Williamson theorem

Template:Short description Template:More footnotes needed 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 $M\in {ℝ}^{2n\times 2n}$, the theorem ensures the existence of a real symplectic matrix $S\in 𝐒𝐩(2n,ℝ)$, and a diagonal positive real matrix $D\in {ℝ}^{n\times n}$, such that $$SM{S}^T=I_2\otimes D\equiv D\oplus D,$$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 $J\equiv \left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, is well-defined and skew-symmetric.
2. Any skew-symmetric real matrix $A\in {ℝ}^{2n\times 2n}$ can be block-diagonalized via orthogonal real matrices, meaning there is $O\in 𝐎(2n)$ such that $OA{O}^T=I_2\otimes \mathrm{\Lambda }$ with $\mathrm{\Lambda }$ a real positive-definite diagonal matrix containing the singular values of $A$.
3. For any orthogonal $O\in 𝐎(2n)$, the matrix $S=(I\otimes \sqrt{D})O{M}^{-1/2}$ is such that $SM{S}^T=I_2\otimes D$.
4. If $O\in 𝐎(2n)$ diagonalizes $M^{-1/2}(J\otimes I_n)M^{-1/2}$, meaning it satisfies $$O{M}^{-1/2}(J\otimes I_n)M^{-1/2}{O}^T=I_2\otimes \mathrm{\Lambda },$$then $S=(I\otimes \sqrt{D})O{M}^{-1/2}$ is such that $$S(J\otimes I_n)S^T=J\otimes (D\mathrm{\Lambda }).$$Therefore, taking $D={\mathrm{\Lambda }}^{-1}$, the matrix $S$ is also a symplectic matrix, satisfying $S(J\otimes I_n)S^T=J\otimes I_n$.

See also

• Symplectic geometry
• Symplectic matrices
• Definite matrix