Compute the code that return the matrix M = AT A for a given matrix A - the superscrupt 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. :H # 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 dim1 # the matrix (A^T A) is of dimension dim2 x dim2 dim1 - A.shape[@] 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 superscrupt 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. :H # 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 dim1 # the matrix (A^T A) is of dimension dim2 x dim2 dim1 - A.shape[@] 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)
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ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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![Q5:
Compute the code that return the matrix M = AT A for a given matrix A - the superscrupt 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.
In [ ]: I # 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F507f99da-9967-4c1f-a8fd-4e55d78bcba3%2F9804532e-5e10-4db7-ba8d-1ab9b2e8c427%2F0lnqufh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q5:
Compute the code that return the matrix M = AT A for a given matrix A - the superscrupt 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.
In [ ]: I # 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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