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Linear Algebra and Complex Numbers

  1. Apr 6, 2007 #1
    1. Complex analysis is the study of number z= x+iy where i^2=-1. can you find a way to represent complex numbers as 2x2 matrices



    i honestly have no clue where to start with this one. we are one week through my linear algebra course.

    the only possible thing i can thing of is det (x -yi
    1 1) but that seems really wrong
     
    Last edited: Apr 6, 2007
  2. jcsd
  3. Apr 6, 2007 #2

    Hootenanny

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    Okay, how do we represent the number one in matrix form?
     
  4. Apr 6, 2007 #3

    radou

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    Google gave me some pretty fruitful results on this one.
     
  5. Apr 6, 2007 #4
    the number 1 in matrix form is just [1]
     
  6. Apr 6, 2007 #5

    HallsofIvy

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    No, we're talking about 2 by 2 matrices. 1 is the "multiplicative identity" for the real numbers. What is the multiplicative identity for 2 by 2 matrices?

    Now think about i. Where would you put the 1's so multiplying the matrix by itself will give you the negative of the identity matrix?
     
    Last edited: Apr 6, 2007
  7. Apr 6, 2007 #6
    diagonol 1's i dunno how to use LAtex but its like [1 0;0 1] on matlab prolly the transpose of that for the second question your asking
     
  8. Apr 6, 2007 #7

    Hootenanny

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    I meant in 2x2 form. I'll start for you, we can write the real component as;

    [tex]\Re = x\cdot\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right][/tex]

    In otherwords, the 2x2 identity matrix. Now, for the imaginary part we want a matrix which represent an anti-clockwise rotation by [itex]\pi/2[/itex] about the origin. Can you think of a matrix that does this?

    EDIT: Halls strikes again.
     
  9. Apr 6, 2007 #8
    [0 1
    1 0] ??

    i think that would do it
     
  10. Apr 6, 2007 #9

    Hootenanny

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    That's very close but not quite. Note that
    [tex]\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]\times\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right] = \left[ \begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right] = I_2[/tex]
    You want;
    [tex]\left[ \begin{array}{cc} a & b \\ c & d\end{array}\right]\times\left[ \begin{array}{cc} a & b \\ c & d\end{array}\right] = -\left[ \begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right][/tex]
     
  11. Apr 6, 2007 #10
    [0 -1
    1 0]


    ok got it i think
     
  12. Apr 6, 2007 #11

    Hootenanny

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    Looks good to me :approve: (Nice name by-the-way :wink:)
     
  13. Apr 6, 2007 #12
    so the answer would be
    x [1 0;0 1] + y[0 -1; 1 0] ???????


    thans jay and silent bob are fantanstci
     
  14. Apr 6, 2007 #13

    Hootenanny

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    Yep, your correct. However, you can represent it as a single matrix thus;

    [tex]\left[ \begin{array}{cc} x & -y \\ y & x\end{array}\right][/tex]

    Both forms are correct.
     
  15. Apr 6, 2007 #14
    You guys are absolute geniuses and i owe you my life. ty
     
  16. Apr 6, 2007 #15

    HallsofIvy

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    Okay, I'll send you my bill!
     
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