Question

NEED HELP WITH PROBLEM 1 AND 2 OF THIS LAB. I NEED TO PUT IT INTO PYTHON CODE! THANK YOU!

LAB 9 - ITERATIVE METHODS FOR EIGENVALUES AND MARKOV CHAINS 1. POWER ITERATION The power method is designed to find the dominant' eigenvalue and corresponding eigen- vector for an n x n matrix A. The dominant eigenvalue is the largest in absolute value. This means if a 4 x 4 matrix has eigenvalues -4, 3, 2,-1 then the power method will converge to = -4 and identify the corresponding eigenvector as well. If, on the other hand, the eigenval- ues are -4, 3, 2, 4 then there is no dominant' eigenvalue, and so power iteration will likely not converge or may converge to either = 4 or = -4 depending on the initial guess for the eigenvector. 1.1. Standard power method. The basic idea of the power method is that if the matrix A is diagonalizable (something that we will assume going forward) and we suppose that its eigenval- ues satisfy 11 212 213 > ... > In, with corresponding eigenvectors V1, V2, 03,..., Ung then any initial guess so can be written as: (1) 20 = Civi+ C202 + C303 + ... + Cnun for some ci, ..., Cn ER. Power iteration simply applies matrix multiplication to the vector co repetitively, i.e. x] = Axo, x2 = Axı = A XO, X3 = Ax2 = Axo, Xk = Axo. Note, by using (1) we can see Ck = Akzo = citivi+c2102 +...+ent un = *(00+ 0 (**ouston+en (3)«»). If X1 + 0, then since 11 is the dominant eigenvalue < 1 for all j #1 and hence all of these terms in the expansion of xk above will get smaller and smaller for each value of k. All the terms, that is except, the first one. This means that as k + then Ik → v so long as ci + 0. Notice that cv is a scalar multiple of the eigenvector v1, and hence it is an eigenvector as well with the same eigenvalue i.e. it is a dominant eigenvector). This leads us to the first version of the power method

Answer 1

import numpy as np
A=np.array([[1,1],[2,0]])#creating A matrix
def evect_approx1(x_0,k):
   x_vect=0
   for j in range(1,k+1):#loop from j=1 to k
       x_vect=np.dot(A,x_0) #eigen vector formual for next iteration
       x_0=x_vect
   return x_vect #return final result
print(evect_approx1(np.array([1,9]),10))

import numpy as np
A=np.array([[1,1],[2,0]])#creating A matrix
def evect_approx1(x_0,k):
   lambda_1=0
   for j in range(1,k+1):#loop from j=1 to k
       x_vect=np.dot(A,x_0)
       x_0=x_vect
       lambda_1=(np.dot(A,x_vect)/x_0)[0]#eigen values can be 2 in this case we are considering only one to get desired output
   return lambda_1
print(evect_approx1(np.array([1,9]),10))

follow indention to get desired output

[8] import numpy as np A=np. array([[1,1],[2,0]]) #creating A matrix def evect_approx1(x_,k): X_vect=0 for j in range(1, k+1): #loop from j=1 to k X_vect=np.dot(A,X_O) #eigen vector formual for next iteration X_O=x_vect return x_vect #return final result print(evect_approx1(np.array([1,9]),10)) [3752 3760] import numpy as np A=np.array([[1,1],[2,0]])#creating A matrix def evect_approx1(x_@,k): lambda_1=0 for j in range(1,K+1): #loop from j=1 to k X_vect=np. dot(A,X_O) x_@=x_vect lambda_1=(np.dot(A,x_vect)/x_)[@]#eigen values can be 2 in this case we return lambda_1 print(evect_approx1(np.array([1,9]),10)) considering only one to get desired output - 2.002132196162047