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Linear Algebra: linear transformation

  • #1
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Homework Statement


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

Homework Equations


find the matrix A

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
 
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Answers and Replies

  • #2
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Homework Statement


let A be the matrix corresponding to the linear transformation from R^3 to R^3 that is rotation of 90 degrees about the axis
About which axis in particular ? :-)
 
  • #3
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About which axis in particular ? :)
about the x-axis, sorry
 
  • #4
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about the x-axis, sorry
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 ?
 

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  • #5
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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 ?
T(e3)=(0,0,1)
 
  • #6
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70
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] ?
T(e3)=(0,0,1)
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?
 

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  • #7
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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?
I got it. thank you so much
 
  • #8
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I got it. thank you so much
Great! Are you able to complete the problem from there?
 
  • #9
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Great! Are you able to complete the problem from there?
yes,T(e1)=(1,0,0),T(e2)=(0,-1,0),T(e3)=(0,1,0)
 
  • #10
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yes,T(e1)=(1,0,0),T(e2)=(0,-1,0),T(e3)=(0,1,0)
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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