http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Wed, 10 Jun 2026 20:17:56 GMT Wed, 10 Jun 2026 20:17:56 GMT The latest comments added to the Post, How Do I Find a Particular Theorem? http://www.mapleprimes.com/images/mapleprimeswhite.jpg http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment73962 Obviously, if your ring R has an identity, there is a square matrix with determinant A, namely a diagonal matrix with one A and all other diagonal entries 1. So I suppose your ring R doesn't have an identity. But then your result is not true, e.g. if R is the commutative ring over the integers generated by x (i.e. polynomials in x with integer coefficients and no constant term) take <maple>M = RTABLE(156754120,MATRIX([[x, 0], [0, x^2]]),Matrix)</maple>. Its determinant is <maple>x^3</maple> so we can take <maple>A = x</maple>, <maple>B = x^2</maple>. But the determinant of any 2 x 2 matrix over R is divisible by <maple>x^2</maple>. The latest comments added to the Post, How Do I Find a Particular Theorem? 73962 Wed, 23 Jan 2008 07:27:55 Z Robert Israel Robert Israel http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment84373 I did mean also to add: my ring R has an identity. Let char(R)=0, too. Feel free to do the char(R)=p>0 case for me, too :). Correct me if I did not set up my conjecture properly, but, your example: M factors as M = A*B where A =[[x,0],[0,1]] and B =[[1,0],[0,x^2]] det M = x^3 det A = x det B = x^2 so det M factors over R into two non-units, and M factors over GL2(R) into two matrices with these factors as determinants. So, at the moment, I still don't see a counterexample. Oh, also, the matrices must be size N by N where N > 1. The result is trivially true for N=1. I started a sketch of a proof in the N=2 case, but I realized it was wrong as I began to fill in the details. The latest comments added to the Post, How Do I Find a Particular Theorem? 84373 Wed, 23 Jan 2008 13:50:47 Z resolvent resolvent http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment73948 I hope I have this straight now. R is a commutative ring with identity, you have an n x n matrix M over R such that det(M) = rs where r and s are non-units in R, and you want to write M = AB where A and B are n x n matrices over R with det(A) = r and det(B) = s. I think I have a counterexample. Let R be the ring with identity generated by one indeterminate x with <maple>x^2 = 0</maple>, let <maple>M = Matrix([[x,0],[0,x]])</maple>, and take r=s=0. Let's try to find A and B using Maple. Start with two general 2x2 matrices over R. The coefficients e[i,j], f[i,j], g[i,j] and h[i,j] must be integers. <pre> > with(LinearAlgebra): M:= DiagonalMatrix([x,x]); A:= Matrix(2,2,(i,j)->e[i,j]+f[i,j]*x); B:= Matrix(2,2,(i,j) -> g[i,j]+h[i,j]*x); </pre> To satisfy our requirements, the coefficients of x^0 and x^1 in the entries of A.B-M and in det(A) and det(B) must be 0. <pre> > P:= A.B-M; L:= map(t -> (coeff(t,x,0),coeff(t,x,1)), [seq(seq(P[i,j],i=1..2),j=1..2), Determinant(A), Determinant(B)]); </pre> Use Groebner[Solve] to see if there is a solution to these equations. <pre> > with(Groebner): Solve(L,indets(L)); </pre> <maple>{}</maple> So, if Maple is to be believed, there is no solution. There's probably a nicer proof than that. The latest comments added to the Post, How Do I Find a Particular Theorem? 73948 Wed, 23 Jan 2008 23:15:09 Z Robert Israel Robert Israel http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment73946 do you allow the matrices to be of arbitrary dimension? The latest comments added to the Post, How Do I Find a Particular Theorem? 73946 Thu, 24 Jan 2008 01:56:05 Z Axel Vogt Axel Vogt http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment73940 Note that the original question did NOT have M=A*B, but merely that det(M)=A*B and exists matrices U,V such that det(U)=A and det(V)=B, and no other relations. In other words, U.V does not have to satisfy any relation at all - unlike Robert's last post. If you allow U and V to be of any dimension, then this is easy - take them both to be of dimension 1. The more interesting case is when dim(M) = n = degree(det(M)). Then we can find matrices U and V with dimension that of the factors A and B - basically these are just the companion matrices of A and B! The latest comments added to the Post, How Do I Find a Particular Theorem? 73940 Thu, 24 Jan 2008 08:13:56 Z JacquesC JacquesC http://www.mapleprimes.com/posts/40202-How-Do-I-Find-A-Particular-Theorem?ref=Feed:MaplePrimes:How Do I Find a Particular Theorem?:Comments#comment84367 Here's a proof of my counterexample without using Maple. Suppose A and B are 2 x 2 matrices over R so that det(A) = det(B) = 0 and A B = M = x I. Let Adj be the classical adjoint operator, so <maple>Adj(Matrix([[a,b],[c,d]]))= Matrix([[d,-b],[-c,a]])</maple>. Note that Adj(A) A = det(A) I = 0 so Adj(A) x = Adj(A) A B = 0. This means that all matrix elements of Adj(A), and therefore of A, are multiples of x. Similarly from x Adj(B) = A B Adj(B) = 0 we see that all matrix elements of B are multiples of x. But that implies that A B = 0, contradiction. This proof relies heavily on the fact that the matrices are 2 x 2, so that Adj is a linear operator. I don't see how to get a counterexample in n x n matrices with n > 2. The most simpleminded generalization doesn't work: for n=4 the matrices <maple>A = Matrix([[x,0,0,0],[0,x,0,0],[0,0,1,0],[0,0,0,1]])</maple> and <maple>B = Matrix([[1,0,0,0],[0,1,0,0],[0,0,x,0],[0,0,0,x]])</maple> over R satisfy A B = x I with det(A) = det(B) = 0. The latest comments added to the Post, How Do I Find a Particular Theorem? 84367 Fri, 25 Jan 2008 00:46:35 Z Robert Israel Robert Israel