Re: math -- f(A)*g(B) = I => AB = BA?

On 12 avr, 09:16, quasi <qu...@xxxxxxxx> wrote:
Question:

If A,B are n x n matrices with complex coefficients such that

   f(A)*g(B) = I

for some nonconstant polynomials f,g in C[x], must A,B commute?

Remarks:

I suspect the answer is no.

If the answer is no, would it still be no if f = g?

quasi

Bonjour,

an other interesting case :
A = P1(C) , B = p2(C) , p1,)p2 polynomials , C a matrix .
then A*B =B*A = P1(C)*p2(C) .

In the case f(A)*g(B) = I ,f and g invertible functions
we may work with B = g^-1(I/f(A)) , put I/f(A) = C ,C regular
and consider the case B = g^-1(C) , A = f^-1(1/C) ,

Alain
.

Relevant Pages

  • Re: Linear algebra with pitfall..
    ... E is identity matrix. ... B are n x n matrices with complex coefficients such that ... must A commute with B? ... for some nonconstant polynomials f,g in C, must A and B commute? ...
    (sci.math)
  • Re: Linear algebra with pitfall..
    ... B are n x n matrices with complex coefficients such that ... must A commute with B? ... for some nonconstant polynomials f,g in C, must A and B commute? ... As a minimum, the functions f,g besides being non-constant, must such ...
    (sci.math)
  • math -- f(A)*g(B) = I => AB = BA?
    ... If A,B are n x n matrices with complex coefficients such that ... for some nonconstant polynomials f,g in C, must A,B commute? ... Remarks: ...
    (sci.math)
  • Re: math -- f(A)*g(B) = I => AB = BA?
    ... In article, quasi ... If A,B are n x n matrices with complex coefficients such that ... for some nonconstant polynomials f,g in C, must A,B commute? ... I suspect the answer is no. ...
    (sci.math)
  • Re: Linear algebra with pitfall..
    ... E is identity matrix. ... B are n x n matrices with complex coefficients such that ... must A commute with B? ...
    (sci.math)