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.

Solution 1:

Rotations preserve volume and orientation.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.