Compute the code that return the matrix M = A" 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. # 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[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

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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Complete the following Code

Compute the code that return the matrix M = A" 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.
: # 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
Transcribed Image Text:Compute the code that return the matrix M = A" 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. : # 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
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