Finding the Frobenius Norm of a given matrix

Last Updated : 24 Jun, 2022

Given an M * N matrix, the task is to find the Frobenius Norm of the matrix. The Frobenius Norm of a matrix is defined as the square root of the sum of the squares of the elements of the matrix.
Example:

Input: mat[][] = {{1, 2}, {3, 4}}
Output: 5.47723
sqrt(1² + 2² + 3² + 4²) = sqrt(30) = 5.47723
Input: mat[][] = {{1, 4, 6}, {7, 9, 10}}
Output: 16.8226

Approach: Find the sum of squares of the elements of the matrix and then print the square root of the calculated value.
Below is the implementation of the above approach:

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int row = 2, col = 2;

// Function to return the Frobenius
// Norm of the given matrix
float frobeniusNorm(int mat[row][col])
{

    // To store the sum of squares of the
    // elements of the given matrix
    int sumSq = 0;
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
            sumSq += pow(mat[i][j], 2);
        }
    }

    // Return the square root of
    // the sum of squares
    float res = sqrt(sumSq);
    return res;
}

// Driver code
int main()
{
    int mat[row][col] = { { 1, 2 }, { 3, 4 } };

    cout << frobeniusNorm(mat);

    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
    
    final static int row = 2, col = 2; 
    
    // Function to return the Frobenius 
    // Norm of the given matrix 
    static float frobeniusNorm(int mat[][]) 
    { 
    
        // To store the sum of squares of the 
        // elements of the given matrix 
        int sumSq = 0; 
        for (int i = 0; i < row; i++) 
        { 
            for (int j = 0; j < col; j++) 
            { 
                sumSq += (int)Math.pow(mat[i][j], 2); 
            } 
        } 
    
        // Return the square root of 
        // the sum of squares 
        float res = (float)Math.sqrt(sumSq); 
        return res; 
    } 
    
    // Driver code 
    public static void main (String[] args)
    { 
        int mat[][] = { { 1, 2 }, { 3, 4 } }; 
    
        System.out.println(frobeniusNorm(mat)); 
    
    } 
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import sqrt
row = 2
col = 2

# Function to return the Frobenius
# Norm of the given matrix
def frobeniusNorm(mat):

    # To store the sum of squares of the
    # elements of the given matrix
    sumSq = 0
    for i in range(row):
        for j in range(col):
            sumSq += pow(mat[i][j], 2)

    # Return the square root of
    # the sum of squares
    res = sqrt(sumSq)
    return round(res, 5)

# Driver code

mat = [ [ 1, 2 ], [ 3, 4 ] ]

print(frobeniusNorm(mat))

# This code is contributed by mohit kumar 29

// C# implementation of the approach 
using System;

class GFG
{
    
    static int row = 2, col = 2; 
    
    // Function to return the Frobenius 
    // Norm of the given matrix 
    static float frobeniusNorm(int [,]mat) 
    { 
    
        // To store the sum of squares of the 
        // elements of the given matrix 
        int sumSq = 0; 
        for (int i = 0; i < row; i++) 
        { 
            for (int j = 0; j < col; j++) 
            { 
                sumSq += (int)Math.Pow(mat[i, j], 2); 
            } 
        } 
    
        // Return the square root of 
        // the sum of squares 
        float res = (float)Math.Sqrt(sumSq); 
        return res; 
    } 
    
    // Driver code 
    public static void Main ()
    { 
        int [,]mat = { { 1, 2 }, { 3, 4 } }; 
    
        Console.WriteLine(frobeniusNorm(mat)); 
    } 
}

// This code is contributed by AnkitRai01

JavaScript

<script>

// JavaScript implementation of the approach

    let row = 2, col = 2;
    
    // Function to return the Frobenius
    // Norm of the given matrix
    function frobeniusNorm(mat)
    {
    
        // To store the sum of squares of the
        // elements of the given matrix
        let sumSq = 0;
        for (let i = 0; i < row; i++)
        {
            for (let j = 0; j < col; j++)
            {
                sumSq += parseInt(Math.pow(mat[i][j], 2));
            }
        }
    
        // Return the square root of
        // the sum of squares
        let res = parseFloat(Math.sqrt(sumSq));
        return res;
    }
    
    // Driver code
    
        let mat = [[ 1, 2 ], [ 3, 4 ]];
    
        document.write(frobeniusNorm(mat).toFixed(5));
    
// This code is contributed by sravan kumar

</script>

Output:

5.47723

Time Complexity: O(M*N)

Auxiliary Space: O(1)

Analysis of Algorithms

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