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Compute the code that return the matrix M = AT A for a given matrix A- the superscript T denoted the transpose. I have started the code for you, which gets the dimension of the matrix, and creates the zero matrix of the correct size. I have also provided some of the loops involved.

Compute the code that return the matrix M = AT A for a given matrix A- the superscript T denoted the transpose. I have started the code for you, which gets the dimension of the matrix, and creates the zero matrix of the correct size. I have also provided some of the loops involved.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Compute the code that return the matrix M = AT A for a given matrix A - the superscript T denoted the transpose. I have started the code for you, which gets the dimension of the matrix, and creates the zero matrix of the correct size. I have also provided some of the loops involved. # perform and return the multiplication of $A^TA$ import numpy as np def multiply_At_A(A): # these Lines set up the correct dimensions of the returned matrix. # the matrix A is of dimension dim1 x dim2 - # the matrix A^T (transpose of A) is dim2 x dim1 # the matrix (A^T A) is of dimension dim2 x dim2 dim1= A.shape [0] dim2 = A.shape [1] matrix = np.zeros([dim2, dim2]) for i in range (dim2): for j in range (dim2): # complete the final Loop to compute matrix[i,j] # YOUR CODE HERE return matrix
Transcribed Image Text:Compute the code that return the matrix M = AT A for a given matrix A - the superscript T denoted the transpose. I have started the code for you, which gets the dimension of the matrix, and creates the zero matrix of the correct size. I have also provided some of the loops involved. # perform and return the multiplication of $A^TA$ import numpy as np def multiply_At_A(A): # these Lines set up the correct dimensions of the returned matrix. # the matrix A is of dimension dim1 x dim2 - # the matrix A^T (transpose of A) is dim2 x dim1 # the matrix (A^T A) is of dimension dim2 x dim2 dim1= A.shape [0] dim2 = A.shape [1] matrix = np.zeros([dim2, dim2]) for i in range (dim2): for j in range (dim2): # complete the final Loop to compute matrix[i,j] # YOUR CODE HERE return matrix
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