Exercises 7.4 - 7.5

Chapter 7. Dimension and Structure

7.4 The Dimension Theorem and Its Implications

Exercise 7.5. (Rank and Nullity)

1. Use the MATLAB command rank and the Formula (2) in Section 7.4 to find the nullity of the matrix

   \[A = \left[\begin{array}{rrrrr} 3& \hspace{1mm} 2& \hspace{1mm} 1&\hspace{1mm} 3&\hspace{1mm} 5\\ 6 & 4 & 3 & 5 & 7 \\ 9 & 6 & 5 & 7 & 9 \\ 3 & 2 & 0 & 4 & 8 \end{array} \right].\]

2. Confirm that the result obtained in (a) is consistent with the number of basis vectors which are obtained by using the MATLAB command null.

Solutions.

1. % Set A.
   A = [3 2 1 3 5; 6 4 3 5 7; 9 6 5 7 9; 3 2 0 4 8]; 
   
   % Find the rank of A by using the command rank.
   rank_A = rank(A); 
   
   % Size of the matrix A.
   [m, n] = size(A); 
   % m = the number of rows of A, n = the number of columns of A.
   
   % By (2) in section 7.4, rankA + nullA = n.
   null_A = n - rank_A;
   
   disp('The nullity of A is'); disp(null_A);
   

   MATLAB results.

   The nullity of A is
        3
   

2. % Set A.
   A = [3 2 1 3 5; 6 4 3 5 7; 9 6 5 7 9; 3 2 0 4 8]; 
   
   % Find a basis for the null space of A.
   nullA = null(A,'r'); 
   % null(A,'r') returns a matrix 
   % whose columns are a basis for the null space of A.
   
   [m, n] = size(nullA);
   % Since the number of columns of nullA is n,
   % thus, n = the number of basis vectors of the null space of A.
   
   disp('The nullity of A is'); disp(n);
   

   MATLAB results.

   The nullity of A is
        3
   

7.5 The Rank Theorem and Its Implications

Exercise 7.6. (Need the Symbolic toolbox)

Note that the rank of a nonzero matrix \(A\) is equal to the order of the largest square submatrix of \(A\) (formed by deleting rows and columns of \(A\)) whose determinant is nonzero. In this problem, we make a function file CheckRank.m to find the rank of the given matrix using this fact. We want to obtain the execution results as follows:

>> A=[1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
>> rankA=CheckRank(A)
rankA =
     2

For this, you may start with the largest square matrices to be found in \(A\) and a search is started for the first submatrix with a nonzero determinant. Use the MATLAB command nchoosek to select all the combinations of rows and columns needed in the search process and you may use the several MATLAB commands if you need. Complete the m-file below and check the determinant of the matrices \(A\), \(B\), and \(C\) given in the Exercise \(7.3\), (2). Also, compare the results using the MATLAB command rank.

Solutions.

%--- function file 'CheckRank.m' ---%
function [rank_A]= CheckRank(A)
[m, n]=size(A); % size of given matrix
flg=1;          % flag for while loop
if m > n        % if (# of row) > (# of col)
    A = A';
end
A=sym(A);       % Set A as a symbolic object

K = min(m,n);   N = max(m,n);   % k : row number, N: col number
k = K;          % from the largest size of submatrix
while flg == 1
    comb_row=nchoosek(1:K, k);  % combinations of row
    comb_col=nchoosek(1:N, k);  % combinations of columns
    for ii=1:size(comb_row)
        selected_A=A(comb_row(ii,:),:); % selected row index
        for jj=1:size(comb_col)
            sub_A=selected_A(:,comb_col(jj, :));   % selected col index
            if det(sub_A)~=0    % if non zeros determinant appears
                rank_A=k;       % the size at that time <- rank
                flg=0;          % stop the while loop.
            end
        end
    end
    % if all submatrices of size k have a zero determinant,
    % reduce the size of submatrix.
    k=k-1;
end
end

To check the determinant of the matrices \(A\), \(B\), and \(C\) given in the Exercise \(7.3\), you execute the followings:

A = [3 2 1 3 5; 6 4 3 5 7; 9 6 5 7 9; 3 2 0 4 8];
B = [3 -1 3 2 5; 5 -3 2 3 4; 1 -3 -5 0 -7; 7 -5 1 4 1];
C = [1 3 2 1; -2 -6 0 -6 ;3 9 1 8; -1 -3 -3 -6; 1 3 2 1; 4 12 1 11];

fprintf('my rank(A): %.5f, MATLAB rank(A): %.5f \n', CheckRank(A), rank(A));
fprintf('my rank(B): %.5f, MATLAB rank(B): %.5f \n', CheckRank(B), rank(B));
fprintf('my rank(C): %.5f, MATLAB rank(C): %.5f \n', CheckRank(C), rank(C));

MATLAB results.

my rank(A): 2.00000, MATLAB rank(A): 2.00000
my rank(B): 3.00000, MATLAB rank(B): 3.00000
my rank(C): 3.00000, MATLAB rank(C): 3.00000

Those are the same results as given in Exercise \(7.3\).