1
$\begingroup$

Let's define "preserve orientation" in the following way (I am not sure it is right, pls point out if there is something wrong):

>[![enter image description here][1]][1]

For a linear transformation, we only need to check non-parallel vectors, because parallel vectors naturally come out parallel under linear transformation.

A rotation is a transformation that preserves length, angle and orientation. Any transformation preserves length and angle must preserves the dot product, and by then it is a linear transformation, and the transformation matrix has to be orthogonal by definition & properties of orthogonal matrix.

Now my question is how to prove a rotation is exactly the orthogonal matrix of determinant 1, i.e. only such orthogonal matrix with determinant 1 preserves orientation?

I have a wild guess that a linear transformation $\bf T$ preserves orientation iff it has positive determinant, so we immediately have the above claim. But I don't know how to show it. This is a wild guess and very likely to be wrong. Thank you for you guys' help!

$\endgroup$
  • $\begingroup$ Sorry if a dumb question, I guess $(x,y) , (f(x), f(y))$ are the matrices formed by $x,y ; f(x),f(y)$ as column vectors , e.g., for $ x=(x_1, x_2), y=(y_1, y_2) , (x,y)=( x_1, y_1 :x_2, y_2 )$? where $ x_1, y_1$ are in the top row ;$ x_2, y_2$ are in the bottom one, i.e., If $M=(x,y) =(m_{ij}) $then $m_{11}=x_1, m_{12}=y_1, m_{21}=x_2, m_{22}=y_2$? $\endgroup$ – MSIS Jan 25 at 2:22
0
$\begingroup$

For the case where it's over $\mathbb R^2$ it's quite trivial. The images of $e_x$ and $e_y$ is the columns of the matrix. According to the requirement these has to be of unit length and orthogonal - therefor we have the requirement that the matrix has to be orthonormal.

In addition to preserve orientation we have that the images of $e_y$ has to be the image of $e_x$ rotated $\pi/2$ counter clock wise (given positively oriented coordinate system). That is if the image of $e_x$ is $(u,v)$ the image of $e_y$ has to be $(-v, u)$ which means that the matrix will have positive determinant.

The reverse can as easily be seen as being true.

For higher dimension you will eventually run into the very definition of orientation being dependent on the determinant. However one could define rotations as images composed of "trivial" rotations around the axes for example, in this case one can see in similar way that we will end up with exactly those matrices as well.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Well...sort of. There are "rotations" in 4-space that have no single "fixed vector", i.e., no "axis of rotation." So that last paragraph isn't true as written. $\endgroup$ – John Hughes Sep 29 '19 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.