Linear Algebra: linear transformation

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The discussion focuses on finding the matrix A that represents a linear transformation corresponding to a 90-degree rotation about the x-axis in R^3. Participants clarify that the transformation involves determining the images of the standard basis vectors e1, e2, and e3. After some back-and-forth, it is established that T(e1) remains (1, 0, 0), T(e2) becomes (0, 0, -1), and T(e3) transforms to (0, 1, 0). The correct matrix is constructed using these vectors as columns, ensuring accurate representation of the transformation. The conversation emphasizes understanding the rotation's effect on each basis vector to derive the correct matrix.
HaLAA
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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
 
Last edited:
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HaLAA said:

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 ? :-)
 
slider142 said:
About which axis in particular ? :)
about the x-axis, sorry
 
HaLAA said:
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 T(e_1), T(e_2), and T(e_3) ?

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 e_3 points right and the positive y-axis, and thus e_2, points up. Rotation of vectors by positive angles occurs counterclockwise. Take e_3, and rotate it counterclockwise by 90 degrees. It should end up on the y-axis. This is T(e_3). What is it in component form ?
 

Attachments

  • Rotation.png
    Rotation.png
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slider142 said:
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 T(e_1), T(e_2), and T(e_3) ?

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 e_3 points right and the positive y-axis, and thus e_2, points up. Rotation of vectors by positive angles occurs counterclockwise. Take e_3, and rotate it counterclockwise by 90 degrees. It should end up on the y-axis. This is T(e_3). What is it in component form ?
T(e3)=(0,0,1)
 
slider142 said:
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 T(e_1), T(e_2), and T(e_3) ?
HaLAA said:
T(e3)=(0,0,1)
Hmm, that can't be true, since e_3 = (0, 0, 1) 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 e_1 = (1, 0, 0), e_2 = (0, 1, 0), and e_3 = (0, 0, 1).
Now, the first attached image is meant to be a viewpoint where the positive x-axis, and thus e_1, 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 e_3 = (0, 0, 1) points right, while the positive y-axis, and thus e_2 = (0, 1, 0) 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 e_3 counterclockwise by 90 degrees in this plane, it coincides with e_2. So the components of the new vector T(e_3) are (0, 1, 0). Does that make sense so far?
 

Attachments

  • StartRotate.png
    StartRotate.png
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  • EndRotate.png
    EndRotate.png
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slider142 said:
Hmm, that can't be true, since e_3 = (0, 0, 1) 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 e_1 = (1, 0, 0), e_2 = (0, 1, 0), and e_3 = (0, 0, 1).
Now, the first attached image is meant to be a viewpoint where the positive x-axis, and thus e_1, 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 e_3 = (0, 0, 1) points right, while the positive y-axis, and thus e_2 = (0, 1, 0) 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 e_3 counterclockwise by 90 degrees in this plane, it coincides with e_2. So the components of the new vector T(e_3) are (0, 1, 0). Does that make sense so far?
I got it. thank you so much
 
HaLAA said:
I got it. thank you so much
Great! Are you able to complete the problem from there?
 
slider142 said:
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
HaLAA said:
yes,T(e1)=(1,0,0),T(e2)=(0,-1,0),T(e3)=(0,1,0)
Not quite! When you rotate e_2, 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 -e_3 or (0, 0, -1). So T(e_2) = (0, 0, -1). 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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