# Linear Algebra: linear transformation

Tags:
1. Nov 6, 2014

### HaLAA

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. Nov 6, 2014

### slider142

About which axis in particular ? :-)

3. Nov 6, 2014

4. Nov 6, 2014

### slider142

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 ?

#### Attached Files:

• ###### Rotation.png
File size:
1.4 KB
Views:
65
Last edited: Nov 6, 2014
5. Nov 6, 2014

### HaLAA

T(e3)=(0,0,1)

6. Nov 7, 2014

### slider142

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?

File size:
1.9 KB
Views:
68
File size:
2.5 KB
Views:
65
7. Nov 7, 2014

### HaLAA

I got it. thank you so much

8. Nov 7, 2014

### slider142

Great! Are you able to complete the problem from there?

9. Nov 7, 2014

### HaLAA

yes,T(e1)=(1,0,0),T(e2)=(0,-1,0),T(e3)=(0,1,0)

10. Nov 7, 2014

### slider142

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.