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Given a symmetric positive definite matrix $X\in\mathbb R^{4\times 4}$ and let $X_d$ be the diagonal matrix with the symplectic eigenvalues of $X$ on its diagonal, how can I find $S$ such that $X = S^T X_d S$ and $S^T J S = J$, where $J = \begin{bmatrix}0&I_2\\-I_2&0\end{bmatrix}$?

Edit: The diagonal entries of $X_d$ are $\lambda_j$, such that $\pm i \lambda_j$ [are the eigenvalues of $JX^{-1}]$, see this.

Zubin
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  • First, the article only refers to the definition of symplectic matrices, not to "symplectic eigenvalues". And second, such a decomposition is obviously only possible for symmetric matrices $X$. – amsmath Aug 01 '21 at 12:49
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Aug 01 '21 at 13:09
  • Your "Edit:..." is confusing (and I cannot access that link). It would be helpful if you could edit this into the question (rather than putting it at the end). Also, you should say where you came across the question - where you just reading the book you linked to? If so, say "I was reading the book X (link here), and couldn't understand Y.... . This reduces to the following: (state question).". Or something. – user1729 Aug 06 '21 at 11:19

1 Answers1

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You left out the information that $X$ should be real, symmetric, and positive definite.

Let $i\lambda_k$ be the eigenvalues in the upper half-plane of the skew-symmetric matrix $Z = X^{1/2}JX^{1/2}$ (I use $J$ instead of $\Omega$). The symplectic eigenvalues of $X$ are then just the $\lambda_k > 0$. Put them into a diagonal matrix $D_0$ and determine an orthonormal system of complex vectors $z_k = u_k + iv_k$ such that $Zz_k = i\lambda_k z_k$, $k=1,\ldots,n$. Then $$ i\lambda_k u_k - \lambda_k v_k = i\lambda_k (u_k + iv_k) = Z(u_k + iv_k) = Zu_k + iZv_k $$ so that \begin{equation}\label{e}\tag{$\star$} Zu_k = -\lambda_k v_k\quad\text{and}\quad Zv_k = \lambda_k u_k. \end{equation} Using the skew-symmetry of $Z$ and the fact that the $z_k$ form an orthonormal system, it is not hard to see that the matrix $$ U := \sqrt 2\cdot[u_1,\ldots,u_n,v_1,\ldots,v_n] $$ is orthogonal (see below). Now, set $$ S := D^{-1/2}U^TX^{1/2}, $$ where $D = \operatorname{diag}(D_0,D_0)$. Then $$ S^TDS = X^{1/2}UD^{-1/2}DD^{-1/2}U^TX^{1/2} = X, $$ as desired. It remains to show that $S$ is symplectic: \begin{align*} S^TJS = J &\Longleftrightarrow\;X^{1/2}UD^{-1/2}JD^{-1/2}U^TX^{1/2} = J\\ &\Longleftrightarrow\;UD^{-1/2}JD^{-1/2}U^T = X^{-1/2}JX^{-1/2}\\ &\Longleftrightarrow\;UD^{1/2}JD^{1/2}U^T = X^{1/2}JX^{1/2}\\ &\Longleftrightarrow\;UD^{1/2}JD^{1/2} = ZU\\ &\Longleftrightarrow\;U\begin{bmatrix}0&D_0\\-D_0&0\end{bmatrix} = ZU, \end{align*} which is exactly \eqref{e}.

Concerning the orthogonality of $U$:

Let $\lambda = \lambda_k$, $u = u_k$, and $v = v_k$ for some $k$. Then $$ 1 = \|u+iv\|^2 = \|u\|^2 + iv^Tu - iu^Tv + \|v\|^2 = \|u\|^2+\|v\|^2. $$ Moreover, $$ \lambda\|u\|^2 = (\lambda u)^Tu = (Zv)^Tu = -v^TZu = v^T(\lambda v) = \lambda\|v\|^2, $$ so that $\|u_k\|^2 = \|v_k\|^2 = \frac 12$ for all $k$. Moreover, $\lambda u^Tv = (Zv)^Tv = 0$ as $Z$ is skew-symmetric. Therefore, $u_k^Tv_k = 0$ for all $k$. For $j\neq k$ we get $$ 0 = \langle u_k+iv_k,u_j+iv_j\rangle = u_k^Tu_j + iv_k^Tu_j - iu_k^Tv_j + v_k^Tv_j, $$ which implies $$ u_k^Tu_j + v_k^Tv_j = 0\quad\text{and}\quad v_k^Tu_j = u_k^Tv_j. $$ This yields $$ \lambda_k u_k^Tu_j = (Zv_k)^Tu_j = -v_k^TZu_j = \lambda_j v_k^Tv_j = -\lambda_j u_k^Tu_j. $$ Hence, $(\lambda_k+\lambda_j)u_k^Tu_j = 0$. As $\lambda_k>0$ for all $k$, this implies $u_k^Tu_j = 0$. Similarly, $$ \lambda_k u_k^Tv_j = (Zv_k)^Tv_j = -v_k^TZv_j = -\lambda_j v_k^Tu_j = -\lambda_j u_k^Tv_j $$ so that $(\lambda_k+\lambda_j)u_k^Tv_j = 0$ and thus $u_k^Tv_j=0$.

amsmath
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  • @Zubin The question is: what is $|iJX|$? Please upvote my answer if it helped you. It is important that you add "positive definite" in your edit. – amsmath Aug 01 '21 at 16:58
  • @Zubin I see. Bad policy of MSE, in my opinion. Thanks for the info. – amsmath Aug 01 '21 at 17:03
  • In what I wrote, $n$ is arbitrary. $X$, $J$, and $D$ are $(2n)\times (2n)$, and $D_0$ is $n\times n$. In your special case we have $n=2$. If you want references, look for "Williamson's theorem". I've found a proof, which I did not quite understand, but also this one: https://math.stackexchange.com/questions/1171842/finding-the-symplectic-matrix-in-williamsons-theorem . Then I picked the stuff from both that I needed to conclude. This theorem was unknown to me before. Very interesting. – amsmath Aug 01 '21 at 22:23
  • @Zubin The motivation really came from the answer in https://math.stackexchange.com/questions/1171842/finding-the-symplectic-matrix-in-williamsons-theorem The motivation for the form of $U$ then came from the reference in the original question of this MSE discussion (Proof of Theorem 27). – amsmath Aug 04 '21 at 11:31