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Linear algebra fun never stops!

  1. Apr 5, 2006 #1
    In each case find a rotation or reflection which equals the given transformation

    Rotation through pi followed by reflection in the X axis

    Now is there a way to actually work this out?

    The text says the answer is simply reflection on the Y axis.
    I tried to visualize by drawing a vector, rotating it pi and then reflecting it on the X axis but it doesnt seem to make sense...

    A rotation thru pi would yield a matrix like this
    [tex] \left(\begin{array}{cc} -1&0 \\ 0&-1 \end{array}\right) [/tex]

    How would one incorporate the reflection on the X Axis.

    For that matter, how would one incorporate a Y axis reflection or a line y = mx reflection onto a matrix?
  2. jcsd
  3. Apr 5, 2006 #2
    if you reflect over the x axis, it only effects the y's. specifically it makes them negative. so that matrix would be
    [tex] \left(\begin{array}{cc} 1&0 \\ 0&-1 \end{array}\right) [/tex]

    when you multiply the two matrices, you get:
    [tex] \left(\begin{array}{cc} -1&0 \\ 0&-1 \end{array}\right) \left(\begin{array}{cc} 1&0 \\ 0&-1 \end{array}\right) = \left(\begin{array}{cc} -1&0 \\ 0&1 \end{array}\right)[/tex]

    when you reflect over the y axis, then you're just making all the x's negative. which, is the matrix we got as our answer.
  4. Apr 5, 2006 #3


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    Here is a neat trick. In any linear transformation, you only need to know where the standard unit vectors go. Let's say you have a linear transformation, and you know from the definition of this transformation that the vector
    [tex] \left(\begin{array}{cc} 1 \\ 0 \end{array}\right) [/tex]
    should go to
    [tex] \left(\begin{array}{cc} 0.5 \\ 0.2 \end{array}\right)[/tex]

    Now you have a matrix
    [tex] \left(\begin{array}{cc} a&b \\ c&d \end{array}\right) [/tex]
    Now what could a and c possibly be so that (1, 0) gets mapped to (0.5, 0.2)? Think about how to multiply a vector by a matrix.
    Last edited: Apr 6, 2006
  5. Apr 6, 2006 #4


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    Orthodontist gave you exactly what you need to know. (1, 0) is rotated 180 degrees counter clockwise so it ends up where? Reflecting that in the x-axis changes it to? That's your first column.
    (0,1) is rotated 180 degrees clockwise so it ends up where? Reflecting that in the x-axis changes it to? That's your second column.
  6. Apr 6, 2006 #5
    so then i would have to multiply

    [tex] \left(\begin{array}{cc} 1 \\ 0 \end{array}\right)^T \left(\begin{array}{cc} a&b \\ c&d \end{array}\right) = \left(\begin{array}{cc} 0.5 \\ 0.2 \end{array}\right)[/tex]


    so what if (1,0) was rotated pi/2? I know it would be (0,1)... but ist here a systematic way of doing it? What about pi/4?
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