Applications of Diagonalization PDF

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3.4 - 3.5 Application of Eigenvectors and Diagonalization If A is diagonalizable, then A = P DP −1 . This helps us compute powers of A simply since Ak = (P DP −1 )k = P DP −1 P DP −1 P DP −1 · · · P DP −1 = P D k P −1

Linear Dynamical Systems In a nuclear reaction, a number of X-particles react with a number of Y -particles. Let xk denote the number of X-particles present after k hours and let yk denote the number of Y -particles present after k hours. Suppose we observe the following relationship: xk+1 = xk + 2yk yk+1 = 3xk + 2yk for k = 0, 1, 2, . . . If the reaction begins with x0 = y0 = 1, how many X and Y particles are present after 10 hours? The amount of particles present after 1 more hour depends on the amount present now. We could calculate x1 , then x2 , then x3 and so on until we eventually obtain x10 OR we can use a the diagonalization process. We will turn our system of equations into the matrix equation      1 2 xk xk+1 = yk+1 3 2 yk or vk+1 ~ = Av~k.      1 xk xk+1 . and v~0 = , then v~k = Since vk+1 ~ = 1 yk yk+1 We have v~1 = Av~0 v~2 = Av~1 = A(Av~0 ) v~3

= A2 v~0 = Av~2 = A(A2 v~0 ) = A3 v~0 .. .

v~k = Ak v~0 To find Ak , we need P and D, since Ak = P DP −1 .     2 −1 4 0 Using the methods from section 3.3, we find P = and D = . Then P −1 = 3 1 0 −1   1 1 1 . 5 −3 2 1

Substituting these matrices into our expression for v~k above, we have v~k = Ak v~0 = (P DP −1 )k v~0     1 2 −1 4k 0 2 = 0 (−1)k −1 5 3 1    1 2 · 4k −(−1)k 2 = −1 5 3 · 4k (−1)k   1 4 · 4k + (−1)k = 5 6 · 4k − (−1)k 

 838 861 Thus, v~10 = . 1 258 291 Population Models: Predator-Prey Relationships Let hk denote the population of hawks and let mk denote the population of mice in a region in year k, k ≥ 0. Suppose the following relationship has been observed in examining the population from each from year to year: 1 1 mk hk + 100 2 50 5 = − hk + m k 4 4

hk+1 = mk+1

Initially, there were 50 hawks and 1600 mice. Compare the population 25 years later. 1 1    hk+1  2 100 . We can write the system of equations as a matrix equation =  50 5 mk+1 − 4 4 or ~vk+1 = A~vk  50 where ~v0 = . 1600 As in the previous example, ~vk = Ak~v0 . We can calculate Ak easily by diagonalizing A. That is, we can find P and D so that A = P DP −1 . 3 It turns out that the characteristic polynomial of A is cA (λ) = (λ− 1)(λ− ) and the basic eigenvector 4     3 1 1 and the basic eigenvector corresponding to λ = is . corresponding to λ = 1 is 50 25 4 # "   1 0 1 1 3 . Then P = ,D= 50 25 0 4

2

We calculate P −1 = −

Then

1  25 −1 and since Ak = P D k P −1 , we have 25 −50 1     1k 0   1 1 1  25 −1 k  Ak = − 3 −50 1 25 50 25 0 4   k   1 3 1 3 k  2 4 − 1 25 − 25 4     =    3 k 3 k  50 − 50 2− 4 4 ~v25 = A25~v0   25  25    3 1 3 1 50 − −1   2 4 25 25    425  =     3 25 3 1600 − 50 2− 50 4 4 " # 14 = 36

So after 25 years, there are 14 hawks and 36 mice.

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