Two operators' matrix representations are known in the |1〉, |2〉, |3〉 basis, where and b are real numbers:  ≐ B̂ ≐ ⎛ a 0 0 ⎞ ⎛ b 0 0 ⎞ ⎜ 0 -a 0 ⎟ ⎜ 0 0 -ιb ⎟ ⎝ 0 0 -a ⎠ and ⎝ 0 ιb 0 ⎠. B exhibits a degenerate spectrum when it has repeated eigenvalues. The eigenvalues of B are obtained from its characteristic equation. |⎛ b-λ 0 0 ⎞| |⎜ 0 -λ -ιb ⎟| = 0, i.e. |⎝ 0 ιb -λ ⎠| (b - λ)(λ² + ι²b²) = (b - λ)(λ² - b²) = (b - λ)(b - λ)(b + λ) = 0. (a) The eigenvalues for this operator are therefore λ = b,b,-b. Since b appears twice, the operator exhibits a degenerate spectrum. To find if A and B commute, their commutator need be evaluated. They commute if the value is 0. The commutator of two operators is defined as [Â,B̂] =  B̂ - B̂ Â. For the given operators, then, the commutator is ⎛ a 0 0 ⎞ ⎛ b 0 0 ⎞ ⎛ b 0 0 ⎞ ⎛ a 0 0 ⎞ ⎜ 0 -a 0 ⎟ ⎜ 0 0 -ιb ⎟ - ⎜ 0 0 -ιb ⎟ ⎜ 0 -a 0 ⎟ ⎝ 0 0 -a ⎠ ⎝ 0 ιb 0 ⎠ ⎝ 0 ιb 0 ⎠ ⎝ 0 0 -a ⎠, which reduces to ⎛ ab 0 0 ⎞ ⎛ ab 0 0 ⎞ ⎜ 0 0 ιab ⎟ - ⎜ 0 0 ιab ⎟ = 0. ⎝ 0 -ιab 0 ⎠ ⎝ 0 -ιab 0 ⎠ (b) Therefore, these operators commute. Since the operators commute, they share a set of common eigenstates. The eigenstates of  are apparent from inspection: |a₁〉 ≐ ⎛1⎞ |a₂〉 ≐ ⎛0⎞ and |a₃〉 ≐ ⎛0⎞ ⎜0⎟ ⎜1⎟ ⎜0⎟ ⎝0⎠, ⎝0⎠, ⎝1⎠. The eigenvalues are already known (λ = b,b,-b.), and using the eigenvalue equations, the eigenstates can be determined. The eigenvalue equation ⎛ b 0 0 ⎞ ⎛ α₁ ⎞ ⎛ b α₁ ⎞ ⎛ α₁ ⎞ ⎜ 0 0 -ιb ⎟ ⎜ β₁ ⎟ = ⎜ -ι b γ₁ ⎟ = b ⎜ β₁ ⎟ ⎝ 0 ιb 0 ⎠ ⎝ γ₁ ⎠ ⎝ ι b β₁ ⎠ ⎝ γ₁ ⎠ reveals -ι γ₁ = β₁, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂.