin K – Yes, it's really not that complicated, and this is a very, very central concept throughout math and physics.
Equivalently, the eigenvalues of A are the complex numbers λ such that A v = λ v (*) for some non-zero vector v . Those vectors v are the eigenvectors with eigenvalue λ . You can write equation (*) as ( λ I n − A ) v = 0 , showing that the matrix λ I n − A is singular and its determinant is zero... thus the condition I gave in my last post.
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This is an interesting equation :
From above equations it is known that r(0) = r(4) & c(0) = c(4) . So, it repeats after a period of 4 and so r ( 1 0 0 1 ) = r ( 1 ) = 1 4 0 .