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Problem finding the transformation in this case

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    give the matrix in the transformation T in each case:
    T: is a reflection in the line y= -x


    2. Relevant equations
    none that i know of for this question


    3. The attempt at a solution
    i know y= -x goes from the top left to bottom right in a cartesian plane, so the reflection of the line with initial point (1,1) in the first quadrant would end up in the third quadrant. i get my answer as:
    -1 0
    0-1
    but the answer in the back of the book is:
    0 -1
    -1 0
    why is their matrix the exact opposite as mine?
     
  2. jcsd
  3. Nov 8, 2008 #2

    gabbagabbahey

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    Well, what does the vector(1,1) become once it is reflected along the line y=-x?
     
  4. Nov 8, 2008 #3
    it becomes (-1,-1)
     
  5. Nov 8, 2008 #4

    gabbagabbahey

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    Good, that means that:

    [tex]T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}[/tex]

    How would you go about solving this for T?
     
  6. Nov 8, 2008 #5
    im not sure with this one,wait would you just rearrange to find t?
     
  7. Nov 8, 2008 #6
    but how does it end up being:
    0 -1
    -1 0
    and not
    -1 0
    0 -1
    ?
     
  8. Nov 8, 2008 #7

    gabbagabbahey

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    There is more than one method, but I would write:

    [tex]T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} T_{11} & T_{12} \\ T_{21} & T_{22} \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}[/tex]

    Then carry out the matrix multiplication and solve for the components of T....if you want to use this method, you will also need another equation of the same form...what does T do to the vector (-1,1)?
     
  9. Nov 8, 2008 #8
    i still dont see how since the point (1,1) is transformed to (-1,-1) it gives
    0 -1
    -1 0
    when i see this it says to me, that y= -1 from the first column when i think it should be x=-1 in the first column and likewise with the second column
     
  10. Nov 8, 2008 #9

    gabbagabbahey

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    Well, what is T suuposed to do to the vector (-1,1)....does your T actually do that?

    Since there are 4 unique elements to T, you need four equations to solve for them...looking at the effect of T on a single vector only gives you two equations, so you need to consider the effect on two vectors.
     
  11. Nov 8, 2008 #10

    HallsofIvy

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    If A is a linear transformation from vector space V to itself and {e1, e2, ..., en} is a basis for V, you can find the matrix representation for A by applying to each of the basis vectors in turn. For example, suppose the matrix representation of A, in that basis, is
    [tex]\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right][/tex]
    then, since the array representation of the basis vectors is just [1, 0, 0], [0, 1, 0], [0, 0 , 1], Applying A to e1 is just
    [tex]\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\left]= \left[\begin{array}{c} a_{11} \\ a_{21} \\ a_{31}\end{array}\right][/tex]
    That is, applying A to e1 tells us what the first column is, e2, the second column, etc.

    The standard basis for R2 is {[1 0], [0 1]}. Applying A to [1 0] gives [0 -1] and applying A to [0 1] is [0 -1] and to [0 1], [-1 0] so the matrix is
    [tex]\left[\begin{array}{cc} 0 & -1 \\-1 & 0\end{array}\right][/tex]

    However, if we use basis {[0 1],[1 0]}, that is the same vectors but in different order, that swaps the columns for A:
    [tex]\left[\begin{array}{cc} -1 & 0 \\0 & -1\end{array}\right][/tex]

    In the first example, e1= [1 0], e2= [0 1]. In the second example, e1= [0 1], e2= [1 0].

    Same vector space, same linear transformation, same vectors in each basis but, because the order or basis vectors is different, different bases and so different matrices.
     
    Last edited: Nov 8, 2008
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