Singular values of matrices which preserve the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$

Question

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.

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

Answer 1

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 $\theta$.