Program for Rank of Matrix

30 Jun, 2022
What is rank of a matrix? Rank of a matrix A of size M x N is defined as 
Maximum number of linearly independent column vectors in the matrix or Maximum number of linearly independent row vectors in the matrix. 
Maximum number of linearly independent column vectors in the matrix or 
Maximum number of linearly independent row vectors in the matrix. 
Example:
Input:  mat[][] = {{10,   20,   10},
                   {20,   40,   20},
                   {30,   50,   0}}
Output: Rank is 2
Explanation: Ist and IInd rows are linearly dependent.
             But Ist and  3rd or IInd and IIIrd  are
             independent.   


Input:  mat[][] = {{10,   20,   10},
                   {-20, -30,   10},
                   {30,   50,   0}}
Output: Rank is 2
Explanation: Ist and IInd rows are linearly independent.
             So rank must be atleast 2. But all three rows
             are linearly dependent (the first is equal to
             the sum of the second and third) so the rank 
             must be less than 3. 
In other words rank of A is the largest order of any non-zero minor in A where order of a minor is the side-length of the square sub-matrix of which it is determinant.So if M < N then maximum rank of A can be M else it can be N, in general rank of matrix can’t be greater than min(M, N).The rank of a matrix would be zero only if the matrix had no non-zero elements. If a matrix had even one non-zero element, its minimum rank would be one. 
How to find Rank? The idea is based on conversion to Row echelon form. 
1) Let the input matrix be mat[][].  Initialize rank equals
   to number of columns

// Before we visit row 'row',  traversal of previous 
// rows make sure that mat[row][0],....mat[row][row-1]
// are 0.
2) Do following for row = 0 to rank-1.

  a) If mat[row][row] is not zero, make all elements of
     current column as 0 except the element mat[row][row]
     by finding appropriate multiplier and adding a the 
     multiple of row 'row'
        
  b) Else (mat[row][row] is zero). Two cases arise:
       (i) If there is a row below it with non-zero entry in 
           same column, then swap current 'row' and that row.
       (ii) If all elements in current column below mat[r][row] 
            are 0, then remove this column by swapping it with
            last column and  reducing number of rank by 1.
     Reduce row by 1 so that this row is processed again.

3) Number of remaining columns is rank of matrix.
Example: 
Input:  mat[][] = {{10,   20,   10},
                   {-20, -30,   10},
                   {30,   50,   0}}

row = 0:
Since mat[0][0] is not 0, we are in case 2.a of above algorithm.
We set all entries of 0'th column as 0 (except entry mat[0][0]).
To do this, we subtract R1*(-2) from R2, i.e., R2 --> R2 - R1*(-2)
         mat[][] = {{10,   20,   10},
                   { 0,    10,   30},
                    {30,   50,   0}}
            And subtract R1*3 from R3, i.e., R3   --> R3 - R1*3  
         mat[][] = {{10,   20,   10},
                    { 0,   10,   30},
                    { 0,  -10,  -30}}

row = 1:
Since mat[1][1] is not 0, we are in case 2.a of above algorithm.
We set all entries of 1st column as 0 (except entry mat[1][1]).
To do this, we subtract R2*2 from R1, i.e., R1 --> R1 - R2*2
         mat[][] = {{10,    0,  -50},
                    { 0,   10,   30},
                    { 0,  -10,  -30}}
           And subtract R2*(-1) from R3, i.e., R3   --> R3 - R2*(-1)
         mat[][] = {{10,   0,   -50},
                    { 0,   10,   30},
                    { 0,   0,     0}}

row = 2:
Since Since mat[2][2] is 0, we are in case 2.b of above algorithm.
Since there is no row below it swap. We reduce the rank by 1 and 
keep row as 2. 

