Let \(n\) be a positive integer, \(I\) the \(n \times n\) identity matrix over \(C\), and \(J\) the \(2n \times 2n\) matrix given by \[ J = \begin{bmatrix} 0 & I \\ -I & 0 \end{bmatrix}. \] Let \(M\) be a \(2n \times 2n\) matrix over \(C\) of the form \[ M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \] where \(A\), \(B\), \(C\), \(D\) are \(n \times n\) matrices over \(C\). Find necessary and sufficient conditions on \(A\), \(B\), \(C\), \(D\) in order that \(M^tJM = J\).
Solution
First compute \[ JM =\begin{pmatrix}0&I\\-I&0\end{pmatrix} \begin{pmatrix}A&B\\C&D\end{pmatrix} =\begin{pmatrix}C&D\\-A&-B\end{pmatrix}. \] Since \[ M^t=\begin{pmatrix}A^t&C^t\\B^t&D^t\end{pmatrix}, \] we obtain \[ M^tJM = \begin{pmatrix} A^tC-C^tA & A^tD-C^tB\\ B^tC-D^tA & B^tD-D^tB \end{pmatrix}. \]
Therefore \[ M^tJM=J \] if and only if \[ A^tC-C^tA=0, \] \[ B^tD-D^tB=0, \] \[ A^tD-C^tB=I, \] and \[ B^tC-D^tA=-I. \]
The last equation is redundant: transposing \[ A^tD-C^tB=I \] gives \[ D^tA-B^tC=I, \] which is equivalent to \[ B^tC-D^tA=-I. \]
Hence the necessary and sufficient conditions can be written compactly as \[ \boxed{ A^tC=C^tA,\qquad B^tD=D^tB,\qquad A^tD-C^tB=I.} \] Equivalently, \(A^tC\) and \(B^tD\) are symmetric and the mixed block equation \(A^tD-C^tB=I\) holds.