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Toán Cao Cấp A2_ Group 3
Transcript of Toán Cao Cấp A2_ Group 3
Ho Thi Ha Tien.
Nguyen Thanh Nhat.
Truong Ngo Truong Huy.
Nguyen Quang Vinh. Eigenvalue and Eigenvectors We return to the Leslie model of population growth. Example:
Find the steady state growth rate and the corresponding ratios between the age classes for the Leslie matrix above. So we must solve : We conclude our discussion of regular Markov chains by proving that the steady state vector x is independent of the initial state. The proof is easily adapted to cover the case of state vectors whose components sum to an arbitrary constant-say. The corresponding eigenvectors are in the null space of L-1.5I, which we find by row reduction: Solution:
We need to find all positive eigenvalues and corresponding eigenvectors of L. The characteristic polynomial of L is Hence, the steady state growth rate is 1.5, and when this rate has been reached, the age classes are in the ratio 18:6:1,as we saw before. In the long run, what proportion of its time will the rat spend in each compartment? We saw that for the Leslie matrix: In this section, we will explore several applications of eigenvalues and eigenvectors. iterates of the population vectors began to approach a multiple of the vector: In other words, the three age classes of this population eventually ended up in the ratio 18:6:1. Moreover, once this state is reached, it is stable, since the ratios for the following year are given by: 1.5 represents the growth rate of this population when it
has reached its steady state. We can now recognize that x is an eigenvector of L corresponding to the eigenvalue A = 1.5. Every Leslie matrix has exactly one positive eigenvalue and a corresponding eigenvector with positive components. Applications Markov Chains A psychologist places a rat in a cage with three compartments, as shown in figure. The rat has been trained to select a door at random whenever a bell is rung and to move through it into the next compartment. Example 4.41 Example 4.43