Simple forms for perplectic and symplectic normal matrices (Q2020689)

For each integer \(n\geq 1\) let us define the \(n\times n\) matrix \(R_{n}:=[r_{ij}]\), where \(r_{ij}=1\) if \(i+j=n+1\) and \(r_{ij}=0\) otherwise, and the \(2n\times 2n\) matrix \[J_{2n}:=\left[ \begin{array}{cc} 0 & I_{n} \\ -I_{n} & 0 \end{array} \right], \] where \(I_{n}\) is the identity matrix. If \(B=R_{n}\) or \(J_{2n}\) then a complex matrix \(A\) is defined to be \(B\)-normal if \(AA^{\ast }=A^{\ast }A\) where the adjoint \(A^{\ast }:=B^{-1}A^{H}B\) (here \(A^{H}\) is the Hermitian conjugate). Moreover, \(Q\) is \(B\)-unitary if \(Q\) is \(B\)-normal and \( Q^{H}BQ=B\); \(R_{n}\)-unitary matrices are also called perplectic. In general classifying \(B\)-normal matrices is a difficult problem (see [\textit{C. Mehl}, Electron. J. Linear Algebra 15, 50--83 (2006; Zbl 1095.15010)]). However, the authors show that such a classification is possible for these specific values of \(B\) under the extra condition that \(A\) is diagonalizable. Theorem. If \(A\) is an \(R_{2m}\)-normal matrix and \(A\) is diagonalizable, then there exists a perplectic matrix \(P\) such that \[ P^{-1}AP=\left[ \begin{array}{cc} D_{11} & R_{m}D_{12} \\ R_{m}D_{21} & R_{22} \end{array} \right], \] where each block \(D_{ij}\) is an \(m\times m\) diagonal matrix. This form is not uniquely determined by \(A\). There is a similar theorem for diagonalizable \(J_{2m}\)-normal matrices. The authors point out that for each of these classes of \(B\)-normal matrices the diagonalizable matrices form a dense subset.