Linear transformations and rotations

[phy]

Linear transformations and rotations...

Hi everyone. I need some help getting started on this question.

Let R: R3 ---> R3 be a rotation of pi/4 around the axix in R3. Find the matrix [R]E that defines the linear transformation R in the standard basis E={e1, e2, e3} of R3. Find R(1,2,1)

The problem I'm having is just I don't know how to handle the question since I'm not given an equation for R nor am I given some sort of vector to start with. Or am I supposed to put vectors e1, e2, and e3 as the colums of a matrix and do something like that? I'm confused so any help would be greatly appreciated. Thanks.

 

[Dr Transport]

Since R3 is most likely the z-axis, the rotation is in the x-y plane. Think about the rotation of a vector in that plane, the rows of the transformation matrix would correspond to the coefficients of the transformation \vec{R}' = A \vec{R}. A hint, the 3rd row of the matrix will be (0 0 1).

 

[phy]

Hmmmmm, I'm not quite sure I understand. Would the first row be (0 0 1) and the second (0 1 0)?

 

[phy]

Ooops I meant (1 0 0) and (0 1 0)

 

[Dr Transport]

No, you are rotating about the z-axis, x' = x \cos(\pi/4) + y \sin(\pi/4) and y' = -x \sin(\pi/4) + y \sin(\pi/4) check my signs, but I think thay may be correct. the 3rd row is as above. The initial vector is (1,2,1).

 

[Fredrik]

A rotation is a linear transformation that doesn't change the length of any vector. This means that

x^tx=(Rx)^t(Rx)

for all x. This fact, together with the condition that any vector in the 3 direction is left unchanged by left action of R, is enough to completely determine the components of R.

 

[Dr Transport]

This is a rotation, not magnitude change in the vector, only direction.

 

[Fredrik]

That's what I said.