It has eigenvalues -2 and 4. If the characteristic equation does not have N distinct roots, we may be able to find more than one linearly independent eigenvector associated with a multiple root and so still find N linearly independent eigenvalues. But we may also find that only one eigenvector is associated with a multiple root and that the operator does not have a basis of eigenvectors. The eigenvalues of a Hermitian operator are real. Take the adjoint.
Every Hermitian operator has at least one basis of orthonormal eigenvectors.
The matrix of the operator is diagonal in this basis and has its eigenvalues as its diagonal entries. The characteristic equation has at least one root, call it w 1 . If the eigenvalues are degenerate, then there exist many bases of eigenvectors that diagonalize W . Summary We have proofed that in a finite dimensional vector space a basis of eigenvectors of a Hermitian operator can be found. Consider an operator whose matrix in that basis is .