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## Homework Statement

The problem is as follows:

Let f : R^2 map to R^2 be rotation through an angle of theta radians about the origin.

Prove that f is an isomorphism.

## Homework Equations

Let f :[itex] R^2 \rightarrow R^2 [/itex]

## The Attempt at a Solution

I know that the rotation can be expressed as the 2 x 2 matrix

cos(theta) -sin(theta)

Sin(theta) cos(theta)

And its inverse I believe is

cos(theta) -sin(theta)

-sin(theta) cos(theta)

Do I first show that f :[itex] R^2 \rightarrow R^2 [/itex] is a linear transformation

by closure of addition and scaler multiplication by using x,y elements of R

^{2}and some scaler k

Say let the 2x2 matrix:

cos(theta) -sin(theta)

Sin(theta) cos(theta) = A

We need to show

A(x+y) = A(x) + A(y)

cos(theta)x -sin(theta)y = cos(theta)x + -sin(theta)y = cos(theta)x -sin(theta)y

sin(theta)x cos(theta)y sin(theta)x + cos(theta)y sin(theta)x cos(theta)y

likewise

A(kx+ky) = K A(x+y)

cos(theta)kx + -sin(theta)ky = cos(theta)xk -sin(theta)yk

sin(theta)kx + cos(theta)ky sin(theta)xk cos(theta)yk

Do I need to use the inverse or can I use an assumption some how?

Basically I’m having trouble with knowing how to prove that the function is 1-1 and surgective.

As well as constructing the proof logically and stating the proof clearly.

If anyone can help that would be great.

Regards