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I have reasons to believe that for any four square matrices $A, B, C, D \in \mathcal{M}_{n}(\mathbb{R})$, and $\lambda \neq 0$, the following is true: \begin{equation} \left.\ \begin{aligned} A^TD-C^TB &=\lambda I_n \\ A^TC &=C^TA\\ B^TD&=D^TB \end{aligned} \right\} \quad \iff \quad \left\{ \begin{aligned} AD^T-BC^T &=\lambda I_n \\ AB^T &=BA^T\\ CD^T&=DC^T \end{aligned} \right.\ \end{equation}

I have tried to tackle a proof or find a counterexample, but I haven't been able to do either. I would appreciate any indications or help. Thank you in advance :)

For context, both conditions have appeared when I tried to study whether the following linear transformation: \begin{equation} \left\{ \begin{aligned} Q&=Aq+Bp\\ P&=Cq+Dp \end{aligned} \right.\ \end{equation} where $p,q,Q,P \in \mathbb{R}^n$ are supposed to represent old and new coordinates and momenta, respectively, is canonical in the sense of classical mechanics. Specifically, I obtained the first set of conditions after forcing $\sum_{k}(\lambda p_k dq_k-P_k dQ_k)$ to be an exact differential, while the second one appears if one sets $[Q_i,Q_j]=[P_i,P_j]=0, [Q_i,P_j]=\lambda\delta_{ij}$ for the Poisson brackets of the new coordinates. I have studied that both these conditions characterize a canonical transformation, hence the question.

Josemi
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    The two sets of equations simply mean $\pmatrix{A^T&C^T\ B^T&D^T}\pmatrix{D&-C\ -B&A}=\lambda I_{2n}$ and $\pmatrix{D&-C\ -B&A}\pmatrix{A^T&C^T\ B^T&D^T}=\lambda I_{2n}$ respectively. They are equivalent because for any pair of linear operators $X$ and $Y$ on a finite-dimensional vector space, $XY=I$ if and only if $YX=I$. – user1551 Nov 18 '23 at 19:15
  • @user1551 Cool! So either condition can be expressed as $\begin{pmatrix}A & B \C & D\end{pmatrix}^{-1}=\lambda^{-1}\begin{pmatrix}D & -C \-B & A\end{pmatrix}^T$, which is really the inverse rule for a $2 \times 2$ matrix, much more beautiful than what I had. Thank you so much :)) – Josemi Nov 18 '23 at 19:34

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