Is Every State an Eigenstate?

For any given superposition of eigenstates of an observable A, is there always an observable B of which the given superposition is an eigenstate? For the two-state and three-state systems the answer is yes. We develop two methods for constructing the matrix representation of B. One works directly from the eigenvalue condition, but solves first for the matrix and then for its eigenvalues. The other uses on the diagonal matrix representing A the inverse of the unitary transformation which diagonalizes a matrix. For the two-level system we obtain two commuting matrices with different eigenvalue spectra. For the three-level system the inverse unitary transform yields a continuous infinity of non-commuting matrices, all with the same eigenvalue spectrum, whereas the other method yields a matrix which is none of these: it does not commute with any of them, and its eigenvalue spectrum is different from theirs.