The loop doesn't iterate next time because loop termination condition
row <= rank-1 returns false.
Below is the implementation of above idea.
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find rank of a matrix#include <bits/stdc++.h>using namespace std;#define R 3#define C 3 /* function for exchanging two rows of   a matrix */void swap(int mat[R][C], int row1, int row2,          int col){    for (int i = 0; i < col; i++)    {        int temp = mat[row1][i];        mat[row1][i] = mat[row2][i];        mat[row2][i] = temp;    }} // Function to display a matrixvoid display(int mat[R][C], int row, int col); /* function for finding rank of matrix */int rankOfMatrix(int mat[R][C]){    int rank = C;     for (int row = 0; row < rank; row++)    {        // Before we visit current row 'row', we make        // sure that mat[row][0],....mat[row][row-1]        // are 0.         // Diagonal element is not zero        if (mat[row][row])        {           for (int col = 0; col < R; col++)           {               if (col != row)               {                 // This makes all entries of current                 // column as 0 except entry 'mat[row][row]'                 double mult = (double)mat[col][row] /                                       mat[row][row];                 for (int i = 0; i < rank; i++)                   mat[col][i] -= mult * mat[row][i];              }           }        }         // Diagonal element is already zero. Two cases        // arise:        // 1) If there is a row below it with non-zero        //    entry, then swap this row with that row        //    and process that row        // 2) If all elements in current column below        //    mat[r][row] are 0, then remove this column        //    by swapping it with last column and        //    reducing number of columns by 1.        else        {            bool reduce = true;             /* Find the non-zero element in current                column  */            for (int i = row + 1; i < R;  i++)            {                // Swap the row with non-zero element                // with this row.                if (mat[i][row])                {                    swap(mat, row, i, rank);                    reduce = false;                    break ;                }            }             // If we did not find any row with non-zero            // element in current column, then all            // values in this column are 0.            if (reduce)            {                // Reduce number of columns                rank--;                 // Copy the last column here                for (int i = 0; i < R; i ++)                    mat[i][row] = mat[i][rank];            }             // Process this row again            row--;        }        // Uncomment these lines to see intermediate results       // display(mat, R, C);       // printf("\n");    }    return rank;} /* function for displaying the matrix */void display(int mat[R][C], int row, int col){    for (int i = 0; i < row; i++)    {        for (int j = 0; j < col; j++)            printf("  %d", mat[i][j]);        printf("\n");    }} // Driver program to test above functionsint main(){   int mat[][3] = {{10,   20,   10},                  {-20,  -30,   10},                   {30,   50,   0}};    printf("Rank of the matrix is : %d",         rankOfMatrix(mat));    return 0;}
// Java program to find rank of a matrixclass GFG {         static final int R = 3;    static final int C = 3;         // function for exchanging two rows    // of a matrix    static void swap(int mat[][],          int row1, int row2, int col)    {        for (int i = 0; i < col; i++)        {            int temp = mat[row1][i];            mat[row1][i] = mat[row2][i];            mat[row2][i] = temp;        }    }         // Function to display a matrix    static void display(int mat[][],                     int row, int col)    {        for (int i = 0; i < row; i++)        {                         for (int j = 0; j < col; j++)                             System.out.print(" "                          + mat[i][j]);                                       System.out.print("\n");        }    }         // function for finding rank of matrix    static int rankOfMatrix(int mat[][])    {                 int rank = C;             for (int row = 0; row < rank; row++)        {                         // Before we visit current row            // 'row', we make sure that            // mat[row][0],....mat[row][row-1]            // are 0.                 // Diagonal element is not zero            if (mat[row][row] != 0)            {                for (int col = 0; col < R; col++)                {                    if (col != row)                    {                        // This makes all entries                        // of current column                        // as 0 except entry                        // 'mat[row][row]'                        double mult =                           (double)mat[col][row] /                                    mat[row][row];                                                             for (int i = 0; i < rank; i++)                                                     mat[col][i] -= mult                                       * mat[row][i];                    }                }            }                 // Diagonal element is already zero.            // Two cases arise:            // 1) If there is a row below it            // with non-zero entry, then swap            // this row with that row and process            // that row            // 2) If all elements in current            // column below mat[r][row] are 0,            // then remove this column by            // swapping it with last column and            // reducing number of columns by 1.            else            {                boolean reduce = true;                     // Find the non-zero element                // in current column                for (int i = row + 1; i < R; i++)                {                    // Swap the row with non-zero                    // element with this row.                    if (mat[i][row] != 0)                    {                        swap(mat, row, i, rank);                        reduce = false;                        break ;                    }                }                     // If we did not find any row with                // non-zero element in current                // column, then all values in                // this column are 0.                if (reduce)                {                    // Reduce number of columns                    rank--;                         // Copy the last column here                    for (int i = 0; i < R; i ++)                        mat[i][row] = mat[i][rank];                }                     // Process this row again                row--;            }             // Uncomment these lines to see        // intermediate results display(mat, R, C);        // printf("\n");        }                 return rank;    }         // Driver code    public static void main (String[] args)    {        int mat[][] = {{10, 20, 10},                       {-20, -30, 10},                       {30, 50, 0}};                                System.out.print("Rank of the matrix is : "                               + rankOfMatrix(mat));    }} // This code is contributed by Anant Agarwal.
# Python 3 program to find rank of a matrixclass rankMatrix(object):    def __init__(self, Matrix):        self.R = len(Matrix)        self.C = len(Matrix[0])             # Function for exchanging two rows of a matrix    def swap(self, Matrix, row1, row2, col):        for i in range(col):            temp = Matrix[row1][i]            Matrix[row1][i] = Matrix[row2][i]            Matrix[row2][i] = temp                 # Function to Display a matrix    def Display(self, Matrix, row, col):        for i in range(row):            for j in range(col):                print (" " + str(Matrix[i][j]))            print ('\n')                 # Find rank of a matrix    def rankOfMatrix(self, Matrix):        rank = self.C        for row in range(0, rank, 1):                         # Before we visit current row            # 'row', we make sure that            # mat[row][0],....mat[row][row-1]            # are 0.                 # Diagonal element is not zero            if Matrix[row][row] != 0:                for col in range(0, self.R, 1):                    if col != row:                                                 # This makes all entries of current                        # column as 0 except entry 'mat[row][row]'                        multiplier = (Matrix[col][row] /                                      Matrix[row][row])                        for i in range(rank):                            Matrix[col][i] -= (multiplier *                                               Matrix[row][i])                                                             # Diagonal element is already zero.            # Two cases arise:            # 1) If there is a row below it            # with non-zero entry, then swap            # this row with that row and process            # that row            # 2) If all elements in current            # column below mat[r][row] are 0,            # then remove this column by            # swapping it with last column and            # reducing number of columns by 1.            else:                reduce = True                                 # Find the non-zero element                # in current column                for i in range(row + 1, self.R, 1):                                         # Swap the row with non-zero                    # element with this row.                    if Matrix[i][row] != 0:                        self.swap(Matrix, row, i, rank)                        reduce = False                        break                                         # If we did not find any row with                # non-zero element in current                # column, then all values in                # this column are 0.                if reduce:                                         # Reduce number of columns                    rank -= 1                                         # copy the last column here                    for i in range(0, self.R, 1):                        Matrix[i][row] = Matrix[i][rank]                                         # process this row again                row -= 1                         # self.Display(Matrix, self.R,self.C)        return (rank) # Driver Codeif __name__ == '__main__':    Matrix = [[10, 20, 10],              [-20, -30, 10],              [30, 50, 0]]    RankMatrix = rankMatrix(Matrix)    print ("Rank of the Matrix is:",           (RankMatrix.rankOfMatrix(Matrix))) # This code is contributed by Vikas Chitturi
// C# program to find rank of a matrixusing System;class GFG {          static  int R = 3;    static  int C = 3;          // function for exchanging two rows    // of a matrix    static void