[math-fun] Polynomials with real roots
Stupid matrix question: 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? If so, is there a construction that takes one from the polynomial (in the form of a vector of coefficients) to the Hermitian matrix? (The construction from the roots themselves is trivial: they form a diagonal matrix that can be rotated by any unitary matrix.) Presumably, such a construction would fail if/when not all of the roots are real.
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
Perhaps there's some condition you're not telling me. Since every real diagonal matrix is Hermitian the answer is obviously yes (just put the roots on the diagonal). There is a related arithmetic question -- Suppose that a monic polynomial with integer coefficients has all real roots, is it the characteristic polynomial of a symmetric integer matrix? The answer is no -- there's a series of papers about this by Ed Bender and Norman Hertzberg around 1970. Victor On Sat, Nov 14, 2009 at 10:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Stupid matrix question:
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?
If so, is there a construction that takes one from the polynomial (in the form of a vector of coefficients) to the Hermitian matrix?
(The construction from the roots themselves is trivial: they form a diagonal matrix that can be rotated by any unitary matrix.)
Presumably, such a construction would fail if/when not all of the roots are real.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
-
Henry Baker -
mcintosh＠servidor.unam.mx -
victor miller