# Prove the orthogonal matrix with determinant 1 is a rotation

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

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!

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.