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Measure, Integration & Real Analysis, 2021a

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304 Chapter 10 Linear Maps on Hilbert Spaces<br />

An operator on a finite-dimensional vector space is left invertible if and only if<br />

it is right invertible. We have seen that this result fails for bounded operators on<br />

infinite-dimensional Hilbert spaces. However, the next result shows that we recover<br />

this equivalency for normal operators.<br />

10.55 invertibility for normal operators<br />

Suppose V is a Hilbert space and T ∈B(V) is normal. Then the following are<br />

equivalent:<br />

(a) T is invertible.<br />

(b) T is left invertible.<br />

(c) T is right invertible.<br />

(d) T is surjective.<br />

(e) T is injective and has closed range.<br />

(f)<br />

T ∗ T is invertible.<br />

(g) TT ∗ is invertible.<br />

Proof Because T is normal, (f) and (g) are clearly equivalent. From 10.29, we know<br />

that (f), (b), and (e) are equivalent to each other. From 10.31, we know that (g),<br />

(c), and (d) are equivalent to each other. Thus (b), (c), (d), (e), (f), and (g) are all<br />

equivalent to each other.<br />

Clearly (a) implies (b).<br />

Suppose (b) holds. We already know that (b) and (c) are equivalent; thus T is left<br />

invertible and T is right invertible. Hence T is invertible, proving that (b) implies (a)<br />

and completing the proof that (a) through (g) are all equivalent to each other.<br />

The next result shows that a normal operator and its adjoint have the same eigenvectors,<br />

with eigenvalues that are complex conjugates of each other. This result can<br />

fail for operators that are not normal. For example, 0 is an eigenvalue of the left shift<br />

on l 2 but its adjoint the right shift has no eigenvectors and no eigenvalues.<br />

10.56 T normal and Tf = α f implies T ∗ f = α f<br />

Suppose T is a normal operator on a Hilbert space V, α ∈ F, and f ∈ V. Then α<br />

is an eigenvalue of T with eigenvector f if and only if α is an eigenvalue of T ∗<br />

with eigenvector f .<br />

Proof Because (T − αI) ∗ = T ∗ − αI and T is normal, T − αI commutes with its<br />

adjoint. Thus T − αI is normal. Hence 10.53 implies that<br />

‖(T − αI) f ‖ = ‖(T ∗ − αI) f ‖.<br />

Thus (T − αI) f = 0 if and only if (T ∗ − αI) f = 0, as desired.<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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