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Thread: Show that A and B are Communitative

  1. #1
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    Show that A and B are Communitative

    I need some guidance on how to start answering this question please:

    Question: Suppose C is a square matrix such that $\displaystyle c^2=I$. Let $\displaystyle CD_1C and B=CD_2C$ where $\displaystyle D_1$ and $\displaystyle D_2$ are diagonal Matrices. Show that A and B are communitative
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by sparky View Post
    Question: Suppose C is a square matrix such that $\displaystyle c^2=I$. Let $\displaystyle CD_1C and B=CD_2C$ where $\displaystyle D_1$ and $\displaystyle D_2$ are diagonal Matrices. Show that A and B are communitative
    I suppose you meant $\displaystyle A=CD_1C$ . Use $\displaystyle C^2=I$ and $\displaystyle D_1D_2=D_2D_1$

    $\displaystyle AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots$

    $\displaystyle BA=\ldots$


    Fernando Revilla
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    Yes you are right, Let A = CD1C, thanks
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    Quote Originally Posted by FernandoRevilla View Post
    I suppose you meant $\displaystyle A=CD_1C$ . Use $\displaystyle C^2=I$ and $\displaystyle D_1D_2=D_2D_1$

    $\displaystyle AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots$

    $\displaystyle BA=\ldots$


    Fernando Revilla
    Thanks for your reply FernandoRevilla.

    Ok, is this correct?

    $\displaystyle AB = CD_1CCD_2C$

    $\displaystyle = CD_1C^2D_2C$

    $\displaystyle = D_1C^4D_2$

    $\displaystyle BA = CD_2CCD_1C$

    $\displaystyle = CD_2C^2D_1C$

    $\displaystyle = D_2C^4D_1$

    $\displaystyle Therefore, AB = BA $

    If this is correct, then what was the purpose of $\displaystyle C^2 = I$? To confuse me? What about the $\displaystyle D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Thanks for your reply FernandoRevilla.

    Ok, is this correct?

    $\displaystyle AB = CD_1CCD_2C$

    $\displaystyle = CD_1C^2D_2C$

    $\displaystyle = D_1C^4D_2$

    $\displaystyle BA = CD_2CCD_1C$

    $\displaystyle = CD_2C^2D_1C$

    $\displaystyle = D_2C^4D_1$

    $\displaystyle Therefore, AB = BA $

    If this is correct, then what was the purpose of $\displaystyle C^2 = I$? To confuse me? What about the $\displaystyle D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?
    You aren't quite done yet. You know that C^2 = I, so therefore C^4 = I. Then we have that AB = D1D2 and BA = D2D1. What do you know about the matrices D1 and D2?

    -Dan
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    I'm not sure how you went from line 2 to line 3, sparky. It looks like you're assuming that $\displaystyle C$commutes with $\displaystyle D_1$ and $\displaystyle D_2$, which you haven't proved (and which, without thinking too much about it, I doubt is true).

    The facts that $\displaystyle C^2=I$ and the $\displaystyle D$'s are diagonal both play key roles in the proof. Can you show that $\displaystyle D_1D_2=D_2D_1$ as topsquark suggests?
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    Thanks for the reply Topsquark

    What I know about D1 and D2 is that they are diagonal matrices, which means that they are square matrices with zeros outside of the diagonal entries, like this:

    $\displaystyle \[
    \begin{matrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
    \end{matrix}
    \]$
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Thanks for the reply Topsquark

    What I know about D1 and D2 is that they are diagonal matrices, which means that they are square matrices with zeros outside of the diagonal entries, like this:

    $\displaystyle \[
    \begin{matrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
    \end{matrix}
    \]$
    Right. Do two such matrices commute?

    -Dan
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  9. #9
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    Quote Originally Posted by topsquark View Post
    Right. Do two such matrices commute?

    -Dan
    Ok, I was not sure, so to answer your question, I worked out a simple example (see below) and yes they do commute:

    $\displaystyle \[
    \begin{pmatrix}
    1 & 0 \\
    0 & 2
    \end{pmatrix}
    \]$$\displaystyle \[
    \begin{pmatrix}
    3 & 0 \\
    0 & 4
    \end{pmatrix}
    \]$

    =$\displaystyle \[
    \begin{pmatrix}
    3 & 0 \\
    0 & 8
    \end{pmatrix}
    \]$

    $\displaystyle \[
    \begin{pmatrix}
    3 & 0 \\
    0 & 4
    \end{pmatrix}
    \]$$\displaystyle \[
    \begin{pmatrix}
    1 & 0 \\
    0 & 2
    \end{pmatrix}
    \]$

    =$\displaystyle \[
    \begin{pmatrix}
    3 & 0 \\
    0 & 8
    \end{pmatrix}
    \]$
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  10. #10
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    Thanks for your reply LoblawsLawBlog.

    I think I now understand. Correct me if I'm wrong:

    Since $\displaystyle C^2 = I$ and I is an identity matrix, therefore I is a diagonal matrix. C is also equal to I, therefore C is also diagonal. And since $\displaystyle D_1$ and $\displaystyle D_2$ are also diagonal, therefore since all of them are diagonal, AB = BA
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Thanks for your reply LoblawsLawBlog.

    I think I now understand. Correct me if I'm wrong:

    Since $\displaystyle C^2 = I$ and I is an identity matrix, therefore I is a diagonal matrix. C is also equal to I, therefore C is also diagonal. And since $\displaystyle D_1$ and $\displaystyle D_2$ are also diagonal, therefore since all of them are diagonal, AB = BA
    Hmmmm... If C = -I would this still work? You don't need this anyway.
    $\displaystyle AB = C D_1 C C D_2 C = C D_1 C^2 D_2 C = C D_1 I D_2 C = C D_1 D_2 C$

    Do this for BA. What do you get?

    -Dan
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  12. #12
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    Ok,

    $\displaystyle BA=CD_2CCD_1C$

    $\displaystyle BA=CD_2C^2D_1C$

    $\displaystyle BA=CD_2ID_1C$

    $\displaystyle BA=CD_2D_1C$

    which is the same thing as $\displaystyle AB=CD_1D_2C$

    Thanks!
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  13. #13
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    Quote Originally Posted by sparky View Post
    Thanks for your reply FernandoRevilla.

    Ok, is this correct?

    $\displaystyle AB = CD_1CCD_2C$

    $\displaystyle = CD_1C^2D_2C$

    $\displaystyle = D_1C^4D_2$
    Where did you get this? How did the "C"s from the sides move inside? Better to use $\displaystyle C^2= I$ here and write
    $\displaystyle = CD_1C_2C$.

    $\displaystyle BA = CD_2CCD_1C$

    $\displaystyle = CD_2C^2D_1C$

    $\displaystyle = D_2C^4D_1$
    same thing: $\displaystyle = CD_2D_1C$ and, since $\displaystyle D_2$ and $\displaystyle D_1$ are diagonal matrices they commute: this is the same as $\displaystyle CD_1D_2C$ that you had before.

    $\displaystyle Therefore, AB = BA $

    If this is correct, then what was the purpose of $\displaystyle C^2 = I$? To confuse me? What about the $\displaystyle D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?
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  14. #14
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    Thanks a lot HallsofIvy
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