commutativity of polynomial matrix multiplication
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From: moba on 13 Aug 2010 09:23 Hi, I wonder whether there is a way of proving that the multiplication of two given polynomial matrices is commutative. Let' s say for the point of argument, we have two given polynomial matrices, whose coefficients are given by matrices p1(A) = -1*A^(-1) + I + 2*A + 3*A*A p2(A) = A^(-1) Obviously p1(A)*p2(A) = p2(A)*p1(A) Is there a prove to check for commutativity of polynomial matrix multiplication, especially if we have multivariate polynomials, e.g. p1(A,B)*p2(A) ? Best Regards
From: Chip Eastham on 13 Aug 2010 10:29 On Aug 13, 9:23 am, moba <wish.i.had.an.i...(a)gmail.com> wrote: > Hi, > > I wonder whether there is a way of proving that the > multiplication of two given polynomial matrices is > commutative. Let' s say for the point of argument, > we have two given polynomial matrices, whose coefficients > are given by matrices > > p1(A) = -1*A^(-1) + I + 2*A + 3*A*A > p2(A) = A^(-1) > > Obviously > p1(A)*p2(A) = p2(A)*p1(A) > > Is there a prove to check for commutativity of > polynomial matrix multiplication, especially if > we have multivariate polynomials, e.g. > p1(A,B)*p2(A) ? > > Best Regards Since your definition of p1(A) and p2(A) involves inverses of A, it is not customary to refer to these as polynomials in A. However they are (special cases of) rational functions in A. Two polynomials in matrix A (or two rational functions in A, where defined) will commute under matrix multiplication. You can prove the purely polynomial case by induction on the degrees of the multiplicands. If you introduce a polynomial in two matrices, as with p1(A,B), then it is no longer the case unless you add a condition such as A and B commute. If p1(A,B) = B and p2(A) = A, then of course the commutativity of those requires that A,B commute, so this condition is needed for the most general conclusion. hope this helps, chip
From: Chip Eastham on 13 Aug 2010 18:32
On Aug 13, 10:29 am, Chip Eastham <hardm...(a)gmail.com> wrote: > On Aug 13, 9:23 am, moba <wish.i.had.an.i...(a)gmail.com> wrote: > > > > > Hi, > > > I wonder whether there is a way of proving that the > > multiplication of two given polynomial matrices is > > commutative. Let' s say for the point of argument, > > we have two given polynomial matrices, whose coefficients > > are given by matrices > > > p1(A) = -1*A^(-1) + I + 2*A + 3*A*A > > p2(A) = A^(-1) > > > Obviously > > p1(A)*p2(A) = p2(A)*p1(A) > > > Is there a prove to check for commutativity of > > polynomial matrix multiplication, especially if > > we have multivariate polynomials, e.g. > > p1(A,B)*p2(A) ? > > > Best Regards > > Since your definition of p1(A) and p2(A) involves > inverses of A, it is not customary to refer to > these as polynomials in A. However they are > (special cases of) rational functions in A. > > Two polynomials in matrix A (or two rational > functions in A, where defined) will commute > under matrix multiplication. You can prove > the purely polynomial case by induction on > the degrees of the multiplicands. > > If you introduce a polynomial in two matrices, > as with p1(A,B), then it is no longer the case > unless you add a condition such as A and B > commute. If p1(A,B) = B and p2(A) = A, then > of course the commutativity of those requires > that A,B commute, so this condition is needed > for the most general conclusion. > > hope this helps, chip Let me add that once you've proven polynomials in matrix A commute, it's easy to extend this to rational functions of matrix A. For example suppose we want to show: p(A) * (q(A))^-1 = (q(A))^-1 * p(A) given p,q are polynomial s.t q(A) is invertible. Since we know q(A)*p(A) = p(A)*q(A), it follows: p(A) = (q(A))^-1 * p(A) * q(A) by multiply both sides on the left by (q(A))^-1, and then multiplying both sides on the right by the same inverse of q(A) gives the desired result. Combining this with the well-worn identity that the the inverse of a product is the product of the inverses in opposite order gives a recipe to put every rational function of matrix A into a "single fraction" p(A)/q(A). regards, chip |