Symmetric matrices with respect to sesquilinear forms
Let \(V\) be a finite-dimensional vector space over the complex numbers \({\mathbb C}\) and let \([x,y]\) be a sesquilinear form on \(V.\) A linear transformation \(A\) will be called symmetric with respect to \([x,y]\) if \([Ax,y]=[x,Ay]\) for all \(x,y\in V.\) Let \(\langle\cdot,\cdot\rangle\) be the standard Euclidean inner product. Then \([x,y]=\langle Hx,y\rangle\) for some uniquely determined Hermitian matrix \(H\) and \(A\) is symmetric with respect to \(\langle Hx,y\rangle\) if \(HA=A^\star H.\) In this case, \(A\) is called \(H\)-selfadjoint. Canonical forms of \(H\)-selfadjoint matrices are well known for the case that \(H\) is invertible. The authors study the case where \(H\) is possibly singular. They show, if \(A\in{\mathbb C}^{n\times n}\) is an \(H\)-selfadjoint matrix, then there exists a nonsingular matrix \(P\in{\mathbb C}^{n\times n}\) such that \[ P^{-1}AP=A_{11}\oplus\cdots\oplus A_{kk} \text{and} P^\star HP=H_{11}\oplus\cdots\oplus H_{kk}, \] where for each \(j,\) the blocks \(A_{jj}\) and \(H_{jj}\) have the same sizes and are of one of three types described in the paper. The authors give an example to show that an indecomposable symmetric matrix with respect to a degenerate sesquilinear form may have arbitrarily many Jordan blocks.
scientific article; zbMATH DE number 1790242