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From: SourceForge.net <noreply@so...>  20100413 22:49:37

Bugs item #2983881, was opened at 20100408 15:25 Message generated for change (Comment added) made by stefano_ferri You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2983881&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Stefano Ferri (stefano_ferri) Assigned to: Nobody/Anonymous (nobody) Summary: Triangularize does not preserve rank Initial Comment: I already reported this issue on the mailing list. Here there is an example where triangularize fails with a symbolic matrix. In a nutshell, doing triangularize over a symbolic matrix and then computing rank when the triangularized matrix is evaluated giving a value to a parameter, yields differents results than doing evaluation followed by triangularization. (%i1) display2d:false; (%o1) false (%i2) M:matrix([h1,h,h+2,h+1,1],[1h,2,h2,1,1],[h1,h,1,0,1],[0,h2,0,2,1]); (%o2) matrix([h1,h,h+2,h+1,1],[1h,2,h2,1,1],[h1,h,1,0,1],[0,h2,0,2,1]) (%i3) ev(M,h=1); (%o3) matrix([2,1,1,0,1],[2,2,1,1,1],[2,1,1,0,1],[0,1,0,2,1]) (%i4) rank(%); (%o4) 3 (%i5) triangularize(%o3); (%o5) matrix([2,1,1,0,1],[0,2,0,2,4],[0,0,0,2,6],[0,0,0,0,0]) (%i6) rank(%); (%o6) 3 that is correct, but: (%i7) Mt:triangularize(M); (%o7) matrix([h1,h,1,0,1],[0,h^2h+2,0,22*h,h1], [0,0,h^32*h^2+h+2,h^32*h^2+h+2,0], [0,0,0,h^42*h^3+h^2+2*h,3*h^36*h^2+3*h+6]) (%i8) ev(Mt,h=1); (%o8) matrix([2,1,1,0,1],[0,2,0,4,2],[0,0,0,0,0],[0,0,0,0,0]) (%i9) rank(%); (%o9) 2 %o9 is a wrong result, if h=1, rank of matrix M is 3  >Comment By: Stefano Ferri (stefano_ferri) Date: 20100414 00:49 Message: Maybe I've found something. Triangularize dos not check if the matrix is already reduced, insisting to multiply for the pivot. Here is an example. (%i3) m:matrix([a,b],[c,d]); (%o3) matrix([a,b],[c,d]) (%i4) triangularize(m); (%o4) matrix([a,b],[0,a*db*c]) (%i5) triangularize(%); (%o5) matrix([a,b],[0,a^2*da*b*c]) (%i6) triangularize(%); (%o6) matrix([a,b],[0,a^3*da^2*b*c]) the matrix %o4 is already triangularized, but triangularize doesn't verify that. It continues to multiply for the pivot a in the first row. Maybe the problem is here. I don't understand why triangularize continues with such multiplications. In fact, in the above example, let a be the pivot: we want all the elements under it (only c here) vanish. To do that, we can sum to the 2nd row the first multiplied by c/a, that's what we see in %o4, that is correct. But if triangularization is repeated, triangularize multiplies for the pivot. Altough this behaviour should not change rank (it is a multiplication by a constant), maybe there are some issues related to it giving rise to errors? Surely, they give rise to high order polynomials (see ptriangularize for example, it checks if the matrix is already reduced and produces nicer expressions).  Comment By: Stefano Ferri (stefano_ferri) Date: 20100412 16:43 Message: Determinant and rank are different things: to reduce a matrix, one can multiply a row for a constant, or exchange the position of two rows: so, in general the determinant is changed, but this is normal. I don't think there is a documentation problem here, since this is a mathematical question. Maybe it could be mentioned in a note, but I think it is not necessary... But the rank must be preserved. This too is mathematics. What I'm seeing as a problem here is that triangularize has some problems handling symbolic matrices. As a consequence, evaluation after reduction yields differents results than reducing after evaluation. And with different results I don't mean a different appearence, but a different rank. Why in your example are you saying R2 < c*R1d*R2 is not rankpreserving?  Comment By: Barton Willis (willisbl) Date: 20100409 20:05 Message: Think about this: (%i7) triangularize(matrix([a,b],[c,d])); (%o7) matrix([a,b],[0,a*db*c]) There is no manifestly nonzero pivot, so triangularize uses a rank non preserving row operation R2 < c*R1d*R2. Even for matrices with rational entries, triangularize doesn't preserve the determinant: (%i13) determinant(triangularize(matrix([2,1],[6,7]))); (%o13) 16 (%i14) determinant(matrix([2,1],[6,7])); (%o14) 8 This too is a documentation problem, not a algorithm bug.  Comment By: Stefano Ferri (stefano_ferri) Date: 20100409 11:16 Message: I think it's not a documentationrelated bug or a lack of feature, but a real bug. What Maxima calls triangularize() is a matrix reduction by row. It doesn't matter how you are doing that, rank has to be preserved. There is a a very simple theorem that states that the rank of a reduced matrix is the same of the original matrix. As a consequence, the rank of a triangular matrix is equal the number of non null rows. Therefore, triangularize is doing something wrong in this case...  Comment By: Barton Willis (willisbl) Date: 20100409 02:47 Message: What do you expect for triangularize(matrix([a,b],[c,d])); A rank preserving triangularize would need to return a conditional statement. The user documentation doesn't say that triangularize preserves the rank. But it does say that triangularize does Gauss elimination, so it's reasonable to think that it is rankpreserving. This is more of a documentation bug than a algorithm bug, I think. You might like to try ptriangularize: (%i5) ptriangularize(matrix([8z, 7z],[1z, 3z]),z); (%o5) matrix([7,4],[0,17/7(3*z)/7]) (%i6) triangularize(matrix([8z, 7z],[1z, 3z]),z); (%o6) matrix([8z,7z],[0,173*z])  Comment By: Stefano Ferri (stefano_ferri) Date: 20100408 15:28 Message: Maxima version is 5.20.1. This problem is present both on Windows version, compiled with GCL, and Linux version, compiled with CLISP (Slackware package).  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2983881&group_id=4933 