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) $ and suppose that $AD_{a,b}=D_{a,b}$. Let $\sigma_2$ be the maximal singular value of $A$.
Question: Is there a way to prove "directly" that $\sigma_2 \le \frac{b}{a}$?
One approach is to characterize all matrices $A$ which preserve $D_{a,b}$. After doing that, one finds that $A$ must be of the form $$ A =A_{\theta}:= \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pmatrix}= \begin{pmatrix} \cos\theta & -\frac ab \sin\theta \\ \frac ba \sin\theta & \cos \theta \end{pmatrix}. $$
From here it is not hard to see that $$X=\{ \sigma_2 \, | \, \sigma_2 \, \, \text{ is the maximal singular value of a matrix } \, A \, \, \text{ satisfying } \, AD_{a,b}=D_{a,b}\}=[1,\frac{b}{a}],$$ which proves the claim.
I wonder if there is a way to show $\sigma_2 \le \frac{b}{a}$ without going through the characterisation of all matrices $A(\theta)$. Here is a start of a direct attempt: Write $A=U\Sigma V^T$ (SVD). Then
$$U\Sigma V^TD_{a,b}=D_{a,b} \iff \Sigma V^TD_{a,b}=U^TD_{a,b}...$$
I don't see how to proceed.
