Exercise 5.6.6

Answers

We have the matrix P = [ 0 0.990 0.99 0 0 0.010.011 ], it has the eigenvector that corresponds to eigenvalue of 1: v1 = [0 0 1 ], which says P = v11T = [000 0 0 0 111 ].

Everyone dies in the end.

import numpy as np 
 
x = np.array([[1, 1], [1, -1]]) 
b = np.array([[1, 0], [0, 0]]) 
a = np.array([[2, 1, 1], [1, 2, 1], [1, 1, 2]])/4 
#r = np.matmul(b, np.matmul(a, np.linalg.inv(b))) 
#np.linalg.norm(r), r 
xi = np.linalg.inv(x) 
np.matmul(x, np.matmul(b, xi)) 
np.linalg.eig(a), 
#x, print(xi)
((array([0.25, 1.  , 0.25]), array([[-0.81649658,  0.57735027,  0.381008  ],
         [ 0.40824829,  0.57735027, -0.81590361],
         [ 0.40824829,  0.57735027,  0.43489561]])),)

x = 2.5 
(x+4/x)/2
2.05

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2020-03-20 00:00
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