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Math Help - A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

  1. #1
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    A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    Dear all, I have checked by computer that for any complex 2\times 2 matrix A,B,C, there we have A(BC-CB)^2-(BC-CB)^2A=0. However, I could not prove it mathematically [without tedious calculation].
    Last edited by xinglongdada; November 3rd 2012 at 07:09 PM.
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    Hey xinglongdada.

    Try expanding all the matrices out and use the fact that (A+B)(C+D) = AC + AD + BC + BD for matrices if they are well defined (dimension wise that is) and if you get zero no matter what, then you're done.
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  3. #3
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    it's not true.

    for a counter-example, let:

    A = \begin{bmatrix}1&1\\0&1 \end{bmatrix};\ B = \begin{bmatrix}1&0\\1&1 \end{bmatrix};\ C = \begin{bmatrix}1&1\\1&0 \end{bmatrix}.

    then:

    BC - CB = \begin{bmatrix}1&1\\2&1 \end{bmatrix} - \begin{bmatrix}2&1\\1&0 \end{bmatrix} = \begin{bmatrix}-1&0\\1&0 \end{bmatrix}.

    so:

    (BC - CB)^2 = \begin{bmatrix}1&0\\-1&0 \end{bmatrix}.

    now:

    A(BC - CB)^2 = \begin{bmatrix}1&1\\0&1 \end{bmatrix}\begin{bmatrix}1&0\\-1&0 \end{bmatrix} = \begin{bmatrix}0&0\\-1&0 \end{bmatrix},

    but:

    (BC - CB)^2A = \begin{bmatrix}1&0\\-1&0 \end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&1\\-1&-1 \end{bmatrix}.

    your "computer check" apparently didn't check some "simple matrices".
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    your calculation of BC-CB is wrong....
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    *sigh*, this happens to me, more often than i'd like to admit. yes, it should be:

    BC - CB = \begin{bmatrix}-1&0\\1&1 \end{bmatrix}

    this means that:

    A(BC - CB) = \begin{bmatrix}1&1\\0&1 \end{bmatrix} \begin{bmatrix}-1&0\\1&1 \end{bmatrix} = \begin{bmatrix}0&1\\1&1 \end{bmatrix}

    and:

    (BC - CB)A = \begin{bmatrix}-1&0\\1&1 \end{bmatrix} \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}-1&-1\\1&2 \end{bmatrix}

    but these 2 still aren't equal.

    *******

    in general, unless B and C commute, BC - CB will be non-zero, and unless BC - CB is of nilpotency degree 2, (BC - CB)2 will also be non-zero.

    in fact, unless (BC - CB)2 is a scalar multiple of the identity, there will always be SOME matrix A that doesn't commute with it.

    so your "theorem" would only be true if (BC - CB)2 = I or 0, which implies BC - CB is either 0, or invertible.

    so all we have to do is find matrices B and C so that BC - CB is singular. this isn't hard to do (and is probably what i SHOULD have done).
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    Quote Originally Posted by Deveno View Post
    *sigh*, this happens to me, more often than i'd like to admit. yes, it should be:

    BC - CB = \begin{bmatrix}-1&0\\1&1 \end{bmatrix}

    this means that:

    A(BC - CB) = \begin{bmatrix}1&1\\0&1 \end{bmatrix} \begin{bmatrix}-1&0\\1&1 \end{bmatrix} = \begin{bmatrix}0&1\\1&1 \end{bmatrix}

    and:

    (BC - CB)A = \begin{bmatrix}-1&0\\1&1 \end{bmatrix} \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}-1&-1\\1&2 \end{bmatrix}

    but these 2 still aren't equal.

    *******

    in general, unless B and C commute, BC - CB will be non-zero, and unless BC - CB is of nilpotency degree 2, (BC - CB)2 will also be non-zero.

    in fact, unless (BC - CB)2 is a scalar multiple of the identity, there will always be SOME matrix A that doesn't commute with it.

    so your "theorem" would only be true if (BC - CB)2 = I or 0, which implies BC - CB is either 0, or invertible.

    so all we have to do is find matrices B and C so that BC - CB is singular. this isn't hard to do (and is probably what i SHOULD have done).
    You didn't square the (BC - CB). I checked it on my TI. The sum is indeed zero.

    -Dan
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  7. #7
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    omg. i am...wrong. just....wrong. how embarrassing.

    it turns out the theorem IS true, and i can prove it.

    it suffices (from a previous post) to prove that (BC - CB)2 is a scalar multiple of the identity (such matrices lie in the center of the ring of 2x2 matrices, and thus will commute with any A).

    let:

    B = \begin{bmatrix}b_1&b_2\\b_3&b_4 \end{bmatrix};\ C =  \begin{bmatrix}c_1&c_2\\c_3&c_4 \end{bmatrix}.

    then:

    BC - CB =  \begin{bmatrix}b_1c_1+b_2c_3&b_1c_2+b_2c_4\\b_3c_1  +b_4c_3&b_3c_2+b_4c_4 \end{bmatrix} - \begin{bmatrix}b_1c_1+b_3c_2&b_2c_1+b_4c_2\\b_1c_3  +b_3c_4&b_2c_3+b_4c_4 \end{bmatrix}

    =\begin{bmatrix}b_2c_3-b_3c_2&b_1c_2-b_2c_1+b_2c_4-b_4c_2\\b_3c_1-b_1c_3+b_4c_3-b_3c_4&b_3c_2-b_2c_3 \end{bmatrix}

    that is, BC - CB is of the form:

    \begin{bmatrix}u&v\\w&-u \end{bmatrix}

    hence (BC - CB)2 is of the form:

    \begin{bmatrix}u&v\\w&-u \end{bmatrix}\begin{bmatrix}u&v\\w&-u \end{bmatrix} = \begin{bmatrix}u^2+vw&0\\0&u^2+vw \end{bmatrix} = (u^2+vw)I

    so sorry
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  8. #8
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    Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

    Thank you very much. This proof is nice.
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