Simple Identity matrix function

Question

Expected Output:

indenitiy_matrix(3)
[[1, 0, 0], [0, 1, 0], [0, 0, 1]]

Actual Output with Error:

indenitiy_matrix(3)
[[1, 1, 1], [1, 1, 1], [1, 1, 1]] 

def identity_matrix(n):
    list_template = [[]]
    list_n = list_template*n

    for sub_l in list_n:
        sub_l.append(0)


    for val in range(n):
        # I have the feeling that the problem lies somewhere around here.
        list_n[val][val]=1


    return(list_n)

Answer 1

list_template*n does not create n copies, instead but all those n copies reference to only one copy. For example see this

a = [[0,0,0]]*2
# Now, lets change first element of the first sublist in `a`. 
a[0][0] = 1
print (a)
# but since both the 2 sublists refer to same, both of them will be changed. 

Output:

[[1, 0, 0], [1, 0, 0]]

Fix for your code

def identity_matrix(n):
    list_n = [[0]*n for i in range(n)]
    for val in range(n):        
        list_n[val][val]=1
    
    return list_n

print (identity_matrix(5))

Output:

[[1, 0, 0, 0, 0],
 [0, 1, 0, 0, 0],
 [0, 0, 1, 0, 0],
 [0, 0, 0, 1, 0],
 [0, 0, 0, 0, 1]]