Trying to construct identity matrix?

Question

I just need to create a function identity which return n identity matrix.

Example:

identity(3)

output's [[1,0,0][0,1,0][0,0,1]]

I also tried the below code:

def identity(n):

matrix=[[0]*n]*n

i=0

while i<n:

    matrix[i][i]=1

    i+=1

return matrix

Also tried with the range function, but it doesn't work.

def identity(n):

    matrix=[[0]*n]*n

    k=matrix[:]

    i=0

    for i in range(1,n):

        matrix[i][i]=1

        i+=1

    return k

print(identity(5))

For n=5, it outputs:

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

Answer 1

To create an identity matrix, you can use the following code:

def identity(n):

    return [[1 if i == j else 0 for j in range(n)] for i in range(n)]

print(identity(5))

This code will generate the desired identity matrix for n=5:

[[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]]

Answer 2

Try the below code:

def identity(n):

    m=[[0 for x in range(n)] for y in range(n)]

    for i in range(0,n):

        m[i][i] = 1

    return m

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Answer 3

You are experiencing an issue with your code because the way you are creating the identity matrix using [[0]*n]*n leads to incorrect references. Modifying one element affects the entire column in the matrix. To fix this, you can use a nested list comprehension or a loop to generate the identity matrix correctly. Here's an updated version of your code:

def identity(n):

    matrix = [[0] * n for _ in range(n)]  # Create an empty matrix

    for i in range(n):

        matrix[i][i] = 1  # Set diagonal elements to 1

    return matrix

print(identity(5))

This code will generate the correct identity matrix for the given input n=5:

[[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]]

Answer 4

You're encountering an issue with your code when trying to create an identity matrix using the approach [[0]*n]*n. The problem arises because this method creates references to the same row, resulting in modifications affecting the entire column. To resolve this, you can use either a nested list comprehension or a loop to construct the identity matrix correctly. Here's an improved version of your code:

def identity(n):

    matrix = [[0] * n for _ in range(n)]  # Create an empty matrix

    for i in range(n):

        matrix[i][i] = 1  # Set diagonal elements to 1

    return matrix

print(identity(5))

By utilizing this code, you'll obtain the desired identity matrix for n=5:

[[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]]