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maddogtheman
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If you had an operator A-hat whose eigenvectors form a complete basis for the Hilbert space has only real eigenvalue how would you prove that is was Hermitian?
Hurkyl said:You wouldn't. That's not even true when restricted to 2x2 matrices whose entries are all real!
(And please don't post the same thing multiple times. We don't tolerate it on this forum)
A diagonal matrix with real eigenvalues is Hermitian. But not necessarily if the matrix is merely diagonalizable with real eigenvalues. Why should [itex](PDP^{-1})^* = PDP^{-1}[/itex]?jostpuur said:If 2x2 matrix is diagonalizable with real eigenvalues, isn't it quite Hermitian then?
jostpuur said:The problem is that [tex]T\mapsto T^{\dagger}[/tex] does not commute with all linear coordinate transformations, so actually we never should speak about a linear mapping being Hermitian. A simple fact, which I had never thought about before. Instead, we should speak about linear mapping being Hermitian with respect to some basis (or with respect to a certain kind of collection of different basis).
A matrix is Hermitian if it is equal to its own conjugate transpose, meaning that the matrix is symmetric and its diagonal elements are real numbers.
To determine if a matrix is Hermitian, you can check if it is equal to its own conjugate transpose. This can be done by taking the transpose of the matrix and then taking the complex conjugate of each element. If the resulting matrix is equal to the original matrix, then it is Hermitian.
Real eigenvalues are numbers that satisfy the characteristic equation of a matrix. They are also the roots of the characteristic polynomial and correspond to the eigenvalues of the matrix.
In quantum mechanics, Hermitian matrices represent observable physical quantities. Real eigenvalues of a Hermitian matrix correspond to real measurements, which are essential for accurate and meaningful predictions in quantum mechanics.
Yes, a matrix can have real eigenvalues but not be Hermitian. For a matrix to be Hermitian, it must not only have real eigenvalues but also be equal to its own conjugate transpose. A non-Hermitian matrix may have real eigenvalues but will not satisfy the condition of being equal to its own conjugate transpose.