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Linear transformations

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    T: R^2-->R2 first rotates points through -3pi/4 radian clockwise and then reflects points through the horizontal x1-axis. Find the standard matrix of T.


    2. Relevant equations
    -


    3. The attempt at a solution
    Vector 1 is:
    (-1sqrt2
    - 1 sqrt2)
    before the reflection, and after the reflection it is:
    (-1 sqrt 2
    1 sqrt 2)
    This I knew since we can use:
    (cos insert radian
    sin radian)
    for v1.

    But for v2, I have some problems.
    If I use:
    (-sin insert radian
    cos insert radian)
    to solve this, my answer doesn't seem to be correct.
    My attempt:
    v2
    (1sqrt 2
    1 sqrt 2)
    before the reflection, since sin is negative here two negatives equals a plus. Cos is plus as we can see.
    After the reflection:
    v2
    (-1sqrt 2
    1 sqrt 2)

    this however, isn't the answer. Can anyone please help me? I'm having enormous troubles.
    /Gramsci
     
  2. jcsd
  3. Sep 21, 2008 #2

    gabbagabbahey

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    Let's call the matrix corresponding to a reflection throught the x-axis, [tex] \overleftrightarrow{A} [/tex]. What does [tex] \overleftrightarrow{A} [/tex] do to the components of a vector [tex]\vec{v} =(v_x,v_y)^T[/tex]?
     
  4. Sep 21, 2008 #3
    It transforms it into:
    (1 0
    0 -1) , right? I can't seem to derive any further info from this though.
     
  5. Sep 21, 2008 #4

    gabbagabbahey

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    No, it transforms [tex]v_x \rightarrow v_x [/tex] and [tex]v_y \rightarrow -v_y[/tex]

    [tex] \Rightarrow \left( \begin{array}{c} v_x \\ -v_y \end{array} \right) = \overleftrightarrow{A} \left( \begin{array}{c} v_x \\ v_y \end{array} \right) = \left( \begin{array}{cc} A_{11} &A_{12} \\ A_{21} &A_{22} \end{array} \right) \left( \begin{array}{c} v_x \\ v_y \end{array} \right)[/tex]

    You should be able to determine [tex] A_{11} [/tex], [tex] A_{12} [/tex], [tex] A_{21} [/tex], and [tex] A_{22} [/tex] from this.
     
  6. Sep 21, 2008 #5
    I think I've got it.
    Since:
    (-sin -3pi/4
    cos -3pi/4)

    can be compared to a linear reflection (can it?) It's supposed to be:
    (1sqrt 2
    - 1 sqrt 2)

    I derived the latter entry from the fact that a linear reflection does the things we stated in the previous post. After that, we transform it back and therefore we can take away the minus from the cos and our final answer is:
    v2=
    (1 sqrt 2
    1 sqrt 2)

    and v1
    -1 sqrt 2
    1 sqrt 2

    It doesn't feel right though. Thank you so much for helping me and excuse me if I am a little slow, I'm not used to these kinds of problems.

    Anything I should think about further on this question? Did I do anything wrong?
    /Gramsci
     
  7. Sep 21, 2008 #6

    gabbagabbahey

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    what are v1 and v2 ??? are they vectors or matrices?
     
  8. Sep 21, 2008 #7
    Vectors
     
  9. Sep 21, 2008 #8

    gabbagabbahey

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    A linear transformation in R2 is a 2X2 matrix, not a vector. A rotation is a linear transformation, and hence also a 2X2 matrix; and the same is true for a reflection.
     
  10. Sep 21, 2008 #9
    Well, then the complete matrix should be:
    (-1 sqrt 2, 1 sqrt 2
    1 sqrt 2, 1 sqrt 2)

    Right?
    Did I do something wrong in how I got these answers?
     
  11. Sep 21, 2008 #10

    gabbagabbahey

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    This is close, but not correct. Let's start with the matrix for just the reflection, [tex]\overleftrightarrow{A}[/tex] , what do you get for that?
     
  12. Sep 21, 2008 #11
    Hm, it seems like I forgot a sign. What I meant was:
    (-1/sqrt 2, 1/ sqrt 2
    1 / sqrt2, 1/ sqrt 2)

    This is however, correct, no?

    However, the reflection matrix should be(?):

    (A11 -A12
    A21 -A22)?
     
  13. Sep 21, 2008 #12

    gabbagabbahey

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    Still incorrect.

    What are the values of A11, A12, A21, A22?
     
  14. Sep 21, 2008 #13
    Are you sure it's incorrect? Because according to my solutions manual, that is the solution.
     
  15. Sep 21, 2008 #14
    However, are the values supposed to be:
    (1, -1
    1, -1)?
    Excuse me if I'm slow.
     
  16. Sep 21, 2008 #15

    gabbagabbahey

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    I get:

    [tex] \overleftrightarrow{T} \rightarrow \left( \begin{array}{cc} \frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) [/tex]
     
  17. Sep 21, 2008 #16

    gabbagabbahey

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    For the matrix representing just the reflection, I get:

    [tex]
    \overleftrightarrow{A} \rightarrow \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)
    [/tex]

    And for just the rotation I get:

    [tex]
    \overleftrightarrow{R} \rightarrow \left( \begin{array}{cc} \frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{array} \right)
    [/tex]
     
  18. Sep 21, 2008 #17
    Hm, according to LAYs "Linear algebra and its applications" it's supposed to be what I previously wrote. Strange. Do you have any form of gmail or msn by the way, it would be easier to discuss through this. Care to show how you came to your answer?
     
  19. Sep 21, 2008 #18
    I thought I wrote that the reflection matrix was:

    (1, 0
    0, -1)

    up there?
     
  20. Sep 21, 2008 #19

    gabbagabbahey

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    Yes, I thought you were saying that that was what the vector became though.

    I just realized that the rotation is by an angle -3pi/4, I thought it was +3pi/4, so your T is correct.
     
  21. Sep 21, 2008 #20
    Otherwise, did I do everything right in how I got to the answer?
     
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