Let $A$ a symmetric positive definite real matrix of dimension $2n\times 2n$ and $J$ the standard symplectic matrix, with block representation \begin{gather} J= \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \end{gather} with $I$ diagonal matrix of order $n \times n$. Consider the product $U=JA$, then there exists a basis that diagonalizes $U$ and have only imaginary eigenvalues.

How can I prove that?

  • $\begingroup$ Why the algebra-precalculus tag? $\endgroup$ – user1551 Apr 6 '16 at 14:35

It's simply because $JA$ is similar to $A^{1/2}JA^{1/2}$, which is skew-Hermitian.

  • $\begingroup$ how can you prove the similarity? $\endgroup$ – Lorban Apr 6 '16 at 15:28
  • $\begingroup$ @Lorban Do you know that every skew-Hermitian matrix (or more generally, every normal matrix) is unitarily diagonalisable? This is taught in most introductory linear algebra courses. $\endgroup$ – user1551 Apr 6 '16 at 15:55

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