“…Let ϵ = 1 andA = A 1 ⊕ A 2 ⊕ A 3 , where A 1 = By setting E = E 1 ⊕E 2 ⊕E 3, where E 1 = ∥E∥≤ 1 . Since A 2 and (A + E) 2 are Hermitian, by[1, Theorem 4.2], we obtain thatV 4 (A) = σ(A) = {1, 0, i, √ 3, − √ 3} and V 4 (A+E) = σ(A+E) = {1, − 1 ∈ V 4 (A) + D(0, 1). Thus, V 4 (A) + D(0, 1) ̸ = V 4 1 (A).…”