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Math Help - Diagonal Matrix question

  1. #1
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    Question Diagonal Matrix question

    The question is that D1 and D2 are any two diagonal 3x3 matrices and

    P = AD1Ainv
    Q = AD2Ainv

    Show that PQ = QP.

    Now, I know that D1 and D2 commute, but im not sure how I can show this algebraically. Any ideas?

    BTW: A is a 3 x 3 matrix and is the inverse of itself. So, really, P = AD1A and Q = AD2A.

    Many thanks

    MB
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mister_bluesman View Post
    The question is that D1 and D2 are any two diagonal 3x3 matrices and

    P = AD1Ainv
    Q = AD2Ainv

    Show that PQ = QP.

    Now, I know that D1 and D2 commute, but im not sure how I can show this algebraically. Any ideas?

    BTW: A is a 3 x 3 matrix and is the inverse of itself. So, really, P = AD1A and Q = AD2A.

    Many thanks

    MB
    Well:
    PQ = (AD1A)(AD2A) = A(D1D2)A <-- Since AA = 1
    QP = (AD2A)(AD1A) = A(D2D1)A

    Now, D1 and D2 commute since they are both diagonal, so D2D1 = D1D2.

    Thus
    QP = A(D2D1)A = A(D1D2)A = PQ

    (Note: The condition that A^{-1} = A is not actually needed here.)

    -Dan
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  3. #3
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    Hi there. Thans for that. But could you explain how you got:

    PQ = (AD1A)(AD2A) = A(D1D2)A <-- Since AA = 1
    QP = (AD2A)(AD1A) = A(D2D1)A

    I'm not sure how you got PQ = A(D1D2)A and QP = A(D2D1)A. What did you do to do that? Rearrange, multiply out?

    Thanks
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mister_bluesman View Post
    Hi there. Thans for that. But could you explain how you got:

    PQ = (AD1A)(AD2A) = A(D1D2)A <-- Since AA = 1
    QP = (AD2A)(AD1A) = A(D2D1)A

    I'm not sure how you got PQ = A(D1D2)A and QP = A(D2D1)A. What did you do to do that? Rearrange, multiply out?

    Thanks
    Let me be more specific, and let me use A^{-1} since this is a bit more clear.

    PQ = (AD_1A^{-1})(AD_2A^{-1})

    PQ = AD_1(A^{-1}A)D_2A^{-1}

    PQ = AD_1 \cdot 1 \cdot D_2A^{-1} <-- A^{-1}A = 1

    PQ = A(D_1D_2)A^{-1}

    You can do QP the same way and get
    QP = A(D_2D_1)A^{-1}

    And the rest of the argument goes through as before.

    (In case it's an issue, matrix multiplication is associative, that is (AB)C = A(BC), so as long as we don't alter the order of the factors we may insert parenthesis wherever we wish.)

    -Dan
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  5. #5
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    Ah ok... but, please i dont mean to be stupid, but doesnt Ainv A = I, not 1? Surely multiplying two matrices results in another matrix and not a scalar?

    Many thanks (again!)

    Andy
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mister_bluesman View Post
    Ah ok... but, please i dont mean to be stupid, but doesnt Ainv A = I, not 1? Surely multiplying two matrices results in another matrix and not a scalar?

    Many thanks (again!)

    Andy
    You aren't being stupid. My own personal habit is to write a "1" for any (multiplicative) identity element. (In rare cases I use an "e.") I won't say this is standard for Physicists, but I've seen it in a number of Physics texts and adopted the habit. My apologies for the confusion!

    -Dan
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