If $A\geq S^TS$ and $A\geq \lambda$ for symplectic $S$, is $A\geq R^TR\geq \lambda$ for some symplectic $R$?

142 Views Asked by At

Consider a $2n$-by-$2n$ real matrix $A$ such that there exists symplectic matrix $S \in Sp(2n,\mathbb{R})$ with $A \geq S^TS$ and $A \geq \lambda \mathbb{1}$ for some $\lambda \in [0,1]$.

Does there exist symplectic matrix $R \in Sp(2n,\mathbb{R})$ such that $A \geq R^T R \geq \lambda \mathbb{1}$?

Note that this is equivalent to asking: For PSD $A$ with symplectic eigenvalues greater than $1$ and eigenvalues greater than $\lambda$, does there exist PSD $B$ with unit symplectic spectrum and $A \geq B \geq \lambda \mathbb{1}$?

For more on symplectic eigenvalues, see e.g. On Symplectic eigenvalues... by Bhatia and Jain.