swap(int [,]mat,          int row1, int row2, int col)    {        for (int i = 0; i < col; i++)        {            int temp = mat[row1,i];            mat[row1,i] = mat[row2,i];            mat[row2,i] = temp;        }    }          // Function to display a matrix    static void display(int [,]mat,                     int row, int col)    {        for (int i = 0; i < row; i++)        {                          for (int j = 0; j < col; j++)                              Console.Write(" "                          + mat[i,j]);                                        Console.Write("\n");        }    }          // function for finding rank of matrix    static int rankOfMatrix(int [,]mat)    {                  int rank = C;              for (int row = 0; row < rank; row++)        {                          // Before we visit current row            // 'row', we make sure that            // mat[row][0],....mat[row][row-1]            // are 0.                  // Diagonal element is not zero            if (mat[row,row] != 0)            {                for (int col = 0; col < R; col++)                {                    if (col != row)                    {                        // This makes all entries                        // of current column                        // as 0 except entry                        // 'mat[row][row]'                        double mult =                           (double)mat[col,row] /                                    mat[row,row];                                                              for (int i = 0; i < rank; i++)                                                      mat[col,i] -= (int) mult                                     * mat[row,i];                    }                }            }                  // Diagonal element is already zero.            // Two cases arise:            // 1) If there is a row below it            // with non-zero entry, then swap            // this row with that row and process            // that row            // 2) If all elements in current            // column below mat[r][row] are 0,            // then remove this column by            // swapping it with last column and            // reducing number of columns by 1.            else            {                bool reduce = true;                      // Find the non-zero element                // in current column                for (int i = row + 1; i < R; i++)                {                    // Swap the row with non-zero                    // element with this row.                    if (mat[i,row] != 0)                    {                        swap(mat, row, i, rank);                        reduce = false;                        break ;                    }                }                      // If we did not find any row with                // non-zero element in current                // column, then all values in                // this column are 0.                if (reduce)                {                    // Reduce number of columns                    rank--;                          // Copy the last column here                    for (int i = 0; i < R; i ++)                        mat[i,row] = mat[i,rank];                }                      // Process this row again                row--;            }              // Uncomment these lines to see        // intermediate results display(mat, R, C);        // printf("\n");        }                  return rank;    }          // Driver code    public static void Main ()    {        int [,]mat = {{10, 20, 10},                       {-20, -30, 10},                       {30, 50, 0}};                                 Console.Write("Rank of the matrix is : "                          + rankOfMatrix(mat));    }}  // This code is contributed by nitin mittal
<?php// PHP program to find rank of a matrix $R = 3;$C = 3; /* function for exchanging two rows ofa matrix */function swap(&$mat, $row1, $row2, $col){    for ($i = 0; $i < $col; $i++)    {        $temp = $mat[$row1][$i];        $mat[$row1][$i] = $mat[$row2][$i];        $mat[$row2][$i] = $temp;    }}  /* function for finding rank of matrix */function rankOfMatrix($mat){    global $R, $C;    $rank = $C;     for ($row = 0; $row < $rank; $row++)    {        // Before we visit current row 'row', we make        // sure that mat[row][0],....mat[row][row-1]        // are 0.         // Diagonal element is not zero        if ($mat[$row][$row])        {            for ($col = 0; $col < $R; $col++)            {                if ($col != $row)                {                    // This makes all entries of current                    // column as 0 except entry 'mat[row][row]'                    $mult = $mat[$col][$row] / $mat[$row][$row];                    for ($i = 0; $i < $rank; $i++)                        $mat[$col][$i] -= $mult * $mat[$row][$i];                }            }        }         // Diagonal element is already zero. Two cases        // arise:        // 1) If there is a row below it with non-zero        // entry, then swap this row with that row        // and process that row        // 2) If all elements in current column below        // mat[r][row] are 0, then remove this column        // by swapping it with last column and        // reducing number of columns by 1.        