Editing Normal operator (section)

==Properties in finite-dimensional case==

If a normal operator ''T'' on a ''finite-dimensional'' real{{clarify|reason=Normal operators were not defined for real Hilbert spaces although the definition is similar and perhaps should be given.|date=May 2015}} or complex Hilbert space (inner product space) ''H'' stabilizes a subspace ''V'', then it also stabilizes its orthogonal complement ''V''<sup>⊥</sup>. (This statement is trivial in the case where ''T'' is self-adjoint.)

''Proof.'' Let ''P<sub>V</sub>'' be the orthogonal projection onto ''V''. Then the orthogonal projection onto ''V''<sup>⊥</sup> is '''1'''<sub>''H''</sub>−''P<sub>V</sub>''. The fact that ''T'' stabilizes ''V'' can be expressed as ('''1'''<sub>''H''</sub>−''P<sub>V</sub>'')''TP<sub>V</sub>'' = 0, or ''TP<sub>V</sub>'' = ''P<sub>V</sub>TP<sub>V</sub>''. The goal is to show that ''P<sub>V</sub>T''('''1'''<sub>''H''</sub>−''P<sub>V</sub>'') = 0.

Let ''X'' = ''P<sub>V</sub>T''('''1'''<sub>''H''</sub>−''P<sub>V</sub>''). Since (''A'', ''B'') ↦ tr(''AB*'') is an [[inner product]] on the space of endomorphisms of ''H'', it is enough to show that tr(''XX*'') = 0. First it is noted that
:<math>\begin{align}
XX^* &= P_V T(\boldsymbol{1}_H - P_V)^2 T^* P_V \\
&= P_V T(\boldsymbol{1}_H - P_V) T^* P_V \\
&= P_V T T^* P_V - P_V T P_V T^* P_V.
\end {align}</math>

Now using properties of the [[Trace (linear algebra)|trace]] and of orthogonal projections we have:

:<math>\begin{align}
\operatorname{tr}(XX^*) &= \operatorname{tr} \left ( P_VTT^*P_V - P_VTP_VT^*P_V \right ) \\
&= \operatorname{tr}(P_VTT^*P_V) - \operatorname{tr}(P_VTP_VT^*P_V) \\
&= \operatorname{tr}(P_V^2TT^*) - \operatorname{tr}(P_V^2TP_VT^*) \\
&= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VTP_VT^*) \\
&= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(TP_VT^*) && \text{using the hypothesis that } T \text{ stabilizes } V\\
&= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VT^*T) \\
&= \operatorname{tr}(P_V(TT^*-T^*T)) \\
&= 0.
\end{align}</math>

The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of the [[Hilbert-Schmidt inner product]], defined by tr(''AB*'') suitably interpreted.<ref>{{cite journal|author=Andô, Tsuyoshi|year=1963|title=Note on invariant subspaces of a compact normal operator|journal=[[Archiv der Mathematik]]|volume=14|pages=337–340|doi=10.1007/BF01234964|s2cid=124945750}}</ref> However, for bounded normal operators, the orthogonal complement to a stable subspace may not be stable.<ref name=Garrett>{{cite web|author=Garrett, Paul|year=2005|title=Operators on Hilbert spaces|url=http://www.math.umn.edu/~garrett/m/fun/Notes/04a_ops_hsp.pdf|access-date=2011-07-01|archive-date=2011-09-18|archive-url=https://web.archive.org/web/20110918213400/http://www.math.umn.edu/~garrett/m/fun/Notes/04a_ops_hsp.pdf|url-status=live}}</ref> It follows that the Hilbert space cannot in general be spanned by eigenvectors of a normal operator. Consider, for example, the [[bilateral shift]] (or two-sided shift) acting on <math>\ell^2(\mathbb{Z})</math>, which is normal, but has no eigenvalues.

The invariant subspaces of a shift acting on Hardy space are characterized by [[Beurling's theorem]].