Introduction to Identity Matrix :
The dictionary definition of an Identity Matrix is a square matrix in which all the elements of the principal or main diagonal are 1’s and all other elements are zeros. In the below image, every matrix is an Identity Matrix.
In linear algebra, this is sometimes called as a Unit Matrix, of a square matrix (size = n x n) with ones on the main diagonal and zeros elsewhere. The identity matrix is denoted by “ I “. Sometimes U or E is also used to denote an Identity Matrix.
A property of the identity matrix is that it leaves a matrix unchanged if it is multiplied by an Identity Matrix.
Examples:
Input : 2 Output : 1 0 0 1 Input : 4 Output : 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 The explanation is simple. We need to make all the elements of principal or main diagonal as 1 and everything else as 0.
Program to print Identity Matrix :
The logic is simple. You need to the print 1 in those positions where row is equal to column of a matrix and make all other positions as 0.
Python3
# Python code to print identity matrix # Function to print identity matrix def Identity(size): for row in range ( 0 , size): for col in range ( 0 , size): # Here end is used to stay in same line if (row = = col): print ( "1 " , end = " " ) else : print ( "0 " , end = " " ) print () # Driver Code size = 5 Identity(size) |
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
Time complexity: O(R*C) where R and C is no of rows and column in matrix respectively
Space Complexity: O(1) as no extra space has been used.
Program to check if a given square matrix is Identity Matrix :
Python3
# Python3 program to check # if a given matrix is identity MAX = 100 ; def isIdentity(mat, N): for row in range (N): for col in range (N): if (row = = col and mat[row][col] ! = 1 ): return False ; elif (row ! = col and mat[row][col] ! = 0 ): return False ; return True ; # Driver Code N = 4 ; mat = [[ 1 , 0 , 0 , 0 ], [ 0 , 1 , 0 , 0 ], [ 0 , 0 , 1 , 0 ], [ 0 , 0 , 0 , 1 ]]; if (isIdentity(mat, N)): print ( "Yes " ); else : print ( "No " ); # This code is contributed # by mits |
Output:
Yes
Time complexity: O(N2) where N is number of rows and columns of matrix
Auxiliary Space: O(1)
Using Numpy:
Note: Before running the code please install the Numpy library using the command below
pip install numpy
Use the numpy library to check if a matrix is an identity matrix. This can be done by using the numpy.allclose function to compare the matrix with the identity matrix of the same size.
For example, the following code snippet checks if a matrix is an identity matrix:
Python3
import numpy as np def is_identity(matrix): # Get the size of the matrix size = matrix.shape[ 0 ] # Create the identity matrix of the same size identity = np.eye(size) # Check if the matrix is equal to the identity matrix using numpy.allclose return np.allclose(matrix, identity) # Example usage matrix = np.array([[ 1 , 0 , 0 , 0 ], [ 0 , 1 , 0 , 0 ], [ 0 , 0 , 1 , 0 ], [ 0 , 0 , 0 , 1 ]]) print (is_identity(matrix)) # prints True #This code is contributed by Edula Vinay Kumar Reddy |
Output: True
This approach has the advantage of being more concise and easier to read, and it also takes advantage of the optimized array operations provided by numpy. The time complexity of this approach will depend on the complexity of the numpy.allclose function, which is generally O(N^2) for dense matrices and space complexity is O(N^2)