Measure, Integration & Real Analysis · Problem 10A.12

Linear Maps on Hilbert Spaces — Sheldon Axler

Mathematics — Measure Theory · left-shift-spectrum-invertibility

Problem

Define T:ℓ2→ℓ2T: \ell^2 \to \ell^2 by T(a1,a2,a3,…)=(a2,a3,a4,…)T(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots). Suppose α∈F\alpha \in \mathbf{F}. (a) Prove that T−αIT - \alpha I is injective if and only if ∣α∣≥1|\alpha| \geq 1. (b) Prove that T−αIT - \alpha I is invertible if and only if ∣α∣>1|\alpha| > 1. (c) Prove that T−αIT - \alpha I is surjective if and only if ∣α∣≠1|\alpha| \neq 1. (d) Prove that T−αIT - \alpha I is left invertible if and only if ∣α∣>1|\alpha| > 1.
Intuitive explanation

The shift operator TT has a special structure: it "forgets" the first coordinate, making it non-injective when we can hide information in that first slot. The key insight is that T−αIT - \alpha I behaves very differently depending on whether ∣α∣|\alpha| is less than, equal to, or greater than 1—this controls whether we can scale and compensate for the shift's destructive action. Injectivity asks if we can recover a sequence from its shifted version; surjectivity asks if every sequence can be written as (T−αI)(T - \alpha I) applied to something; invertibility requires both. The threshold ∣α∣=1|\alpha| = 1 is critical because it determines whether the "decay" or "growth" from the −αI-\alpha I term can balance the information loss from the shift.

Intuitive explanation

The shift operator TT has a special structure: it "forgets" the first coordinate, making it non-injective when we can hide information in that first slot. The key insight is that T−αIT - \alpha I behaves very differently depending on whether ∣α∣|\alpha| is less than, equal to, or greater than 1—this controls whether we can scale and compensate for the shift's destructive action. Injectivity asks if we can recover a sequence from its shifted version; surjectivity asks if every sequence can be written as (T−αI)(T - \alpha I) applied to something; invertibility requires both. The threshold ∣α∣=1|\alpha| = 1 is critical because it determines whether the "decay" or "growth" from the −αI-\alpha I term can balance the information loss from the shift.

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