From eigenvalues to non-invertibility
Fix a bounded linear operator T on a complex Banach space X. A scalar λ is an eigenvalue if Tx = λx for some nonzero x — equivalently, T − λI has nontrivial kernel. In finite dimensions that is the only way T − λI can fail to be invertible, so the eigenvalues are exactly the bad λ. The honest infinite-dimensional definition therefore drops eigenvectors and keeps the real culprit: λ is in the spectrum σ(T) precisely when T − λI is not invertible as a bounded operator. Its complement is the resolvent set ρ(T).
Three ways to fail
Invertibility of S = T − λI can break for three reasons, and they carve σ(T) into pieces. If S is not injective, λ lies in the point spectrum — these are the genuine eigenvalues. If S is injective with dense but not closed range, no bounded inverse exists; λ sits in the continuous spectrum. If S is injective but its range is not even dense, λ is in the residual spectrum. The shift operator below shows that the continuous and residual cases are not exotic curiosities.
Right shift S on l^2: S(x_1, x_2, x_3, ...) = (0, x_1, x_2, ...)
Claim: 0 is in sigma(S) but is NOT an eigenvalue.
Injective? Sx = 0 forces (0, x_1, x_2, ...) = 0, so every x_k = 0.
Hence ker S = {0}: 0 is NOT in the point spectrum.
Surjective? Range of S = { y in l^2 : y_1 = 0 }.
The vector e_1 = (1,0,0,...) is not in the range.
So S is not onto, hence S - 0*I = S is not invertible.
Conclusion: 0 in sigma(S), but 0 is not an eigenvalue.
(Range is closed but proper => 0 lies in the residual spectrum.)
Contrast: ||S x|| = ||x||, so S is an isometry, ||S|| = 1,
and one shows sigma(S) = closed unit disk { |lambda| <= 1 }.The spectrum is compact and nonempty
Two structural facts hold for every bounded T on a complex Banach space. First, σ(T) is closed and bounded — indeed contained in the disk |λ| ≤ ‖T‖, because for |λ| > ‖T‖ the Neumann series Σ λ⁻ⁿ⁻¹ Tⁿ converges in operator norm and inverts T − λI. Second, σ(T) is never empty: if it were, the resolvent λ ↦ (T − λI)⁻¹ would be a bounded entire function vanishing at infinity, hence zero by Liouville's theorem — absurd.