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For a given matrix $A$ the question "is there a vector (or are there vectors) whose direction remain the same when it is multiplied with the matrix" leads to the well-known eigenvalue problem $A\mathbf{v}=\lambda\mathbf{v}$, where $\mathbf{v}$ is an eigenvector of the matrix. Since a vector is fully described by its direction and amplitude, one can also ask the other obvious question:

For a given matrix $A$ is there a vector (or are there vectors) whose amplitude remains the same when it is multiplied with the matrix?

This would amount to finding "eigenrotations" that satisfy $\left\Vert A\mathbf{r}\right\Vert =\left\Vert \mathbf{r}\right\Vert .$ The eigenrotations are thus characterized by $\mathbf{r}^{T}\left( A^{T}A-I\right) \mathbf{r}=0,$ and under some conditions on the coefficients of $A$ there exist such vectors.

Question: In what contexts do "eigenrotations" come up? (other than a mathexchange question)

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Looks like singular value decomposition, except you focus on singular value 1, which might not always exist. – Raskolnikov Nov 28 '10 at 17:51
up vote 5 down vote accepted

Allow me to generalize the question a bit. It's a very natural operation to multiply a matrix by a scalar. Most properties of matrices, including the eigenvectors and eigenvalues, are either not affected by scalar multiplication or are multiplied by the same scalar. But $\lVert Ax \rVert = \lVert x \rVert$ does not remain true for the same vector $x$. So it's more natural to look at all the sets $\lVert Ax \rVert = c \lVert x \rVert$ as a function of a scalar $c$. These are just the level sets of $f(x) := \lVert Ax \rVert / \lVert x \rVert$. Of course, unlike the eigendecomposition, where you get a discrete set of (at most) $n$ vectors out of it, you will generically have infinitely many solutions for any $c$. Since all nonzero scalar multiples of $x$ have the same value of $f(x)$, we can without loss of generality restrict to the surface of the unit sphere $\lVert x \rVert = 1$. Then we are talking about solutions to the equations $\lVert x \rVert = 1$, $\lVert Ax \rVert = c$. Clearly, the solutions form a $(n-2)$-dimensional manifold.

Here is the only context in which I have seen such a construction. I think it's called Poinsot's ellipsoid. Consider a rigid body with moment of inertia $I$ tossed into the air with some angular velocity $\omega$. Since no external forces are acting on the body, its angular momentum $L = I\omega$ is conserved, and so is its rotational kinetic energy $T = \frac{1}{2}\omega^T I\omega$. Take a reference frame attached to the body. Then $I$ becomes a constant matrix. $L$ is no longer constant, but its magnitude still must be. So we get two equations: $\lVert L \rVert = \mathrm{const}$, and $\frac{1}{2} L^T I^{-1} L = \mathrm{const}$, which has almost the same form as your equations. This blog post called "On tossing stuff into the air" has some nice pictures, as well as discussion about how this tells you why things wobble if you spin them about the intermediate moment of inertia, and what happens when you toss a bag of peas into the air.

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