Measure, Integration & Real Analysis, 2021a

Section 10B Spectrum 307
The equivalence between (a) and (c) in the previous result shows that every unitary
operator is an isometry.
Next we have a result giving conditions that are equivalent to being a unitary
operator. Notice that parts (d) and (e) of the previous result refer to orthonormal
families, but parts (f) and (g) of the following result refer to orthonormal bases.
10.61 unitary operators and their adjoints are isometries
Suppose T is a bounded operator on a Hilbert space V. Then the following are
equivalent:
(a) T is unitary.
(b) T is a surjective isometry.
(c) T and T ∗ are both isometries.
(d) T ∗ is unitary.
(e) T is invertible and T −1 = T ∗ .
(f)
{Te k } k∈Γ is an orthonormal basis of V for every orthonormal basis {e k } k∈Γ
of V.
(g) {Te k } k∈Γ is an orthonormal basis of V for some orthonormal basis {e k } k∈Γ
of V.
Proof The equivalence of (a), (d), and (e) follows easily from the definition of
unitary.
The equivalence of (a) and (c) follows from the equivalence in 10.60 of (a) and (c).
To prove that (a) implies (b), suppose (a) holds, so T is unitary. As we have
already noted, this implies that T is an isometry. Also, the equation TT ∗ = I implies
that T is surjective. Thus (b) holds, proving that (a) implies (b).
Now suppose (b) holds, so T is a surjective isometry. Because T is surjective and
injective, T is invertible. The equation T ∗ T = I [which follows from the equivalence
in 10.60 of (a) and (c)] now implies that T −1 = T ∗ . Thus (b) implies (e). Hence at
this stage of the proof, we know that (a), (b), (c), (d), and (e) are all equivalent to
each other.
To prove that (b) implies (f), suppose (b) holds, so T is a surjective isometry.
Suppose {e k } k∈Γ is an orthonormal basis of V. The equivalence in 10.60 of (a) and (d)
implies that {Te k } k∈Γ is an orthonormal family. Because {e k } k∈Γ is an orthonormal
basis of V and T is surjective, the closure of the span of {Te k } k∈Γ equals V. Thus
{Te k } k∈Γ is an orthonormal basis of V, which proves that (b) implies (f).
Obviously (f) implies (g).
Now suppose (g) holds. The equivalence in 10.60 of (a) and (e) implies that T
is an isometry, which implies that the range of T is closed. Because {Te k } k∈Γ is an
orthonormal basis of V, the closure of the range of T equals V. Thus T is a surjective
isometry, proving that (g) implies (b) and completing the proof that (a) through (g)
are all equivalent to each other.
Measure, Integration & Real Analysis, by Sheldon Axler