How does the following code work for Matrix Multiplication in Python

Question

I'm trying to do a matrix multiplication in python I have found the following code that I am trying to understand. (I know how to multiply matrices by hand and I want to understand how the following code performs the same action and by that i mean the first element in BA (row 1 column 1) is calculated by doing (1*1 + 3*3 + 3*3 + 1*1) etc.

from numpy import array
A= array([[ 1, 4, 5 ],[ 3, 2, 9], [ 3,6, 2], [ 1,6, 8]])
B=A.T

BA= array([[ 0, 0, 0 ],[ 0,0, 0], [ 0,0, 0] ])

for i in range(len(B)):
   for j in range(len(A[0])):
       for k in range(len(A)):
           BA[i][j] += B[i][k] * A[k][j]

I know that the length command for a list returns how many elements there are in that list. I am not sure how it works here since B is a matrix, I assume it returns how many rows there are.

range of len(B) would be (0,3) corresponding to row 1,2 and 3. for i in range would correspond to i=0, i=1, i= 2

next confusing thing is for j in range len(A[0]) The first element of A is the first row, the length here would thus correspond how many elements there are in the first element of A.

Basically I have a basic understanding of what range and len etc put out for this example but I would like to get a better understand of each value of i, j, k as a result of these as well as the last line which I really don't understand.

BA[i][j] += B[i][k] * A[k][j]

Please explain as basic as possible because I am new to programming and so at this point nothing is trivial to me. Thank you for your time to help others :)

Answer 1

Here is the actual result from your code:

   B    *   A   =      AB
1 3 3 1   1 4 5     20 34 46
4 2 6 6   3 2 9     34 92 98
5 9 2 8   3 6 2     46 98 174
          1 6 8

Assuming i = 0 and j = 0 lets calculate the BA[0][0] , which is the first element from matrix BA.

BA[0][0] = B[0][k] * A[k][0]

B[0][k] means the line 0 from matrix B. As k is iterating over all lines of A, which is the same size as the number of columns in B.
A[k][0] means the column 0 from matrix A.

The loop for k in range(len(A)): will reproduce:

B[0][0]*A[0][0] + B[0][1]*A[1][0] + B[0][2]*A[2][0] + B[0][3]*A[3][0]

Resulting in:

1×1 + 3×3 + 3×3 + 1×1 = 20

Which is the value for BA[0][0] resulted from your code.

The following nested loops will iterate over all columns of A as j for every line of B as i in order to perform the multiplication for all (line) x (column) pairs:

for i in range(len(B)):
   for j in range(len(A[0])):

Answer 2

Consider here the array as just a more convenient representation of the list you gave to the function.

A is built on top of the list [[ 1, 4, 5 ],[ 3, 2, 9], [ 3,6, 2], [ 1,6, 8]] , a list of length 4.

range(start, stop) is a function which returns a generator which produces a sequence of integers, from the start to the stop point, stop not being included. If not provided, start defaults to 0.

B has a length of 4 rows so range(len(B)) will be like range(0, 3) , which will produce 0,1 and 2 integers, when "asked" by the for loop. i will subsequently be 0,1 and 2.

A[0] returns the first row of A, which has a length of 3, so the same way, j will subsequently be 0, 1 and 2 (in this case) ; and k will subsequently be 0, 1, 2 and 3 as A has a length of 4.

BA[i] returns the row of index i . Which can also be indexed by j So BA[i][j] is the element of row i and column j, which we increments by the product of the element of row i and index k of matrix B; and the element of row k and index j of matrix A.

Answer 3

In you code sample a matrix is represented aa list of sublists, where each sublist is a row.

So the most outer loop goes over the rows of B:

for i in range(len(B)):

A[0] is the first row of A, and the number of elements in it is the number of A's columns.

So the second loop goes over the columns of A:

for j in range(len(A[0])):

The most inner loop simply sums the products of the elements in the j-th row of B and the i-th row of A.

BA[i][j] += B[i][k] * A[k][j]

This add to BA[i][j] the product. += adds its right argument to its left one.