The unilateral shift operator on the Hilbert space l² ( with its standard orthonormal basis (e_n)) given by Te_n = e_n+1, n Î N, has no eigenvalue; since obviously 0 Ï eig(T) and if 0 ¹ l Î eig(T) and Tx = lx for some x = å_{n = 1}^¥a_n e_n ¹ 0, then å_{n = 1}^¥a_n e_n+1 = å_{n = 1}^¥la_n e_n and hence a_n = 0 for all n, i.e. x = 0, a contradiction.
Next observe that 0 Î sp(T); otherwise T would be invertible so T(T^-1(e₁)) = e₁ , but < T(T^-1e₁) , e₁ > = 0 by the definition of T, that is impossible.