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Find all 2x2 matrices X such that AX=XA for all 2x2 matrices

  1. Jan 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all 2x2 matrices X such that AX=XA for all 2x2 matrices


    3. The attempt at a solution
    Let A =
    a b
    c d

    and X =
    w x
    y z

    Then AX = XA ==>

    aw+by=wa+xc .........(1)
    ax+bz=wb+xd .........(2)
    cw+dy=ya+zc .........(3)
    cx+dz=yb+zd .........(4)

    (1) ==> by = xc, which holds for all b and c only if x=y=0.
    (2) ==> bz = wb, which hods for all b only if z=w.
    (3) ==> w=z
    (4) ==> x=y=0

    So the answer is x=y=0, and w=z. That is, X = k*I with k being any real number and I is the identity matrix.



    How does this look?
     
  2. jcsd
  3. Jan 28, 2015 #2

    Simon Bridge

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    I see the algebra for (1) ... not the others. Perhaps you should be more explicit.
    But your result works. Does k have to be a real number?
     
  4. Jan 28, 2015 #3
    Hello. Thank you for the reply. The algebra for (2)-(4) were based on the results of (1).

    For (2), we have ax+bz=wb+xd; in (1) we decided that x=y=0, thus this expression becomes bz=wb
    For (3), if x=y=0, then w=z
    For (4), we have cx+dz=yb+zd=> xc=yb, thus x=y=0.


    I am not too confident with what I did for part 1. "(1) ==> by = xc, which holds for all b and c only if x=y=0. "
    Alternatively, could I have also said "(1) ==> by = xc, which holds for all x and y only if b=c=0. "
     
  5. Jan 28, 2015 #4

    Simon Bridge

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    I think I see what you mean. It kinda looks like you are saying that x=y=0 as a strong assertion - you got there by reason more than algebra - and that feels weak right?

    In the second case, then you have just observed that for any A where b=c=0, the values of x and y in X don't matter.
    But your problem was to find the values of x and y (and the rest) that work when b and c are other numbers too.

    If you want a more rigorous approach, that depends less on direct reasoning, then try simplifying each line independently, building a matrix of coefficients, then using echelon reduction to arrive at the relationships.
     
  6. Jan 28, 2015 #5
    Ah I see what you are saying. I had forgotten the essence of the problem statement.

    Regarding the rigorous approach, it seems there are too many unknowns for that to come out nicely?
     
  7. Jan 31, 2015 #6

    Simon Bridge

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    No, it just means the solution is not unique...
     
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