 Solution

Hermiticity supplies the reverse matrix elements, also equal to the real number $t$. In an orthonormal basis for the degenerate eigenspace, the perturbation iswhere ${\mathbf{J}}_3$ is the all-ones matrix. Its symmetric vector $(1,1,1)/\sqrt{3}$ has eigenvalue $2t$. Every vector whose components sum to zero has eigenvalue $-t$; that subspace has dimension two. For example, $(1,-1,0)/\sqrt{2}$ and $(1,1,-2)/\sqrt{6}$ form an orthonormal basis for it.

The first-order energies are thereforeThe trace of the shifts is $2t-t-t=0$, agreeing with the zero diagonal. The ordering depends on the sign of $t$; the formula does not assume that the symmetric state is the lowest. Diagonalizing the perturbation inside the degenerate space is essential: the original diagonal entries alone would incorrectly predict no splitting.