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Why is the determinant of a rotation matrix equal to 1? I would like a geometric interpretation of this. Just curious.

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5 Answers 5

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Rotations preserve volume and orientation.

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    $\begingroup$ What do you mean orientation? The whole point is to change orientation, right? $\endgroup$
    – chris
    Commented Mar 10, 2021 at 23:50
  • $\begingroup$ @chris, the orientation of a basis is the sign of the determinant of the matrix that is composed of the basis vectors stacked columnwise, so if that sign is positive, the new rotated basis will also have a positive sign of the determinant. $\endgroup$
    – IntegrateThis
    Commented Jul 2, 2021 at 19:09
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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.

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  • $\begingroup$ You showed that $|det(A)|=1$, but I wonder if your argument can be modified to show that $det(A)=1$. $\endgroup$
    – IntegrateThis
    Commented Jan 26, 2020 at 0:21
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    $\begingroup$ In order to get $det(A) = 1$, you can decompose the rotation about an axis into two rotations by half the rotation angle, say $B$, then $A=BB$, and $det(A) = det(B)^2>0$ . $\endgroup$
    – IntegrateThis
    Commented Jul 2, 2021 at 19:13
  • $\begingroup$ @IntegrateThis You implicitly assume that the determinant of the rotation matrix is a constant. This is however not yet demonstrated in the answer. The determinant can be for example $e^{ic\phi}$, where $\phi$ is the (real) rotation angle and $c$ is some real constant. $\endgroup$
    – user
    Commented May 15, 2023 at 8:33
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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.

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    $\begingroup$ Your first claim is only true for $2$ dimensional matrices. $\endgroup$
    – Potato
    Commented Sep 24, 2013 at 1:35
  • $\begingroup$ @Potato In higher dimensional matrices, the determinant doesn't still calculate to cos^2+sin^2? There may be more columns and rows, but they are all filled with zero except for one entry which is one, so wouldn't the determinant be 1*cos^2 + 1*sin^2? $\endgroup$
    – john doe
    Commented Sep 24, 2013 at 1:41
  • $\begingroup$ A rotation matrix is an orthogonal matrix with determinant 1. It's easy to produce examples of these that don't fit the form you give. $\endgroup$
    – Potato
    Commented Sep 24, 2013 at 2:35
  • $\begingroup$ @Potato all these matrices can be decomposed into ones john telling us about. $\endgroup$
    – Yola
    Commented Feb 16, 2016 at 17:59
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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.

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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!

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