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Isomorphism Proof

  • Thread starter beetle2
  • Start date
  • #1

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 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


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.


Answers and Replies

  • #2
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
Yes It is a linear Transformation.

\[ \left( \begin{array}{ccc}
cos(\theta) & -sin(\theta) \\
sin(\theta) & cos(\theta) \end{array} \right)\] = A.\\


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 [itex]f:G\rightarrow W there must be a function

f:W\rightarrow G = f(x)^{-1}= w(f(x)) [/itex]
[itex] x [/itex] is an element of G

  • #4
Science Advisor
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.
  • #5
I Did the following on my calculator

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

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

[itex]\[ \left( \begin{array}{ccc}cos(\theta)x & sin(\theta)y \\-sin(\theta)x & cos(\theta)y \end{array} \right)\][/itex]

which gave me


\[ \left( \begin{array}{ccc}x & \\y \end{array} \right)\][/itex]

Is that right ?

As both functions have dim(2) have I done enough to show that the function is isomorphic ?
  • #6
Science Advisor
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:
[tex]\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}[/tex].

Of course the opposite of "rotate through angle [itex]\theta[/itex]" is "rotate through angle [itex]-\theta[/itex]". So the inverse of
[tex]\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}[/tex]
[tex]\begin{bmatrix}cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-\theta)\end{bmatrix}= \begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}[/tex]
since "sine" is an odd function and "cosine" is an even function.
  • #7
Thanks for all your help guys.