Re: [Scilab-users] a linear equation
Dear all I found a solution. I get the column echelon form of A by X=rref(A')'. Then the rows' numbers with the non-zero pivot of X are the independent rows' numbers of A. In this problem 26th row and 27th row are redundant. 26th rows are represented by linear combinations of 24th row and 1st-23rd rows. 27th rows are represented by a linear combination of 25th row and 1st-23rd rows. Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Dear Wescot I know how to approximate A with SVD, but I do not know how to find independent row vectors of A. How can I do? Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Sorry, I was werong with the constraint matrix. The right values are as follows. A=zeros(27,27) A(1,10)=1 A(1,10)=1 A(2,5)=1 A(3,14)=1 A(4,23)=1 A(5,7)=1 A(6,16)=1 A(7,25)=1 A(8,17)=1 A(9,1)=0,A(9,18)=1 A(10,2)=1,A(10,3)=1 A(11,4)=1,A(11,5)=1,A(11,6)=1 A(12,7)=1,A(12,8)=1,A(12,9)=1 A(13,10)=1,A(13,11)=1,A(13,12)=1 A(14,13)=1,A(14,14)=1,A(14,15)=1 A(15,19)=1,A(15,20)=1,A(15,21)=1 A(16,22)=1,A(16,23)=1,A(16,24)=1 A(17,1)=0,A(17,25)=1,A(17,26)=1,A(17,27)=1 A(18,4)=1,A(18,7)=1 A(19,2)=1,A(19,5)=1,A(19,8)=1 A(20,10)=1,A(20,13)=1,A(20,16)=1 A(21,11)=1,A(21,14)=1,A(21,17)=1 A(22,19)=1,A(22,22)=1,A(22,25)=1 A(23,1)=0,A(23,20)=1,A(23,23)=1,A(23,26)=1 A(24,10)=1,A(24,19)=1 A(25,2)=1,A(25,11)=1,A(25,20)=1 A(26,4)=1,A(26,13)=1,A(26,22)=1 A(27,8)=1,A(27,17)=1,A(27,26)=1 Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Is there a reason not to do SVD, and throw out the singular values that are too small? On Sun, 2018-12-02 at 09:56 -0700, fujimoto2005 wrote: > This problem is an economic problem. The i-th row of the square > constraint > matrix A with m dimension expresses certain economic constraints. > The elements of the constraint matrix are either 0 or 1. > Suppose the rank of A is r and by changing the row number a_1, ..., > a_r are > linearly independent. > I guess the coefficient cj (1 ≤ j ≤ r) in a_i = c1 * a_1 + ... cr * > a_r for > any m>=i> r is found to be 0, -1, 1 for some economic reasons. > I would like to find such a_1, ..., a_r pairs. > > The following matrix is the constraint matrix which I am dealing > with. > > A=zeros(27,27) > > A(1,10)=1 > A(2,5)=1 > A(3,14)=1 > A(4,23)=1 > A(5,9)=1 > A(6,18)=1 > A(7,27)=1 > A(8,17)=1 > A(9,1)=0,A(9,18)=1 > A(10,2)=1,A(10,3)=1 > A(11,4)=1,A(11,5)=1,A(11,6)=1 > A(12,7)=1,A(12,8)=1,A(12,9)=1 > A(13,10)=1,A(13,11)=1,A(13,12)=1 > A(14,13)=1,A(14,14)=1,A(14,15)=1 > A(15,19)=1,A(15,20)=1,A(15,21)=1 > A(16,22)=1,A(16,23)=1,A(16,24)=1 > A(17,1)=0,A(17,25)=1,A(17,26)=1,A(17,27)=1 > A(18,4)=1,A(18,7)=1 > A(19,2)=1,A(19,5)=1,A(19,8)=1 > A(20,10)=1,A(20,13)=1,A(20,16)=1 > A(21,11)=1,A(21,14)=1,A(21,17)=1 > A(22,19)=1,A(22,22)=1,A(22,25)=1 > A(23,1)=0,A(23,20)=1,A(23,23)=1,A(23,26)=1 > A(24,4)=1,A(24,7)=1 > A(25,2)=1,A(25,5)=1,A(25,8)=1 > A(26,10)=1,A(26,13)=1,A(26,16)=1 > A(27,20)=1,A(27,23)=1,A(27,26)=1 > > Best regards. > > > > -- > Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists- > Archives-f2602246.html > ___ > users mailing list > users@lists.scilab.org > http://lists.scilab.org/mailman/listinfo/users -- Tim Wescott www.wescottdesign.com Control & Communications systems, circuit & software design. Phone: 503.631.7815 Cell: 503.349.8432 ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Hi mottelet. Thank you for the question. I'm happy if all possible solutions are available. Since the rank is known to be 22, the number of solutions is 27Conb22 = 80730 or less, but if the number is small, I think I can find "basic" constraints that is economically meaningful. If it is impossible, if even one solution is found, then it may be possible to follow a chain from the economic relationship to the basic constraints. Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Please I want to ask a question. Please I have been trying to export a graphic from scilab to an MS word document. How do I do that? Please help. Thanks Sent from Yahoo Mail on Android On Sun, Dec 2, 2018 at 5:57 PM, fujimoto2005 wrote: This problem is an economic problem. The i-th row of the square constraint matrix A with m dimension expresses certain economic constraints. The elements of the constraint matrix are either 0 or 1. Suppose the rank of A is r and by changing the row number a_1, ..., a_r are linearly independent. I guess the coefficient cj (1 ≤ j ≤ r) in a_i = c1 * a_1 + ... cr * a_r for any m>=i> r is found to be 0, -1, 1 for some economic reasons. I would like to find such a_1, ..., a_r pairs. The following matrix is the constraint matrix which I am dealing with. A=zeros(27,27) A(1,10)=1 A(2,5)=1 A(3,14)=1 A(4,23)=1 A(5,9)=1 A(6,18)=1 A(7,27)=1 A(8,17)=1 A(9,1)=0,A(9,18)=1 A(10,2)=1,A(10,3)=1 A(11,4)=1,A(11,5)=1,A(11,6)=1 A(12,7)=1,A(12,8)=1,A(12,9)=1 A(13,10)=1,A(13,11)=1,A(13,12)=1 A(14,13)=1,A(14,14)=1,A(14,15)=1 A(15,19)=1,A(15,20)=1,A(15,21)=1 A(16,22)=1,A(16,23)=1,A(16,24)=1 A(17,1)=0,A(17,25)=1,A(17,26)=1,A(17,27)=1 A(18,4)=1,A(18,7)=1 A(19,2)=1,A(19,5)=1,A(19,8)=1 A(20,10)=1,A(20,13)=1,A(20,16)=1 A(21,11)=1,A(21,14)=1,A(21,17)=1 A(22,19)=1,A(22,22)=1,A(22,25)=1 A(23,1)=0,A(23,20)=1,A(23,23)=1,A(23,26)=1 A(24,4)=1,A(24,7)=1 A(25,2)=1,A(25,5)=1,A(25,8)=1 A(26,10)=1,A(26,13)=1,A(26,16)=1 A(27,20)=1,A(27,23)=1,A(27,26)=1 Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
This problem is an economic problem. The i-th row of the square constraint matrix A with m dimension expresses certain economic constraints. The elements of the constraint matrix are either 0 or 1. Suppose the rank of A is r and by changing the row number a_1, ..., a_r are linearly independent. I guess the coefficient cj (1 ≤ j ≤ r) in a_i = c1 * a_1 + ... cr * a_r for any m>=i> r is found to be 0, -1, 1 for some economic reasons. I would like to find such a_1, ..., a_r pairs. The following matrix is the constraint matrix which I am dealing with. A=zeros(27,27) A(1,10)=1 A(2,5)=1 A(3,14)=1 A(4,23)=1 A(5,9)=1 A(6,18)=1 A(7,27)=1 A(8,17)=1 A(9,1)=0,A(9,18)=1 A(10,2)=1,A(10,3)=1 A(11,4)=1,A(11,5)=1,A(11,6)=1 A(12,7)=1,A(12,8)=1,A(12,9)=1 A(13,10)=1,A(13,11)=1,A(13,12)=1 A(14,13)=1,A(14,14)=1,A(14,15)=1 A(15,19)=1,A(15,20)=1,A(15,21)=1 A(16,22)=1,A(16,23)=1,A(16,24)=1 A(17,1)=0,A(17,25)=1,A(17,26)=1,A(17,27)=1 A(18,4)=1,A(18,7)=1 A(19,2)=1,A(19,5)=1,A(19,8)=1 A(20,10)=1,A(20,13)=1,A(20,16)=1 A(21,11)=1,A(21,14)=1,A(21,17)=1 A(22,19)=1,A(22,22)=1,A(22,25)=1 A(23,1)=0,A(23,20)=1,A(23,23)=1,A(23,26)=1 A(24,4)=1,A(24,7)=1 A(25,2)=1,A(25,5)=1,A(25,8)=1 A(26,10)=1,A(26,13)=1,A(26,16)=1 A(27,20)=1,A(27,23)=1,A(27,26)=1 Best regards. -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] a linear equation
Hello Do you just want one of the solutions, e.g. of minimum norm, or do you want more precision on the nullspace ? S. > Le 2 déc. 2018 à 09:43, fujimoto2005 a écrit : > > I am trying to solve linear equations with 30 variables. Since the > determinant of the coefficient matrix is 0, I can tell that some row vectors > are linearly dependent on other row vectors. I want to solve the problem by > deleting the linearly dependent rows while simultaneously changing the > variables to exogenous variables. > But I can't find which row vector expresses redundant constraints. > Is there any way to find out which row is linearly dependent? > > Best regards. > > > > -- > Sent from: > https://antispam.utc.fr/proxy/1/c3RlcGhhbmUubW90dGVsZXRAdXRjLmZy/mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html > ___ > users mailing list > users@lists.scilab.org > https://antispam.utc.fr/proxy/1/c3RlcGhhbmUubW90dGVsZXRAdXRjLmZy/lists.scilab.org/mailman/listinfo/users ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
