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 $A^TA$¹ 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] # You have to use the Loop method, instead of using other methods # YOUR CODE HERE for n in range (dim1): matrix[i,j]=matrix[i,j]+(A[n,j]*A[n,i]) return matrix ## Check your code below using print command ## For example, assign A with a random matrix A = np.array([[5,6], [4,6], [4,5]]) print (multiply_At_A(A)) Cell In [50], line 20 for n in range (dim1): 回个 早
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 $A^TA$¹ 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] # You have to use the Loop method, instead of using other methods # YOUR CODE HERE for n in range (dim1): matrix[i,j]=matrix[i,j]+(A[n,j]*A[n,i]) return matrix ## Check your code below using print command ## For example, assign A with a random matrix A = np.array([[5,6], [4,6], [4,5]]) print (multiply_At_A(A)) Cell In [50], line 20 for n in range (dim1): 回个 早
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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i can't get my python code to work for this:
![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 $A^TA$¹
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]
# You have to use the Loop method, instead of using other methods
# YOUR CODE HERE
for n in range (dim1):
matrix[i,j]=matrix[i,j]+(A[n,j]*A[n,i])
return matrix
## Check your code below using print command
## For example, assign A with a random matrix
A = np.array([[5,6], [4,6], [4,5]])
print (multiply_At_A(A))
Cell In [50], line 20
for n in range (dim1):
回个
早](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b995fb6-0105-410c-a398-9564236d4729%2F26452a54-9653-43c3-9b8c-daffbf479262%2F5s8pr8_processed.png&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 $A^TA$¹
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]
# You have to use the Loop method, instead of using other methods
# YOUR CODE HERE
for n in range (dim1):
matrix[i,j]=matrix[i,j]+(A[n,j]*A[n,i])
return matrix
## Check your code below using print command
## For example, assign A with a random matrix
A = np.array([[5,6], [4,6], [4,5]])
print (multiply_At_A(A))
Cell In [50], line 20
for n in range (dim1):
回个
早
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