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 Author: rambiz [ Tue Jun 07, 2016 3:27 pm ] Post subject: Reduced Row Echelon Form Hi all, how can one get the "Reduced Row Echelon Form" of a matrix in Singular, please? example: (input is a 4x3 matrix)input=[1 2 3 4 5 6 7 8 9 10 11 12]desired output=[1 0 -1 0 1 2 0 0 0 0 0 0]

 Author: rambiz [ Wed Jun 08, 2016 12:09 am ] Post subject: My code for the Reduced Row Echelon Form Hi all, I have written a code to implement conversion to the reduced row echelon form.I used the Groebner Basis approach, which is most probably not the smartest way. One needs to set the correct number of ring variables manually in the code, and of course, enter the matrix. In case the number of rows of output is smaller than the number of rows of the matrix, those rows are all zero.Code:ring r=0,(x(1..4)),dp;    //number of variables must be the same as the number of columns of the matrix for which we want the reduced row echelon formmatrix m[4][4]=16,2,3,13,5,11,10,8,9,7,6,12,4,14,15,1; //enter an arbitrary matrix hereprint("m=");print(m);int nr=nrows(m);int nc=ncols(m);int j;int k;for(j=1;j<=nr;j=j+1){for(k=1;k<=nc;k=k+1){m[j,k]=m[j,k]*x(k);}};kill j;kill k;matrix ONES[nc][1];for(int j=1;j<=nc;j=j+1){ONES[j,1]=1;};kill j;ideal i=m*ONES;option(redSB);ideal g=std(i);int L=size(g);for(int j=1;j<=L;j=j+1) {g[j]=g[j]/leadcoef(g[j]);} // set the coefficients of the leading terms to unitykill j;matrix RREF[L][nc];int k;int j;for (j=L;j>0;j=j-1){for (k=1;k<=nc;k=k+1){RREF[j,k]=jet(g[L+1-j]/x(k),0);}};print("RREF=");print(RREF);print("RREF is the Reduced Row Echelon Form of the matrix m")A) Do you think this code is correct? I have double-checked a few matrices with MATLAB and it seems to be OK. B)Any suggestion how to improve my code?

 Author: hannes [ Thu Jun 09, 2016 10:27 am ] Post subject: Re: Reduced Row Echelon Form see http://www.singular.uni-kl.de/Manual/4-0-3/sing_338.htm

 Author: rambiz [ Thu Jun 09, 2016 4:29 pm ] Post subject: ludecomp Hi Hannes, I had actually had a look at the function you suggested before I posted my original message. However, I don't know how that decomposition can be used to get the reduced row echelon form. Can you please elaborate? Thanks.

 Author: hannes [ Fri Jun 10, 2016 11:05 am ] Post subject: Re: Reduced Row Echelon Form Code:ring R=0,x,dp;matrix m[4][3]=1,2,3,4,5,6,7,8,9,10,11,12;list L=ludecomp (m);print(L[3]);1,2, 3, 0,-3,-6,0,0, 0, 0,0, 0

 Author: rambiz [ Sat Jun 11, 2016 4:42 pm ] Post subject: Re: Reduced Row Echelon Form According to my definition in a reduced row echelon form the leading terms should all be one, and the remaining entries in the same column should be zero. In the case of your example: matrix m[4][3]=1,2,3,4,5,6,7,8,9,10,11,12; Desired output is: 1 0 -10 1 20 0 00 0 0

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