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Math Help - 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 c^2=I. Let CD_1C and B=CD_2C where D_1 and 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 c^2=I. Let CD_1C and B=CD_2C where D_1 and D_2 are diagonal Matrices. Show that A and B are communitative
    I suppose you meant A=CD_1C . Use C^2=I and D_1D_2=D_2D_1

    AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots

    BA=\ldots


    Fernando Revilla
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  3. #3
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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 A=CD_1C . Use C^2=I and D_1D_2=D_2D_1

    AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots

    BA=\ldots


    Fernando Revilla
    Thanks for your reply FernandoRevilla.

    Ok, is this correct?

    AB = CD_1CCD_2C

    = CD_1C^2D_2C

    = D_1C^4D_2

    BA = CD_2CCD_1C

    = CD_2C^2D_1C

    = D_2C^4D_1

    Therefore, AB = BA

    If this is correct, then what was the purpose of C^2 = I? To confuse me? What about the 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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  5. #5
    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?

    AB = CD_1CCD_2C

    = CD_1C^2D_2C

    = D_1C^4D_2

    BA = CD_2CCD_1C

    = CD_2C^2D_1C

    = D_2C^4D_1

    Therefore, AB = BA

    If this is correct, then what was the purpose of C^2 = I? To confuse me? What about the 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 Ccommutes with D_1 and D_2, which you haven't proved (and which, without thinking too much about it, I doubt is true).

    The facts that C^2=I and the D's are diagonal both play key roles in the proof. Can you show that D_1D_2=D_2D_1 as topsquark suggests?
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  7. #7
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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:

    \[<br />
 \begin{matrix}<br />
  1 & 0 & 0 \\<br />
  0 & 1 & 0 \\<br />
  0 & 0 & 1<br />
 \end{matrix}<br />
\]
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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:

    \[<br />
 \begin{matrix}<br />
  1 & 0 & 0 \\<br />
  0 & 1 & 0 \\<br />
  0 & 0 & 1<br />
 \end{matrix}<br />
\]
    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:

    \[<br />
 \begin{pmatrix}<br />
  1 & 0 \\<br />
  0 & 2   <br />
 \end{pmatrix}<br />
\] \[<br />
 \begin{pmatrix}<br />
  3 & 0 \\<br />
  0 & 4   <br />
 \end{pmatrix}<br />
\]

    = \[<br />
 \begin{pmatrix}<br />
  3 & 0 \\<br />
  0 & 8   <br />
 \end{pmatrix}<br />
\]

    \[<br />
 \begin{pmatrix}<br />
  3 & 0 \\<br />
  0 & 4   <br />
 \end{pmatrix}<br />
\] \[<br />
 \begin{pmatrix}<br />
  1 & 0 \\<br />
  0 & 2   <br />
 \end{pmatrix}<br />
\]

    = \[<br />
 \begin{pmatrix}<br />
  3 & 0 \\<br />
  0 & 8   <br />
 \end{pmatrix}<br />
\]
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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 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 D_1 and 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 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 D_1 and 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.
    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,

    BA=CD_2CCD_1C

    BA=CD_2C^2D_1C

    BA=CD_2ID_1C

    BA=CD_2D_1C

    which is the same thing as 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?

    AB = CD_1CCD_2C

    = CD_1C^2D_2C

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

    BA = CD_2CCD_1C

    = CD_2C^2D_1C

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

    Therefore, AB = BA

    If this is correct, then what was the purpose of C^2 = I? To confuse me? What about the 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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