Why is the determinant of a rotation matrix equal to 1? Why is the determinant of a rotation matrix equal to 1? I would like a geometric interpretation of this. Just curious.
 A: For a given vector $\vec v$, we rotate it to $\vec w$. Conventionally, after the rotation, $||\vec w||_2 = ||\vec v||_2 $, Let A a rotation matrix, so we $$||A\vec v||_2 = 1 $$ for all $\vec v$. For the simplicity, if you pick an eigenvector $\vec x$, $\lambda$ is the corresponding eigenvalue of A, you will see $$||A\vec x|| =|\lambda|||\vec x|| = ||\vec x||$$
so all eigenvvaluess of A have $|\lambda| = 1$.
So for the rotational matrix, $|det(A)| = |\lambda_1||\lambda_2| ... |\lambda_n| =  1$
The argument works for complex eigenvalues of $||$ is interpreted in C.
A: Using the definition of a determinant you can see that the determinant of a rotation matrix is $\cos^2(\theta) + \sin^2(\theta)$ which equals $1$. A geometric interpretation would be that the area does not change, this is clear because the matrix is merely rotating the picture and not distorting it in any other way.
A: Our usual geometric intuition of a rotation corresponds to what is called a  proper rotation, which is represented by an orthogonal matrix with determinant equal to +1. 
Orthogonal matrices with determinant equal to -1 represent rotations combined with a reflection, such as $\left( \begin{matrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{matrix} \right)$, which result in an inverted orientation, unlike proper rotations.
A: Rotations preserve volume and orientation.
A: The determinant of the standard rotation matrix is 1, this can be calculated very easy in 2D (see Answers above) and in 3D as well:
$\begin{bmatrix}
  1&0&0\\
  0&c\Phi&-s\Phi\\
  0&s\Phi&c\Phi\\ 
\end{bmatrix} = R \left( \Phi \right)$
$\begin{bmatrix}
  c\Theta&0&s\Theta\\
  0&1&0\\
  -s\Theta&0&c\Theta\\
\end{bmatrix} = R \left( \Theta \right)$
$\begin{bmatrix}  
  c\Psi&-s\Psi&0\\
  s\Psi&c\Psi&0\\
  0&0&1\\
\end{bmatrix} = 
R \left( \Psi \right)$
e.g.: $|| R \left( \Phi \right) || = 1 * cos^2(\phi) + sin^2(\phi) = 1$
and the others are identical.
Now using det(A*B) = det(A)*det(B) proof it is obvious to see, that any rotation matrix you builded from those three above will have a determinant = 1
edit: Thanks for link to the tutorial!