else        {            $reduce = true;             /* Find the non-zero element in current                column */            for ($i = $row + 1; $i < $R; $i++)            {                // Swap the row with non-zero element                // with this row.                if ($mat[$i][$row])                {                    swap($mat, $row, $i, $rank);                    $reduce = false;                    break ;                }            }             // If we did not find any row with non-zero            // element in current column, then all            // values in this column are 0.            if ($reduce)            {                // Reduce number of columns                $rank--;                 // Copy the last column here                for ($i = 0; $i < $R; $i++)                    $mat[$i][$row] = $mat[$i][$rank];            }             // Process this row again            $row--;        }     // Uncomment these lines to see intermediate results    // display(mat, R, C);    // printf("\n");    }    return $rank;} /* function for displaying the matrix */function display($mat, $row, $col){    for ($i = 0; $i < $row; $i++)    {        for ($j = 0; $j < $col; $j++)            print(" $mat[$i][$j]");        print("\n");    }} // Driver code$mat = array(array(10, 20, 10),                array(-20, -30, 10),                array(30, 50, 0));print("Rank of the matrix is : ".rankOfMatrix($mat)); // This code is contributed by mits?>
<script> // javascript program to find rank of a matrix          var R = 3;    var C = 3;         // function for exchanging two rows    // of a matrix    function swap(mat, row1 , row2 , col)    {        for (i = 0; i < col; i++)        {            var temp = mat[row1][i];            mat[row1][i] = mat[row2][i];            mat[row2][i] = temp;        }    }         // Function to display a matrix    function display(mat,row , col)    {        for (i = 0; i < row; i++)        {                         for (j = 0; j < col; j++)                             document.write(" "                          + mat[i][j]);                                       document.write('<br>');        }    }         // function for finding rank of matrix    function rankOfMatrix(mat)    {                 var rank = C;             for (row = 0; row < rank; row++)        {                         // Before we visit current row            // 'row', we make sure that            // mat[row][0],....mat[row][row-1]            // are 0.                 // Diagonal element is not zero            if (mat[row][row] != 0)            {                for (col = 0; col < R; col++)                {                    if (col != row)                    {                        // This makes all entries                        // of current column                        // as 0 except entry                        // 'mat[row][row]'                        var mult =                           mat[col][row] /                                    mat[row][row];                                                             for (i = 0; i < rank; i++)                                                     mat[col][i] -= mult                                       * mat[row][i];                    }                }            }                 // Diagonal element is already zero.            // Two cases arise:            // 1) If there is a row below it            // with non-zero entry, then swap            // this row with that row and process            // that row            // 2) If all elements in current            // column below mat[r][row] are 0,            // then remove this column by            // swapping it with last column and            // reducing number of columns by 1.            else            {                reduce = true;                     // Find the non-zero element                // in current column                for (var i = row + 1; i < R; i++)                {                    // Swap the row with non-zero                    // element with this row.                    if (mat[i][row] != 0)                    {                        swap(mat, row, i, rank);                        reduce = false;                        break ;                    }                }                     // If we did not find any row with                // non-zero element in current                // column, then all values in                // this column are 0.                if (reduce)                {                    // Reduce number of columns                    rank--;                         // Copy the last column here                    for (i = 0; i < R; i ++)                        mat[i][row] = mat[i][rank];                }                     // Process this row again                row--;            }             // Uncomment these lines to see        // intermediate results display(mat, R, C);        // printf(<br>);        }                 return rank;    }         // Driver code    var mat = [[10, 20, 10],            [-20, -30, 10],            [30, 50, 0]];                         document.write("Rank of the matrix is : "                    + rankOfMatrix(mat)); // This code is contributed by Amit Katiyar</script>
Rank of the matrix is : 2
Time complexity: O(row x col x rank).Auxiliary Space: O(1)
Since above rank calculation method involves floating point arithmetic, it may produce incorrect results if the division goes beyond precision. There are other methods to handle.
nitin mittal
Mithun Kumar
Vikas Chitturi
amit143katiyar
varshagumber28
sweetyty
codewithshinchan
hardikkoriintern
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