# Polar Decomposition for Symplectic Matrices

If $P$ is a $2n \times 2n$ real matrix satisfying $P^t J P = J$ where $J = \left(\begin{array}10 & I \\ -I & 0\end{array}\right)$ and if $P = Q S$ is the polar decomposition of $P$ then show that $Q$ and $S$ also satisfy

$$Q^t J Q = J \quad\text{and}\quad S^t J S = J.$$

Thanks for any help.

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By ( 0 I -I 0 ) I meant the block matrix with diagonal entries nxn null matrix and the upper right entry the nXn identity matix and the lower left entry the additive inverse of the nxn identity matrix – Ester Oct 20 '12 at 15:59
Sorry everyone for the edits – Ester Oct 20 '12 at 16:02

By definition $P$ is a real symplectic matrix and your question is called the polar decomposition for symplectic matrices.
Let $Q=\sqrt{PP^t}$. Then the matrix $Q$ is symmetric and positive definite. Now take $S=Q^{-1}P$ and check that $SS^t=I$, that is, $S$ is orthogonal. Moreover, this decomposition is unique: if $P=Q_1S_1=Q_2S_2$, then $Q_1^2=Q_2S_2S_1^tQ_1=Q_2^2$ and thus, by the uniqueness of the square root for positive symmetric matrices, we get $Q_1=Q_2$ and then $S_1=S_2$. (Nothing new so far: this is the polar decomposition of any invertible matrix.)
From $P^tJP=J$ we get $P=J^{-1}(P^t)^{-1}J$. By using that $P=QS$ we obtain $$P=J^{-1}(S^tQ)^{-1}J=(J^{-1}Q^{-1}J)(J^{-1}(S^t)^{-1}J).$$ The matrix $J^{-1}Q^{-1}J$ is symmetric and positive definite, while $J^{-1}(S^t)^{-1}J$ is orthogonal. By the uniqueness of the polar decompositon we obtain that $Q=J^{-1}Q^{-1}J$ and $S=J^{-1}(S^t)^{-1}J$, therefore $Q$ and $S$ are also symplectic.