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Linear Algebra: linear transformation

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    let A be the matrix corresponding to the linear transformation from R^3 to R^3 that is rotation of 90 degrees about the x-axis

    2. Relevant equations
    find the matrix A

    3. The attempt at a solution
    I got stuck on rotating z component.

    I tried T([e1,e2,e3])=[0 -1 0]
    [1 0 0]
    [0 0 -1]
    I am sure this is not right
     
    Last edited: Nov 6, 2014
  2. jcsd
  3. Nov 6, 2014 #2
    About which axis in particular ? :-)
     
  4. Nov 6, 2014 #3
    about the x-axis, sorry
     
  5. Nov 6, 2014 #4
    Okay. The first step of determining the matrix of a linear transformation in a particular basis is to first determine where each basis vector goes. So we first make a list. What are [itex]T(e_1), T(e_2),[/itex] and [itex]T(e_3)[/itex] ?

    Edit: If this is the part you are having trouble with, maneuver your imaginary coordinate axes so that the positive x-axis is pointing straight out of the page, so we are looking down at the yz-plane. The positive z-axis, and thus [itex]e_3[/itex] points right and the positive y-axis, and thus [itex]e_2[/itex], points up. Rotation of vectors by positive angles occurs counterclockwise. Take [itex]e_3[/itex], and rotate it counterclockwise by 90 degrees. It should end up on the y-axis. This is [itex]T(e_3)[/itex]. What is it in component form ?
     

    Attached Files:

    Last edited: Nov 6, 2014
  6. Nov 6, 2014 #5
    T(e3)=(0,0,1)
     
  7. Nov 7, 2014 #6
    Hmm, that can't be true, since [itex]e_3 = (0, 0, 1)[/itex] in the standard basis. Remember we are going to rotate this vector counterclockwise in the yz-plane, so its image after the transformation will be a different vector. Just to make sure we're on the same page, the standard basis is [itex]e_1 = (1, 0, 0), e_2 = (0, 1, 0),[/itex] and [itex]e_3 = (0, 0, 1)[/itex].
    Now, the first attached image is meant to be a viewpoint where the positive x-axis, and thus [itex]e_1[/itex], is pointing out of the page. So we cannot see the x-axis from this viewpoint: it is directly above the origin of the yz-plane. The positive z-axis, and thus [itex]e_3 = (0, 0, 1)[/itex] points right, while the positive y-axis, and thus [itex]e_2 = (0, 1, 0)[/itex] points up. The diagram shows the counterclockwise direction of the rotation by 90 degrees and its magnitude.
    The second illustration shows that when we rotate [itex]e_3[/itex] counterclockwise by 90 degrees in this plane, it coincides with [itex]e_2[/itex]. So the components of the new vector [itex]T(e_3)[/itex] are (0, 1, 0). Does that make sense so far?
     

    Attached Files:

  8. Nov 7, 2014 #7
    I got it. thank you so much
     
  9. Nov 7, 2014 #8
    Great! Are you able to complete the problem from there?
     
  10. Nov 7, 2014 #9
    yes,T(e1)=(1,0,0),T(e2)=(0,-1,0),T(e3)=(0,1,0)
     
  11. Nov 7, 2014 #10
    Not quite! When you rotate [itex]e_2[/itex], the unit vector for the positive y-axis, by 90 degrees counterclockwise, you should end up on the negative z-axis, which is represented by [itex]-e_3[/itex] or (0, 0, -1). So [itex]T(e_2) = (0, 0, -1)[/itex]. These three vectors form the column vectors of the matrix that represents T with respect to this basis. You can then take the product of the matrix with each unit vector to check that you do get the correct outputs.
     
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