Does the eigenvectors of Hermitian operator constitute a basis Eigenvectors of a Hermitian operator corresponding to different eigenvalues are orthogonal. Even for a degenerate eigenvalue we can produce orthogonal eigenvectors in that eigensubspace.
Does this system of orthogonal vectors necessarily span the whole vector space, i.e., do they constitute a basis?
 A: Yes. Not only the eigenvectors of a Hermitian operator constitute a basis, but it is a complete basis, i.e., and function in the space where the operator acts, can be expanded in terms of this operator eigenfunctions. The latter fact is sometimes stated differently, as the resolution of identity, see here.
Note that the above is true even in case of degenerate spectrum, provided that one appropriately orthogonalizes the eigenfunctions corresponding to the degenerate eigenvalues (which can always be done).
A: I think yes. Say for $N$ dimensional vector space and operator $\hat{A}$ we have $N-2$ distinct eigenvalues and one twice repeating eigenvalue $\lambda$. The $N-2$ eigenvectors corresponding to distinct eigenvalues will be orthogonal and hence, linearly independent. For the degenerate case of degree $2$, we can obtain two orthogonal vectors, each of these will also be orthogonal to the each of $N-2$ eigenvectors . Therefore we have a total of $(N-2)+2=N$ linearly independent vectors in a $N$ dimensional vector space, i.e., a basis is formed.
