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 2 n × 2 n {\displaystyle 2n\times 2n} Hermitian real matrix M ∈ R 2 n × 2 n {\displaystyle M\in \mathbb {R} ^{2n\times 2n}}, the theorem ensures the existence of a real symplectic matrix S ∈ S p ( 2 n , R ) {\displaystyle S\in \mathbf {Sp} (2n,\mathbb {R} )}, and a diagonal positive real matrix D ∈ R n × n {\displaystyle D\in \mathbb {R} ^{n\times n}}, such that S M S T = I 2 ⊗ D ≡ D ⊕ D , {\displaystyle SMS^{T}=I_{2}\otimes D\equiv D\oplus D,}where I 2 {\displaystyle 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 ⊗ I n ) M − 1 / 2 {\displaystyle M^{-1/2}(J\otimes I_{n})M^{-1/2}}, with J ≡ ( 0 1 − 1 0 ) {\displaystyle J\equiv {\begin{pmatrix}0&1\\-1&0\end{pmatrix}}}, is well-defined and skew-symmetric.
  2. For any invertible skew-symmetric real matrix A ∈ R 2 n × 2 n {\displaystyle A\in \mathbb {R} ^{2n\times 2n}}, there is O ∈ O ( 2 n ) {\displaystyle O\in \mathbf {O} (2n)} such that O A O T = J ⊗ Λ {\displaystyle OAO^{T}=J\otimes \Lambda }, where Λ {\displaystyle \Lambda } a real positive-definite diagonal matrix containing the singular values of A {\displaystyle A}.
  3. For any orthogonal O ∈ O ( 2 n ) {\displaystyle O\in \mathbf {O} (2n)}, the matrix S = ( I 2 ⊗ D ) O M − 1 / 2 {\displaystyle S=\left(I_{2}\otimes {\sqrt {D}}\right)OM^{-1/2}} is such that S M S T = I 2 ⊗ D {\displaystyle SMS^{T}=I_{2}\otimes D}.
  4. If O ∈ O ( 2 n ) {\displaystyle O\in \mathbf {O} (2n)} diagonalizes M − 1 / 2 ( J ⊗ I n ) M − 1 / 2 {\displaystyle M^{-1/2}(J\otimes I_{n})M^{-1/2}}, meaning it satisfies O M − 1 / 2 ( J ⊗ I n ) M − 1 / 2 O T = J ⊗ Λ , {\displaystyle OM^{-1/2}(J\otimes I_{n})M^{-1/2}O^{T}=J\otimes \Lambda ,}then S = ( I 2 ⊗ D ) O M − 1 / 2 {\displaystyle S=\left(I_{2}\otimes {\sqrt {D}}\right)OM^{-1/2}} is such that S ( J ⊗ I n ) S T = J ⊗ ( D Λ ) . {\displaystyle S(J\otimes I_{n})S^{T}=J\otimes (D\Lambda ).}Therefore, taking D = Λ − 1 {\displaystyle D=\Lambda ^{-1}}, the matrix S {\displaystyle S} is also a symplectic matrix, satisfying S ( J ⊗ I n ) S T = J ⊗ I n {\displaystyle S(J\otimes I_{n})S^{T}=J\otimes I_{n}}.