Matrix diagonalisation in sesquilinear symplectic spaces (Q6655723)

Consider the symplectic sesquilinear form \N\[\N\left[x,y\right]=\langle x,J_{2n}y\rangle=x^*J_{2n}y, \N\]\Nwhere \N\[\NJ_{2n}=\begin{bmatrix} 0 & I_n \\\N-I_n & 0 \\\N\end{bmatrix}. \N\]\NDenote by \(A^\sharp\) the matrix \(J_{2n}^TA^*J_{2n}\). A matrix is called \(J\)-normal, if \(AA^\sharp=A^\sharp A\). A matrix is Hamiltonian (skew-Hamiltonian) if and only if \(A^\sharp=-A\) (\(A^\sharp=A\)).\N\NThe paper focuses on the problem of diagonalizability of matrices under symplectic similarity. It is known that not all Hamiltonian, skew-Hamiltonian and \(J\)-normal matrices are diagonalizable. Moreover, even if they are diagonalisable, they do not have to be diagonalisable by symplectic similarity. In order to examine this issue the authors introduce the following notion. A \(2n\times 2n\) matrix is called \textit{\(s\)-diagonal}, if it is of the form \N\[\N\begin{bmatrix} D_1 & D_2 \\\N-D_2 & D_3 \\\N\end{bmatrix},\N\]\Nwhere each \(D_j\) (\(1\leqslant j\leqslant 3\)) is a diagonal matrix \(\operatorname{diag}(d_j^{(1)},d_j^{(2)},\ldots,d_j^{(n)})\) such that \(d_1^{(i)}=d_3^{(i)}\) on condition that \(d_2^{(i)}\neq 0\).\N\NThe main result states that a \(2n\times 2n\) complex matrix is \(J\)-normal and diagonalisable if and only if it is symplectically similar to an \(s\)-diagonal matrix. The authors provide also necessary and sufficient conditions for Hamiltonian (skew-Hamiltonian) and diagonalisable matrices, slightly more complicated than the preceding one. Additionally, they prove that a \(2n\times 2n\) matrix \(A\) is normal and \(J\)-normal if and only if there exists a unisymplectic (i.e. unitary and symplectic) matrix \(S\) such that \(S^{-1}AS\) is an \(s\)-diagonal matrix.