How to check that this is an orthogonal linear map with $\det (A) = 1$, so it is a rotation?

Question

$V$ is a $3$-dimensional Euclidean vector space with scalar product. Let $(e_1,e_2,e_3)$ be an ordered orthonormal basis of $V$ and let $A$ be the permutation operator defined by $$A(e_1) = e_2, A(e_2) = e_3, A(e_3) = e_1$$

How to check that this is an orthogonal linear map with $\det (A) = 1$, so it is a rotation?

Answer 1

You see that the matrix of the transformation w.r.t. this basis is $$ M(A)=\left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right). $$ It is easy to verify that the transpose of this matrix is its inverse, so it is orthogonal. Because $\det M(A)=1$, it is a rotation.

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• In a comment under this answer there is a proof for the fact that such a $3\times3$ matrix has eigenvalue $1$, that is, an axis of the rotation.
• Here is a recipe for finding the angle of rotation.

Answer 2

It is clear that $v=e_1+e_2+e_3$ is sent to itself, so the transformation $A$ has the line spanned by $v$ fixed.

If you now compute the other two eigenvalues, you'll find that $A$ is rotation around the $v$-axis

Answer 3

Although I think the best bet is just to write out the corresponding matrix, here's an alternative method. Since $Ae_i = e_{i+1}$ (taking the indices mod $3$), it's easy to verify that the eigenvectors are given by $v_\omega = e_1 + \omega e_2 + \omega^2 e_3$ with $\omega^3 = 1$, with corresponding eigenvalue $\omega$. (There are three such vectors over $\mathbb{C}$, and they're clearly linearly independent.) Thus choosing a primitive $3$rd root of unity $\omega_0 = e^{2\pi i/3}$, we have $\det A = \omega_0.\omega_0^2.\omega_0^3 = \omega_0^6 = 1$. To verify that this map is orthogonal, note that $$\langle{Ae_i, Ae_j\rangle} = \langle{e_{i+1}, e_{j+1}\rangle} = \delta_{ij} = \langle{e_i, e_j\rangle}$$ for all $i$ and $j$.