Normal Operator - Properties in Finite-dimensional Case

Properties in Finite-dimensional Case

If a normal operator on a finite-dimensional real or complex Hilbert space (inner product space) stabilizes a subspace, then it also stabilizes its orthogonal complement . (This statement is trivial in the case where is self-adjoint )

Proof. Denote by the orthogonal projection onto . Then the orthogonal projection onto is . The fact that stabilizes can be expressed as, or . The goal is to show that . Since is an inner product on the space of endomorphisms of, it is enough to show that . This follows from a direct computation, using properties of the trace and of orthogonal projections:

,

= \operatorname{tr}(P_VTT^*P_V) - \operatorname{tr}(P_VTP_VT^*P_V) = \operatorname{tr}(P_V^2TT^*) - \operatorname{tr}(P_V^2TP_VT^*) .

The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of the Hilbert-Schmidt inner product. However, for bounded normal operators orthogonal complement to a stable subspace may not be stable. It follows that such subspaces cannot be spanned by eigenvectors. Consider, for example, the bilateral shift, which has no eigenvalues. The invariant subspaces of the bilateral shift is characterized by Beurling's theorem.

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