Re: reduced row echelon form

[Paul Kienzle]

Here is one that I found.  I don't know if it is correct.

Paul Kienzle
address@hidden

##rref   Reduced row echelon form
##      rref (a, tol) returns the reduced row echelon form of a.
##      tol defaults to eps * max (size (a)) * norm (A, inf)

## This code is in the public domain

## Adapted-by: Paul Kienzle

function A = rref (A, tolerance)

  ## Supress empty list warnings
  ele_save = empty_list_elements_ok;
  unwind_protect
    empty_list_elements_ok = 1;
    
    [rows,cols] = size (A);
    r = 1;
    c = 1;

    if (nargin < 2)
      tolerance = eps * max (r, c) * norm (A, inf);
    endif
    
    while ((r<=rows) & (c<=cols))
      ## Find the pivot row
      [m, pivot] = max (abs (A (r:rows, c)));
      pivot = r + pivot - 1;
      
      if (m <= tolerance)
        ## Zero out the column
        A (r:rows, c++) = zeros (rows-r+1, 1);
      else
        ## Swap row p and row r
        A ([pivot, r], c:cols) = A ([r, pivot], c:cols);
        ## Divide by leading term
        A (r, c:cols) = A (r, c:cols) / A (r, c);
        ## Clear column c
        for i = [1:r-1, r+1:rows]
          A (i, c:cols) = A (i, c:cols) - A (i, c) * A (r, c:cols);
        endfor;
        c++;
        r++;
      endif
    endwhile
    
  unwind_protect_cleanup
    ## Restore state
    empty_list_elements_ok = ele_save;
  end_unwind_protect
endfunction

On Sun, Oct 15, 2000 at 11:43:37PM -0400, ben partan wrote:
> Is there a built-in command for get a reduced row echelon form of a
> matrix? Or is there an easy way to get one?
> Thanks.
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