1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Why the algebra-precalculus tag? $\endgroup$ – user1551 Apr 6 '16 at 14:35
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.