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 diml x dim2 # the matrix A^T (transpose of A) is dim2 x diml # 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
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 diml x dim2 # the matrix A^T (transpose of A) is dim2 x diml # 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
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
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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 diml
dim2 -
# the matrix A^T (transpose of A) is
dim2 x diml
# 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9af9de7-0e45-421f-b306-611991ba84cf%2F26a0fc50-0397-43eb-a380-9dc5d19257f8%2F94nrmr8_processed.jpeg&w=3840&q=75)
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 diml
dim2 -
# the matrix A^T (transpose of A) is
dim2 x diml
# 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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