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Linear Transformation involving pi/2

  1. Jun 9, 2013 #1

    dwn

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    Resource: Linear Algebra (4th Edition) -David C. Lay

    I understand that there are identities associated with transformations, but what I don't understand is when the transformation is rotated about the origin through an angle β. I believe β in this case is [itex]\frac{}{}\pi/2[/itex]

    [itex]\left[1,0\right][/itex] into [cos([itex]\beta[/itex]) , sin([itex]\beta[/itex])]
    [itex]\left[0,1\right][/itex] into [-sin([itex]\beta[/itex]), cos([itex]\beta[/itex])]

    Can someone please explain to me why this is the case? Why do these values suddenly translate to trig identities?

    Thanks!
     
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  3. Jun 9, 2013 #2

    Fredrik

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    I don't understand the question. "...β in this case..." What case?

    Are you asking why (1,0) rotated counterclockwise by an angle of β is (cos β,sin β)? This is a very common way to define sin and cos. Do you want to use another definition?
     
  4. Jun 9, 2013 #3

    dwn

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    I suppose my questions reiterates my confusion...haha.

    There is something I'm not grasping in the definition of this counterclockwise rotation. How am I suppose to know the positive/negative values of the matrix and whether they're sin or cosine..? What if this type of rotation is not [itex]\pi[/itex]/2?
     
  5. Jun 9, 2013 #4

    Fredrik

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    \begin{pmatrix}\cos\beta & -\sin\beta\\ \sin\beta & \cos\beta\end{pmatrix} is the matrix representation of a counterclockwise rotation by an arbitrary angle β. When ##\beta=\pi/2##, we have ##\sin\beta=1## and ##\cos\beta=0##, so the matrix representation of a counterclockwise rotation by ##\pi/2## is
    \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}
     
  6. Jun 9, 2013 #5

    dwn

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    Sometimes its difficult "to see the wood for the forest". That's all I will say.

    Thanks for your clarifying this point for me.
     
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