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Hi. What type of 2n dimensional real symmetric matrices can be diagonalized with symplectic transformations (meaning M->SMS^T, S^T means transpose and S is an element of the 2n dimensional real symplectic group. Usually normal forms of the literature are given as representatives of orthogonal group orbits, but I need to know the symplectic version. Thanks for any help, recommendation of literature etc. Zoltan

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I've retagged the question. –  Marco Golla Sep 29 '11 at 14:26

2 Answers 2

up vote 0 down vote accepted

A $2n\times 2n$ dimensional Hermitian matrix that can be diagonalized by a symplectic transformation can be viewed as an $n\times n$ matrix with elements consisting of $2\times 2$ blocks of the quaternion real form

${\bar{z}\;-\bar{w}}\choose{w\;\; z}$

so if you choose real $z$ and $w$ you have constructed a real symmetric matrix $M$ that can be diagonalized by a symplectic $S$.


@Federico: this is the general form for matrices that commute, $MT=TM$, with

$T=1_{N}\otimes$ ${0\; 1}\choose{-1\; 0}$ $K$

($K$ is the operator of complex conjugation); alternatively, one can take matrices that anticommute, $MT=-TM$; then the $2\times 2$ blocks have the form

${\bar{z}\;\bar{w}}\choose{w\;\; -z}$

and again, for a real $M$ one would choose real $w,z$. these two choices exhaust the possibilities.

In applications to physical systems, the matrix $M$ is a Hamiltonian and $T$ is the operator of time reversal. Then only commuting matrices, $MT=TM$, are permitted.

For a discussion in the physics context, see Section 1.4.2 of Forrester's book, online here:

http://www.ms.unimelb.edu.au/~matpjf/b1.ps

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Are the ones you construct all of them? –  Federico Poloni Sep 30 '11 at 7:01
    
half of them, I added the other half. –  Carlo Beenakker Sep 30 '11 at 10:20

I don't have a prompt answer (though I would guess "all those that are normal with respect to the scalar product induced by $J$"), but I suggest you to take a look at Indefinite linear algebra and applications, Gohberg and Lancaster.

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