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Let $0<a<b$, $ab=1$, and let $$ D_{a,b}=\biggl\{(x,y) \,\biggm | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \biggr\} $$ be the ellipse with diameters $a,b$.

Let $A \in \operatorname{SL}_2(\mathbb R) \setminus \operatorname{SO}(2)$ and suppose that $AD_{a,b}=D_{a,b}$.

Question: Must the singular values of $A$ be $\bigl\{\frac{a}{b},\frac{b}{a}\bigr\}$?

This option is always possible since we can take $A=\begin{pmatrix} \frac{a}{b} & 0 \\\ 0 & \frac{b}{a}\end{pmatrix}R_{\pi/2}$.

Here is a (very) partial attempt:

The condition $AD_{a,b}=D_{a,b}$ implies* that $A$ is similar to an orthogonal matrix, i.e.

$A=CQC^{-1}$, where $C \in \operatorname{SL}_2(\mathbb R) , Q \in \operatorname{SO}(2)$.

$AD_{a,b}=D_{a,b}$ implies that $Q\tilde D=\tilde D$, where $\tilde D=C^{-1}D_{a,b}$.

If $Q$ is an irrational rotation (of infinite order), then $\tilde D$ must be the unit disk, which implies that the singular values of $C$ are $a,b$. This mean that $C=U\Sigma V^T$, where $\Sigma=\operatorname{diag}(\sigma_1,\sigma_2)$.

Thus, $$ A=CQC^{-1}=U\Sigma V^T Q V\Sigma^{-1}U^T, $$ so up to left and right multiplication by rotations, $A=\Sigma R\Sigma^{-1}$, where $R= V^T Q V$.

I don't see how to continue from here.

I also don't know how to start analyzing the case where $Q$ is of finite order.

*You may see this answer and the comments below it.

Arctic Char
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Asaf Shachar
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1 Answers1

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That's not true in general: the singular values of $A$ are by definition the square roots of the eigenvalues of $A^*A$. Set \begin{align*}A(\theta) & = \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pmatrix}\\ &= \begin{pmatrix} \cos\theta & -\frac ab \sin\theta \\ \frac ba \sin\theta & \cos \theta \end{pmatrix} \end{align*}

Then it is clear that there are choices of $\theta$ such that the singular values are not $\{ a/b, b/a\}$ when $a\neq b$ (For example, note that the singular values are continuous functions (see remark) of $\theta$, and one has $\{1, 1\}$ at $\theta = 0$ and $\{ a/b, b/a\}$ at $\theta = \pi/2$).

Remark One can calculate that the eigenvalues of $A^*A$ are

$$ \frac{1}{2} \left(2\cos ^2 \theta + \left( \frac{a^2}{b^2} + \frac{b^2}{a^2} \right) \sin^2\theta \right) \pm \sqrt{\frac 14\left(2\cos ^2 \theta + \left( \frac{a^2}{b^2} + \frac{b^2}{a^2} \right) \sin^2\theta \right)^2 -1}, $$ which are clearly continuous in $\theta$.

Arctic Char
  • 16,007
  • Thank you! this is a perfect answer. I have also seen your nice analysis (which you have edited out) showing that these are the only matrices which preserve the ellipse. It follows that the maximal singular value cannot exceed $\frac{b}{a}$ when $b>a$; it actually obtains any value between $1$ and $\frac{b}{a}$, but not greater than $\frac{b}{a}$. This is interesting, and was not obvious to me a-priori. – Asaf Shachar Aug 03 '20 at 16:53