Linear Algebra: linear transformation

In summary, we are trying to find the matrix A for the linear transformation T from R^3 to R^3 that rotates vectors by 90 degrees about the x-axis. To determine the components of T(e_i) for each basis vector e_i, we need to rotate the vector counterclockwise in the yz-plane and determine its new components. After doing so, we find that T(e_1) = (1, 0, 0), T(e_2) = (0, 0, -1), and T(e_3) = (0, 1, 0). These vectors form the column vectors of the matrix A, and we can check our work by taking the product of A with
  • #1
HaLAA
85
0

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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  • #2
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 ? :-)
 
  • #3
slider142 said:
About which axis in particular ? :)
about the x-axis, sorry
 
  • #4
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 [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
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 [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
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 [itex]T(e_1), T(e_2),[/itex] and [itex]T(e_3)[/itex] ?
HaLAA said:
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
slider142 said:
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
HaLAA said:
I got it. thank you so much
Great! Are you able to complete the problem from there?
 
  • #9
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 [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.
 

Related to Linear Algebra: linear transformation

1. What is a linear transformation?

A linear transformation is a mathematical function that maps a vector space to another vector space in a way that preserves the structure of the original space. In other words, it is a transformation that maintains the linearity of the space.

2. What are some common examples of linear transformations?

Some common examples of linear transformations include scaling, rotation, reflection, and shearing. These transformations can be represented by matrices and are often used in computer graphics, physics, and engineering applications.

3. How is a linear transformation represented mathematically?

A linear transformation is represented mathematically by a matrix, which is a rectangular array of numbers. The dimensions of the matrix correspond to the number of inputs and outputs of the transformation. For example, a 2x2 matrix represents a transformation from a two-dimensional vector space to another two-dimensional vector space.

4. What is the difference between a linear transformation and a nonlinear transformation?

A linear transformation preserves the linearity of a vector space, meaning that straight lines remain straight after the transformation. On the other hand, a nonlinear transformation does not preserve this linearity and can result in curved or distorted shapes. Nonlinear transformations are often used in applications such as computer animation or image manipulation.

5. How is linear algebra used in real-world applications?

Linear algebra is used in a wide range of real-world applications, including computer science, engineering, physics, and economics. It is used to solve problems involving systems of linear equations, optimization, and data analysis. Linear algebra is also essential for understanding and developing advanced mathematical concepts and models.

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