How do i fix the error that i am getting?

My code is

A = [1,1,0,2,0;0,1,1,3,0;2,0,0,0,1;3,1,0,2,1;2,1,1,3,0;1,0,0,2,1]

rrefA = rref(A)

basisRowSpace = A(rrefA(:,end)~=0,:)

pivotColumns = rrefA(:,1:end-1)

basisColumnSpace = A(:,pivotColumns)

This is where it stops working and I dont know how to fix it.

The error that I get is Index in postion 2 is invails. Array indices must be postive integers or logical values.

I would appricate any help to be able to fix this and get it to run.

Thank You.

Answer 1

[Torsten]

Remove the semicolon behind the line

pivotColumns = rrefA(:,1:end-1);

execute the code and see if it makes sense to insert the displayed matrix in

basisColumnSpace = A(:,pivotColumns);

[Taylor]

This still does not seem to fix the code. And even if I try putting just the matrix into the basisColumnSpace code it still gives me the same error. Thanks for try though.

[Taylor]

Though I may not be understanding what you are telling me to do.

[Torsten]

"pivotColumns" must be a vector of length <= 5 with entries between 1 and 5 that identifies the linear independent columns of the matrix A.

What you insert for "pivotColumns" is a matrix consisting of 0's and 1's.

This might be of interest:

https://www.mathworks.com/matlabcentral/fileexchange/77437-extract-linearly-independent-subset-of-matrix-columns

A = [1,1,0,2,0;0,1,1,3,0;2,0,0,0,1;3,1,0,2,1;2,1,1,3,0;1,0,0,2,1];
licols(A)

ans = 6×5

1 1 0 2 0 0 1 1 3 0 2 0 0 0 1 3 1 0 2 1 2 1 1 3 0 1 0 0 2 1

function [Xsub,idx]=licols(X,tol)
%Extract a linearly independent set of columns of a given matrix X
%
%    [Xsub,idx]=licols(X)
%
%in:
%
%  X: The given input matrix
%  tol: A rank estimation tolerance. Default=1e-10
%
%out:
%
% Xsub: The extracted columns of X
% idx:  The indices (into X) of the extracted columns

   if ~nnz(X) %X has no non-zeros and hence no independent columns
       
       Xsub=[]; idx=[];
       return
   end

   if nargin<2, tol=1e-10; end
   

           
     [Q, R, E] = qr(X,0); 
     
     if ~isvector(R)
      diagr = abs(diag(R));
     else
      diagr = abs(R(1));   
     end

     %Rank estimation
     r = find(diagr >= tol*diagr(1), 1, 'last'); %rank estimation

     idx=sort(E(1:r));

     Xsub=X(:,idx);                      
end

Thus all 5 columns of A are linearly independent.

[Taylor]

That is what i tried to insert the answer to pivotColumns that I get is [1,0,0,0;0,1,0,0;0,0,1,0;0,0,0,1;0,0,0,0;0,0,0,0]and when i tried inserting that into the basiscolumnSpace it stil didn't work.

[Torsten]

What is your code supposed to do ? Extract the maximum number of linearly independent columns of the matrix A ? Then use "licols" from above.

[Taylor]

I am supposed to be finding the basis for column space.

[Torsten]

Ok, that's what "licols" does. The maximum number of independent columns of A is a basis for the column space of A.