Williamson theorem

In the context of linear algebra and symplectic geometry, the Williamson theorem concerns the diagonalization of positive definite matrices through symplectic matrices.

More precisely, given a strictly positive-definite $2n\times 2n$ Hermitian real matrix $M\in \mathbb {R} ^{2n\times 2n}$, the theorem ensures the existence of a real symplectic matrix $S\in \mathbf {Sp} (2n,\mathbb {R} )$, and a diagonal positive real matrix $D\in \mathbb {R} ^{n\times n}$, such that $$SMS^{T}=I_{2}\otimes D\equiv D\oplus D,$$where $I_{2}$ denotes the 2x2 identity matrix.

Proof

The derivation of the result hinges on a few basic observations:

1. The real matrix $M^{-1/2}(J\otimes I_{n})M^{-1/2}$, with $J\equiv {\begin{pmatrix}0&1\\-1&0\end{pmatrix}}$, is well-defined and skew-symmetric.
2. Any skew-symmetric real matrix $A\in \mathbb {R} ^{2n\times 2n}$ can be block-diagonalized via orthogonal real matrices, meaning there is $O\in \mathbf {O} (2n)$ such that $OAO^{T}=I_{2}\otimes \Lambda$ with $\Lambda$ a real positive-definite diagonal matrix containing the singular values of $A$.
3. For any orthogonal $O\in \mathbf {O} (2n)$, the matrix $S=(I\otimes {\sqrt {D}})OM^{-1/2}$ is such that $SMS^{T}=I_{2}\otimes D$.
4. If $O\in \mathbf {O} (2n)$ diagonalizes $M^{-1/2}(J\otimes I_{n})M^{-1/2}$, meaning it satisfies $$OM^{-1/2}(J\otimes I_{n})M^{-1/2}O^{T}=I_{2}\otimes \Lambda ,$$then $S=(I\otimes {\sqrt {D}})OM^{-1/2}$ is such that $$S(J\otimes I_{n})S^{T}=J\otimes (D\Lambda ).$$Therefore, taking $D=\Lambda ^{-1}$, the matrix $S$ is also a symplectic matrix, satisfying $S(J\otimes I_{n})S^{T}=J\otimes I_{n}$.