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A manipulative matrix question

  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data

    If A and B are 2 matrices such that AB = A and BA =B, then B2 is equal to
    B
    A
    Zero matrix
    I
    2. Relevant equations
    We can pre or post multiply a matrix on both sides of equation.

    3. The attempt at a solution
    (AB).(BA) = A.B
    AB2A = A.B
    Pre multiply both sides by A-1
    We get B2A = AB
    Implies B2AB = AB as (A = AB)
    Post multiplying by B-1 and A we get B2 = I

    Is this correct?
     
  2. jcsd
  3. May 12, 2015 #2

    SammyS

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    Shouldn't that last equation be as follows?

    B2A = B

    By the way: Are you certain that matrix, A, has an inverse?
    Are you certain that matrix, A, has an inverse?
     
  4. May 12, 2015 #3

    Mark44

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    I don't see anything in the problem that says that either matrix is invertible, so there's no justification for assuming that A-1 exists.

    I found it helpful to start with B2 and work with that.
     
  5. May 12, 2015 #4
    Maybe it was a trick question - nothing involving inverse matrices. Both A and B could be identity matrices. This would satisfy the problem statement in the strictest sense.
     
  6. May 12, 2015 #5

    SammyS

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    I started with A2 .
     
  7. May 12, 2015 #6

    Mark44

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    That seems the long way around if the goal it to find B2, but then I don't know what you did.
     
  8. May 12, 2015 #7

    SammyS

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    Well, I suppose we'll have to see what OP does before we can all compare notes.
     
  9. May 12, 2015 #8
    If you start with ##B^2##,
    Then i got A=I.
    ##B^2=B.B=B.BA=B^2A##
    ##B^2=B^2A##
    A=I satisfies the above equation.
    And if i put it on AB=A, then i got B=I.
     
  10. May 12, 2015 #9

    SammyS

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    While I satisfied that equation, it's not necessary that A = I .
     
  11. May 13, 2015 #10

    Mark44

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    As Sammy said, B isn't necessarily I. Here's a counterexample, using your assumption that A = I.
    $$B = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$
    This matrix satisfies B2 = B2A = B2I, yet ##B \neq I##.
     
  12. May 13, 2015 #11
    So,
    I started with B2 ,

    ## B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B ## Is multiplication of matrices associative which I am doing here?

    Starting with A2.
    Following the above similar pattern , I get A2= A.
    How did you Sammy get B2 value by starting from A2?
     
  13. May 13, 2015 #12

    Dick

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    Yes, multiplication of matrices is certainly associative.
     
  14. May 13, 2015 #13
    How?
     
  15. May 13, 2015 #14

    Dick

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  16. May 13, 2015 #15
  17. May 13, 2015 #16

    Dick

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    It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).
     
  18. May 13, 2015 #17
    Thanks, got it.
     
  19. May 13, 2015 #18

    SammyS

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    Of course, I didn't follow that pattern, but, yes, A2= A.

    I start with:
    A2 = A2

    BAAB = BAAB

    ((BA)A)B = (BA)(AB)

    (BA)B = BA

    BB = B


    Added in Edit:
    Here are two such matrices.
    ##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]##

    ##\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##
     
    Last edited: May 13, 2015
  20. May 13, 2015 #19
    How you have written A2 = BAAB ?
     
  21. May 13, 2015 #20

    SammyS

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    Here are two such matrices.
    ##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]##

    ##\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##
    I didn't say that.

    I said

    BAAB = BAAB


    Now that I look at that, I only needed:

    BAB = BAB

    (BA)B = B(AB)

    BB = BA

    B2 = B
     
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