Square root of Symplectic and Positive Definite Matrices in $M_{2n\times 2n}(\mathbb{R})$

Question

Let $R$ be a symplectic matrix. Then the adjoint $R^*$ of $R$ is also a symplectic matrix. Then $RR^*$ is symplectic and positive definite. I want to know that $(RR^*)^{1/2}$ is also symplectic.

Answer 1

Write $R=AU$, $A$ is symmetric positive definite $U$ is orthogonal.

$R^*$ is symplectic, so $J=R^*JR=U^*(AJA)U$, ie $AJA=UJU^*$, thus, with $J’=UJU^*$ orthogonal, we have $V:= AJ=J’A^{-1}$.

The conclusion then stems from the following lemma: if $X=AU=WB$ is an invertible matrix, $U,W$ are orthogonal, $A,B$ are symmetric positive semidefinite, then $U=W$. Indeed, if the lemma holds, then $J=J’$ and $U$ is symplectic, therefore so is $A=RU^{-1}$.

The proof runs as follows: we then have $X=U(U^*AU)=WB$, where $U,W$ are orthogonal, $A’=U^*AU$ and $B$ are positive semi-definite, so $BA’^{-1}$ is orthogonal, ie $A’^{-1}B^2A’^{-1}=I$, so $B^2=A’^2$, so (symmetric positive definite) $B=A’$ so $U=W$.

Answer 2

I am going to base my answer on the answer to this question :

Finding Euler decomposition of a symplectic matrix

As the answer was aimed at a different question and slightly incomplete, my answer will differ from it even if a large part is common.

Some notations :

$$S:=RR^T, U:=S^{1/2}, V:=U^{-1}R \tag{1}.$$

An essential fact is that V is orthogonal.

Indeed, $VV^T=U^{-1}(RR^T)(U^T)^{-1}=U^{-1}(U^2)(U^{-1})=I$

Therefore

$$R=UV \tag{2},$$

is a version of the polar decomposition ($R$ is equal to the product of a psd matrix and an orthogonal matrix).

Besides, in the question mentionned above, it is recalled a general structure for symplectic matrices :

$$R=\Omega\Delta \Omega'\tag{3}$$

for a certain diagonal matrix $\varDelta=(\lambda_1,\cdots\lambda_n,1/\lambda_1,\cdots 1/\lambda_n)$ and certain orthogonal matrices $\Omega, \Omega'$.

Rewriting (3) under the form :

$$R=(\Omega\varDelta \Omega^T)(\Omega\Omega')\tag{4}$$

Relationship (4) appears as another form of the polar decomposition already found in (2). By unicity of polar decomposition, we can conclude in particular that :

$$U=\Omega\varDelta \Omega^T\tag{5}$$

Therefore, $U:=S^{1/2}$ has the characteristic structure of a symplectic matrix.

Please note the difference between (3) and (5) : the second orthogonal matrix is $\Omega'$ in a case and $\Omega$ in the other.