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How to make a matrix B such that AB=BA

  1. Feb 14, 2007 #1
    B is not the identity matrix or the zero matrix
    suppose both A and B are 3X3
    I dont have clue on this..any hits?
     
  2. jcsd
  3. Feb 14, 2007 #2
    Well write out two 3x3 matrices of variables and calculate AB, and BA set them equal to one another(element by element) and find the conditions on the entries of each matrix that make it true that AB=BA.
     
  4. Feb 14, 2007 #3

    mathwonk

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    how abouit 2Id?
     
  5. Feb 14, 2007 #4
    what do you mean by 2Id? 2xI?
     
  6. Feb 15, 2007 #5

    matt grime

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    There is an obvious choice for B (well infinitely many) that is not Id or 0. What will A always commute with?
     
  7. Feb 15, 2007 #6

    HallsofIvy

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    Is the question "Given B find A such that AB= BA"?
     
  8. Feb 15, 2007 #7
    it is given A find B such that AB=BA..
    what is the obvious choice...
     
  9. Feb 16, 2007 #8

    morphism

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    A will commute with itself and all its powers. Also, A will commute with all multiples of Id.
     
  10. Feb 16, 2007 #9

    dextercioby

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    B=A, or B=I.
     
  11. Feb 17, 2007 #10

    robphy

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    special polynomials in kA, where k is a constant
     
  12. Feb 17, 2007 #11

    matt grime

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    Uh? *any* polynomial in A commutes with A (dunno what the k has to do with anything).
     
  13. Feb 17, 2007 #12

    robphy

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    Hmm... I must have been thinking about something else in particular [which I can't recall now]. Yeah, the k is redundant if I say polynomial.
     
  14. Feb 17, 2007 #13
    Any B sharing all of A's eigenvectors would work too wouldn't it? Or is that pretty much amounting to a polynomial of A anyway?

    Or is there some subtle reason I'm missing why that wouldn't work?
     
  15. Feb 17, 2007 #14

    matt grime

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    It certainly doesn't work. Not unless there is a basis of eigenvectors. It is easy to find matrices A and B with precisely one eigenvector each that contradict your hypothesis.
     
  16. Feb 19, 2007 #15
    B=A^-1

    For any other matrix to work the matrix A must be in a special format. For example, a 2x2 matrix must have the upper right and lower left values equal. It might need to meet another requirement, I'm not too sure.
     
  17. Feb 19, 2007 #16

    matt grime

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    Where does it say A is invertible? I don't understand your 'upper right and lower left' condition at all. Any polynomial in A will do, as has been said many times in this thread. (I think this is about the 3rd time I've said it.)
     
  18. Feb 21, 2007 #17
    So I'm assuming you want a matrix B such that a known matrix A when multiplied by B can commute, that is AB=BA. The idea is to set up a system of linear equations with all of your unknowns. First we have to specify the unknowns. If B is a 3X3 matrix then we will have a matrix containing a,b,c,d,e,f,g,h,i where these letters are the unknowns representitive of the coefficients in the B matrix. Next you want to multiply A times B, and B times A, which should give you 18 different equations. We want to treat a,b,c, etc. as if they were x1, x2, x3, etc. Remember AB=BA, which means AB - BA = 0. Now you can set up and solve for a linear system using elementary row operations. Once the linear system is in reduced row echelon form, you will see the conditions for AB=BA. I hope that helps.
     
  19. Feb 21, 2007 #18

    matt grime

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    Does nobody read the f**king posts before them and try to figure out if they've answered the question or not?
     
  20. Feb 22, 2007 #19
    I read all of the posts and nobody appeared to answer the posters question to satisfaction. Such anger, I would have expected a better show from someone with such credentials as "homework helper" and "science advisor."
     
  21. Feb 22, 2007 #20

    matt grime

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    Several suggestions have been made that lead to infinitely many B, none of which require the setting up or solution of anything as tedious as simultaneous equations.
     
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