Measure, Integration & Real Analysis, 2021a

330 Chapter 10 Linear Maps on Hilbert Spaces
A normal compact operator on a nonzero real Hilbert space might have no eigenvalues
[consider, for example the normal operator T of counterclockwise rotation by
a right angle on R 2 defined by T(x, y) =(−y, x)]. However, the next result shows
that normal compact operators on complex Hilbert spaces behave better. The key idea
in proving this result is that on a complex Hilbert space, the real and imaginary parts
of a normal compact operator are commuting self-adjoint compact operators, which
then allows us to apply the Spectral Theorem for self-adjoint compact operators.
10.107 Spectral Theorem for normal compact operators
Suppose T is a compact operator on a complex Hilbert space V. Then there is an
orthonormal basis of V consisting of eigenvectors of T if and only if T is normal.
Proof One direction of this result has already been proved as part (b) of 10.103.
To prove the other direction, suppose T is a normal compact operator. We can
write
T = A + iB,
where A and B are self-adjoint operators and, because T is normal, AB = BA (see
10.54). Because A =(T + T ∗ )/2 and B =(T − T ∗ )/(2i), the operators A and B
are both compact.
If α ∈ R and f ∈ null(A − αI), then
(A − αI)(Bf)=A(Bf) − αBf = B(Af) − αBf = B ( (A − αI) f ) = B(0) =0
and thus Bf ∈ null(A − αI). Hence null(A − αI) is an invariant subspace for B.
Applying the Spectral Theorem for self-adjoint compact operators [10.106(a)] to
B| null(A−αI) shows that for each eigenvalue α of A, there is an orthonormal basis of
null(A − αI) consisting of eigenvectors of B. The union (over all eigenvalues α of
A) of all these orthonormal bases is an orthonormal family in V (use the set itself as
the index set) because eigenvectors of A corresponding to distinct eigenvalues of A
are orthogonal (see 10.57). The Spectral Theorem for self-adjoint compact operators
[10.106(a)] as applied to A tells us that the closure of the span of this orthonormal
family is V. Hence we have an orthonormal basis of V, each of whose elements is an
eigenvector of A and an eigenvector of B.
If f ∈ V is an eigenvector of both A and B, then there exist α, β ∈ R such
that Af = α f and Bf = β f . Thus Tf =(A + iB)( f )=(α + βi) f ; hence f is
an eigenvector of T. Thus the orthonormal basis of V constructed in the previous
paragraph is an orthonormal basis consisting of eigenvectors of T, completing the
proof.
The following example shows the power of the Spectral Theorem for normal
compact operators. Finding the eigenvalues and eigenvectors of the normal compact
operator V−V ∗ in the next example leads us to an orthonormal basis of L 2 ([0, 1]).
Easy calculus shows that the family {e k } k∈Z , where e k is defined as in 10.112, is
an orthonormal family in L 2 ([0, 1]). The hard part of showing that {e k } k∈Z is an
orthonormal basis of L 2 ([0, 1]) is to show that the closure of the span of this family
is L 2 ([0, 1]). However, the Spectral Theorem for normal compact operators (10.107)
provides this information with no further work required.
Measure, Integration & Real Analysis, by Sheldon Axler