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Isomorphism Proof

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    The problem is as follows:



    Let f : R^2 map to R^2 be rotation through an angle of theta radians about the origin.

    Prove that f is an isomorphism.


    2. Relevant equations

    Let f :[itex] R^2 \rightarrow R^2 [/itex]

    3. The attempt at a solution

    I know that the rotation can be expressed as the 2 x 2 matrix


    cos(theta) -sin(theta)

    Sin(theta) cos(theta)


    And its inverse I believe is


    cos(theta) -sin(theta)

    -sin(theta) cos(theta)



    Do I first show that f :[itex] R^2 \rightarrow R^2 [/itex] is a linear transformation
    by closure of addition and scaler multiplication by using x,y elements of R2 and some scaler k


    Say let the 2x2 matrix:

    cos(theta) -sin(theta)

    Sin(theta) cos(theta) = A


    We need to show


    A(x+y) = A(x) + A(y)


    cos(theta)x -sin(theta)y = cos(theta)x + -sin(theta)y = cos(theta)x -sin(theta)y

    sin(theta)x cos(theta)y sin(theta)x + cos(theta)y sin(theta)x cos(theta)y


    likewise

    A(kx+ky) = K A(x+y)

    cos(theta)kx + -sin(theta)ky = cos(theta)xk -sin(theta)yk

    sin(theta)kx + cos(theta)ky sin(theta)xk cos(theta)yk


    Do I need to use the inverse or can I use an assumption some how?


    Basically Iā€™m having trouble with knowing how to prove that the function is 1-1 and surgective.


    As well as constructing the proof logically and stating the proof clearly.



    If anyone can help that would be great.



    Regards
     
  2. jcsd
  3. Aug 25, 2009 #2
    Try writing it out as in f(a,b)=...for (a,b) in R^2

    To show that it is an isomorphism you should show that the function is 1-1, onto, and that it is a homomorphism. You mentioned that it may be a linear transformation. Did you check that?
     
  4. Aug 25, 2009 #3
    Yes It is a linear Transformation.

    Let
    [itex]
    \[ \left( \begin{array}{ccc}
    cos(\theta) & -sin(\theta) \\
    sin(\theta) & cos(\theta) \end{array} \right)\] = A.\\

    [/itex]
    Than
    [itex]

    A \times\[ \left( \begin{array}{ccc}
    x \\
    0 \end{array} \right)\] + A \times\[ \left( \begin{array}{ccc}
    0 \\
    y \end{array} \right)\]
    = \[ \left( \begin{array}{ccc}
    cos(\theta)x & -sin(\theta)y \\
    sin(\theta)x & cos(\theta)y \end{array} \right)\]


    [/itex]

    Sorry I'm not to good a latex yet.




    I Think the inverse might be:

    [itex]
    \[ \left( \begin{array}{ccc}
    cos(\theta)x & -sin(\theta)y \\
    -sin(\theta)x & cos(\theta)y \end{array} \right)\]

    [/itex]
    For the homomorphism the transformation is to R2 as well so that is shown.

    As of how to show it is 1-1 I'm don't know how. I'm not sure wether I need to use the inverse function or just use the assumption to show for [itex]f:G\rightarrow W there must be a function

    [itex]
    f:W\rightarrow G = f(x)^{-1}= w(f(x)) [/itex]
    [itex] x [/itex] is an element of G

    regards
     
  5. Aug 25, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think the inverse of [[cos(theta),-sin(theta)],[sin(theta),cos(theta)]] might be [[cos(theta),sin(theta)],[-sin(theta),cos(theta)]]. (Each bracketed pair is a row of the matrix - I'm not that hot at latex either). Can you check that? If a linear transformation R^2->R^2 has an inverse, then it's 1-1.
     
  6. Aug 26, 2009 #5
    I Did the following on my calculator


    [itex]A \times\[ \left( \begin{array}{ccc}x \\0 \end{array} \right)\] + A \times\[ \left( \begin{array}{ccc}0 \\y \end{array} \right)\] = \[ \left( \begin{array}{ccc}cos(\theta)x & -sin(\theta)y \\sin(\theta)x & cos(\theta)y \end{array} \right)\][/itex]

    I then multiplied the result by the inverse as you said:

    [itex]\[ \left( \begin{array}{ccc}cos(\theta)x & sin(\theta)y \\-sin(\theta)x & cos(\theta)y \end{array} \right)\][/itex]

    which gave me

    [itex]

    \[ \left( \begin{array}{ccc}x & \\y \end{array} \right)\][/itex]

    Is that right ?

    As both functions have dim(2) have I done enough to show that the function is isomorphic ?
     
  7. Aug 26, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, yes, that is the whole point of a "inverse", isn't it- to "undo" the original operation. But you didn't really need the "x y" matrix:
    [tex]\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}[/tex].

    Of course the opposite of "rotate through angle [itex]\theta[/itex]" is "rotate through angle [itex]-\theta[/itex]". So the inverse of
    [tex]\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}[/tex]
    is
    [tex]\begin{bmatrix}cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-\theta)\end{bmatrix}= \begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}[/tex]
    since "sine" is an odd function and "cosine" is an even function.
     
  8. Aug 26, 2009 #7
    Thanks for all your help guys.
     
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