# Isomorphism Proof

## 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 :$R^2 \rightarrow R^2$

## 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

Related Calculus and Beyond Homework Help News on Phys.org
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?

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

Dick
Homework Helper
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.

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 ?

HallsofIvy
Homework Helper
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.

Thanks for all your help guys.