For another example, consider the Hermitian matrix: [ 0 − − 0.5 − − 0.5 0.5 − − 0.5 − − 2 0 0 − − 0.5 0 0 0 0.5 0 0 2 ] Although there are already some suggestive relationships between the matrix elements, it is not clear what the symmetry of this matrix is at first glance. However, it is easy to demonstrate that by a unitary transformation, this matrix is equivalent to: [ 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 ] Which can be verified directly by numerically (or analytically - see Chebyshev polynomials) diagonalising the sub-matrix formed by removing the first row and column. Rotating the basis defining this sub matrix using the resulting unitary brings the original matrix into the originally stated form. This matrix has a P 4 permutation symmetry, which in this basis is much easier to see, and could constitute a 'hidden' symmetry. In this case, there are no degeneracies in the eigenspectrum. The technical reason for this is that each eigenstate transforms with respect to a different irreducible representation of P 4. If one encountered a case where some group of eigenstates correspond to the same irreducible representation of the 'hidden' symmetry group, a degeneracy would be observed.