Quoting Henry Baker <hbaker1@pipeline.com>:
I know that the characteristic polynomial of a Hermitian matrix must have all real roots (it is diagonalizable into a real diagonal matrix).
But is every polynomial with only real roots the characteristic polynomial of a Hermitian matrix?
Not necessarily. Use the spectral decomposition formula: matrix = sum(eigenvalue x (column eigenvector) x (row e.v.) ) The matrix is real when this is equal to its hermitean congugate: real eigenvalus, row = (column conjugate). If the eigenvectors do not make a biorthogonal system, even though the eigenvalues are real, the matrix is not hermitean. You might still be able to make something with degenerate eigenvalues; after all, the unit matrix is hermitean no matter what the eigenvectors. -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos