# Isomorphism Proof

1. Aug 25, 2009

### beetle2

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Let f :$R^2 \rightarrow R^2$

3. 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 :$R^2 \rightarrow R^2$ is a linear transformation
by closure of addition and scaler multiplication by using x,y elements of R2 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

2. Aug 25, 2009

### VeeEight

Try writing it out as in f(a,b)=...for (a,b) in R^2

To show that it is an isomorphism you should show that the function is 1-1, onto, and that it is a homomorphism. You mentioned that it may be a linear transformation. Did you check that?

3. Aug 25, 2009

### beetle2

Yes It is a linear Transformation.

Let
$$\left( \begin{array}{ccc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{array} \right)$ = A.\$/extract_itex] Than $A \times\[ \left( \begin{array}{ccc} x \\ 0 \end{array} \right)$ + A \times$\left( \begin{array}{ccc} 0 \\ y \end{array} \right)$ = $\left( \begin{array}{ccc} cos(\theta)x & -sin(\theta)y \\ sin(\theta)x & cos(\theta)y \end{array} \right)$$

Sorry I'm not to good a latex yet.

I Think the inverse might be:

$$\left( \begin{array}{ccc} cos(\theta)x & -sin(\theta)y \\ -sin(\theta)x & cos(\theta)y \end{array} \right)$$
For the homomorphism the transformation is to R2 as well so that is shown.

As of how to show it is 1-1 I'm don't know how. I'm not sure wether I need to use the inverse function or just use the assumption to show for $f:G\rightarrow W there must be a function [itex] f:W\rightarrow G = f(x)^{-1}= w(f(x))$
$x$ is an element of G

regards

4. Aug 25, 2009

### Dick

I think the inverse of [[cos(theta),-sin(theta)],[sin(theta),cos(theta)]] might be [[cos(theta),sin(theta)],[-sin(theta),cos(theta)]]. (Each bracketed pair is a row of the matrix - I'm not that hot at latex either). Can you check that? If a linear transformation R^2->R^2 has an inverse, then it's 1-1.

5. Aug 26, 2009

### boneill3

I Did the following on my calculator

$A \times$\left( \begin{array}{ccc}x \\0 \end{array} \right)$ + A \times$\left( \begin{array}{ccc}0 \\y \end{array} \right)$ = $\left( \begin{array}{ccc}cos(\theta)x & -sin(\theta)y \\sin(\theta)x & cos(\theta)y \end{array} \right)$$

I then multiplied the result by the inverse as you said:

$$\left( \begin{array}{ccc}cos(\theta)x & sin(\theta)y \\-sin(\theta)x & cos(\theta)y \end{array} \right)$$

which gave me

$$\left( \begin{array}{ccc}x & \\y \end{array} \right)$$

Is that right ?

As both functions have dim(2) have I done enough to show that the function is isomorphic ?

6. Aug 26, 2009

### HallsofIvy

Staff Emeritus
Well, yes, that is the whole point of a "inverse", isn't it- to "undo" the original operation. But you didn't really need the "x y" matrix:
$$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$.

Of course the opposite of "rotate through angle $\theta$" is "rotate through angle $-\theta$". So the inverse of
$$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}$$
is
$$\begin{bmatrix}cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-\theta)\end{bmatrix}= \begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}$$
since "sine" is an odd function and "cosine" is an even function.

7. Aug 26, 2009

### boneill3

Thanks for all your help guys.