Question

How I can Prove that the function R that transforms R^2 in R^2 when rotate the plane in the opposite direction to rotate clockwise a positive angle "α" is a linear transformation

Answer 1

To prove a rotation is a linear transformation, you need:
1) pick 3 non-colinear points. Let say A, B, C
2) calculate the distance between them
3) Put them under rotation to get their images
4) Calculate the distance between their images
5) Make a comparison between the distances of images and preimages. They must be equal
6) Conclusion: that transformation is linear since it preserve distance and betweeness.

Answer 2

Thanks for your answer :D How can I put in with therms? I mean something like "V=(v1,v2,...) w=(w1,w2,...) v+s=(v1+w1,v2+w2,...) Then that proves the axiom blahblah " , can do you help me with that please please please? :(

Answer 3

oh, this is linear algebra, not geometry. Ah ha, hence, we need another proof. I am sorry. Let me post the proof for linear algebra.

Answer 4

Omg yes linear algebra, thanks thanks thanks :D!!!!

Answer 5

One more question, it is just linear algebra or theoretical linear algebra? They are different and they have different proof also. :) I don't want to make a mistake again.

Answer 6

I assume it is just linear algebra. OK, for R^2, you have just 2 dimensions.

Answer 7

Let \(\vec v =(v_1,v_2)\), \(\vec w=(w_1,w_2)\)

Answer 8

I suppose it's linear algebra because the course it's called like that I'm not sure but the teacher want the answer with demostration, like what i put before somehow, so I'm not sure D:

Answer 9

let k be a scalar, then we need prove T(v+w) = T(v) + T(w) and T(kv) = kT(v)

Answer 10

How is the rotation operation defined in your course?

Answer 11

(This part is from modern geometry)

Answer 12

OMG, I don't know whether you know this or not,
Rotation an angle alpha about origin is define as a matrix
\(\left[\begin{matrix}cos\alpha &-sin\alpha\\sin\alpha&cos\alpha\end{matrix}\right]\)

Answer 13

I didn't even reached that part, the teacher just left homework more advanced that we already know lol :(

Answer 14

That is the rotation transformation. hence, if you apply to v and w, you will see
\(\left[\begin{matrix}cos\alpha &-sin\alpha\\sin\alpha&cos\alpha\end{matrix}\right]\)\(\left[\begin{matrix} v_1\\v_2 \end{matrix}\right]\) = ...

Answer 15

Uhmmm I have one idea, I have another question and I think that's more easy than this one and will have the same punctuation I need to get too much points for this course so... We could left this question by now and better answer this? D: http://postimg.org/image/d961jqv83/

Answer 16

do the same with w
then the same with v+w
that is for part a), part 2, you do kv = (kv1, kv2), then apply that operation to kv and you can see Matrix (kv) = kMatrix (v) and you are done

Answer 17

I got your answers somehow lol:D

Answer 18

wait That's all? :D

Answer 19

yup

Answer 20

That's all but it is not a piece of cake since you need a lot of trig function to get the answer. hehehe...

Answer 21

YOu see, for v, the matrix (v) = (cos (a) * v1 - sin (a)*v2 , cos (a) v1+ sin(a)*v2)
same for matrix(w)
same for matrix (v+w),
then prove them equal. hahaha... good luck.

Answer 22

wow thanks, that's a lot of points :D I will do take the best response! :D