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Commutative 2x2 Matrices

  1. Sep 5, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data

    Let A =
    \begin{bmatrix}
    0 & 1 \\
    1 & 0
    \end{bmatrix}

    Find all 2 x 2 matrices B such that AB = BA.

    2. Relevant equations
    http://euclid.colorado.edu/~roymd/m3130/Exam2sol.pdf

    3. The attempt at a solution

    I let B =
    \begin{bmatrix}
    a & b \\
    c & d
    \end{bmatrix} and set AB=BA.

    From here I see that a and d must be 0, and b=c must be true.

    So the answer will be that all matrices that are commutative will be of form:

    \begin{bmatrix}
    0 & b \\
    b & 0
    \end{bmatrix}

    And there is no other possible commutative matrix outside of this form.

    1. Is this correct?
    2. Is there any further proof of this needed?

    Thank you kindly.
     
  2. jcsd
  3. Sep 5, 2015 #2

    pasmith

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    Homework Helper

    That must be wrong, because the 2x2 identity matrix commutes with every 2x2 matrix but is not of that form.

    What did you actually get for AB and BA? I would suggest double-checking those calculations.

    You have the right idea, but have not executed it correctly.
     
  4. Sep 5, 2015 #3

    RJLiberator

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    Gold Member

    So to get this right:

    if A = \begin{bmatrix}
    0 & 1 \\
    1 & 0
    \end{bmatrix}

    All possible commutative matrices with matrix A should be in the form:

    \begin{bmatrix}
    0 & b \\
    b & 0
    \end{bmatrix}


    This is the wrong answer? It seems to be right when I calculate it. I get the same answer either way. AB = BA
     
  5. Sep 5, 2015 #4

    pasmith

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    Homework Helper

    [itex]\begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix}[/itex] is a subset of the matrices you are looking for. It can't be all of them, because the identity [itex]\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/itex] commutes with every 2x2 matrix.

    Recheck your initial calculations with [itex]B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}[/itex].
     
  6. Sep 5, 2015 #5

    RJLiberator

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    Gold Member

    Ahhh, I see.

    I calculated it out and found that while b=c, also a=d:

    \begin{bmatrix}
    a & b \\
    b & a
    \end{bmatrix}

    This makes sense to me as the original answer was a subset of this.

    Is there any further proof needed to show that is all?
     
  7. Sep 5, 2015 #6

    SammyS

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    Staff Emeritus
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    To follow up on what pasmith is asking you to do:

    What do you get for the product ##\displaystyle \ \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \ \ \ ?##

    What do you get for the product ##\displaystyle \ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \ \ \ ? ##


    (I see you posted you answer just before I posted this.)

    That looks good.
     
  8. Sep 5, 2015 #7

    RJLiberator

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    Gold Member

    Thanks guys for the help here. Greatly appreciated.
     
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