Q4. An application of diagonalization. Suppose we have a population of predators and preys, say geese and UW undergrads. Let Since geese prey on undergrads, the population of geese will grow if there are undergrads around. Conversely, the population of undergrads will be negatively impacted by the popu- lation of geese. Let's model this using the following (very simple-minded) relationships: u(n) = the number of undergrads on day n, g(n) = the number of geese on day n. Let's write (m) = = These relationships give us the population sizes on day n+1 in terms of the population sizes on day n. This is an example of a system of time-evolution equations. Our goal now is to solve this system to obtain the population sizes u(m) and g(m) at an arbitrary day m. We can model our system in matrix form as Notice that u(n + 1) = u(n) - 2g(n) g(n+1)u(n) + 4g(n). [u(m)] Lg(m [u(n+1)] [g(n+ and A = = [u(n)] (1 (n+1) = A(n). so that the above equation takes the form (n+1) = A (n) = A² (n − 1) = ... = A+¹7(0), where (0) is the vector containing the initial population sizes u(0) and g(0). So in order to determine x(m), we would have to compute A" which doesn't seem like a very pleasant task. Below you will see how diagonalization sidesteps this issue./n(a) Diagonalize A, obtaining a diagonal matrix D and an invertible matrix P such that A = PDP-¹. (b) Show that if we put y(n) = P¹7(n) then the system 7(n+1) = A7(n) is transformed into the system (n + 1) = Dy(n). (c) Show that y(m) = Dmy(0), and actually compute D™ explicitly. (d) Suppose that u(0) = 10 and g(0) = 6, so that 7(0) = [16] Determine closed-form expressions for u(m) and g(m). [Hint: Find (0) and then use it to find 7(m).]

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