1 Introduction↩︎

The \(n\)th root problem asks what classes of Hilbert-space operators \(T\kern-1pt\) are such that, if \(T^n\!\) is normal, then \(T\kern-1pt\) is normal.The problem has been investigated for more than six decades and remains under active research these days.Perhaps the most popular example along this line is the celebrated result for hyponormal operators proved by Stampfli in the early sixties [1], which was extended to paranormal operators by Ando a decade afterwards [2]. The problem has recently been extended to \(k\)-paranormal operators and beyond [3] and [4] (and also the references therein).

The main results proved here are Theorems 5.2 and 6.1, which read as follows.

Theorem 5.2 will contribute to the proof of Theorem 6.1.The class of normaloid Hilbert-space operators with normaloid direct summands is quite large, being near to the class of normaloid operators themselves.It includes the \(k\)-paranormal operators (and so, in particular, the classes of paranormal and hyponormal operators).

The paper is split into 7 sections.Basic terminology and notation are posed in Section 2.The class of completely normaloid operators is introduced in Section 3, and the root problem is considered in Section 4.Section 5 discusses in detail a class of normaloid operators with a norm condition on their powers that become multiples of unitary operators.The root problem for completely normaloid operators is investigated in Section 6.Additional results close the paper in Section 7.

2 Basic Terminology and Notation↩︎

By an operator we mean a bounded linear transformation of a normed space \({\mathcal{X}}\) into itself.A subspace \({\mathcal{M}}\) of \({\mathcal{X}}\) is a closed linear manifold of \({\mathcal{X}}\); it is nontrivial if \({\{0\}\ne{\mathcal{M}}\ne{\mathcal{X}}}.\) A subspace \({\mathcal{M}}\) is invariant for an operator \(T\kern-1pt\) (or is \(T\kern-1pt\)-invariant) if \({T({\mathcal{M}})\subseteq{\mathcal{M}}}.\) The induced uniform norm of an operator \(T\kern-1pt\) is denoted by \(\|T\|\), and, if \({\mathcal{X}}\) is a (complex) Banach space, the spectrum of \(T\kern-1pt\) is denoted by \(\sigma(T)\) and its spectral radius by \(r(T).\) If \(T\kern-1pt\) is an operator on a Hilbert space \({\mathcal{H}}\), then the operator \(T^*\kern-1pt\) on \({\mathcal{H}}\) stands for the adjoint of \(T\kern-1pt.\) A subspace \({\mathcal{M}}\) of an operator \(T\kern-1pt\) on a Hilbert space \({\mathcal{H}}\) reduces \(T\kern-1pt\) (or \({\mathcal{M}}\) is a reducing subspace for \(T\kern-1pt\)) if \({\mathcal{M}}\) is invariant for \(T\kern-1pt\) and for \(T^*\!.\) Equivalently, if \({\mathcal{M}}\) and \({\mathcal{M}}^\perp\!\) are both \(T\kern-1pt\)-invariant, where \({\mathcal{M}}^\perp\!\) stands for the orthogonal complement of \({\mathcal{M}}\) in \({\mathcal{H}}.\) The restriction \(T|_{\mathcal{M}}\!\) of \(T\kern-1pt\) to a reducing subspace \({\mathcal{M}}\) is referred to as a direct summand of \(T\kern-1pt\) or as a part of \(T\kern-1pt.\) (Sometimes the term “part” is defined as the restriction of an operator to an invariant subspace, but here we use the term as a synonym of direct summand.) A part \(T|_{\mathcal{M}}\) of an operator \(T\kern-1pt\) is nontrivial if \({\mathcal{M}}\) is a nontrivial reducing subspace for \(T\kern-1pt\), and \(T\kern-1pt\) is reducible if it has a nontrivial part; otherwise it is irreducible.The symbol \(\oplus\) stands for the orthogonal direct sum of subspaces, as in \({{\mathcal{H}}=\kern-1pt{\mathcal{M}}\oplus{\mathcal{M}}^\perp\!}\), or of operators, as in \({T\kern-1pt=T|_{\mathcal{M}}\oplus T|_{{\mathcal{M}}^\perp\!}}\!\) if \({\mathcal{M}}\) reduces \(T\kern-1pt\), where \(T|_{\mathcal{M}}\) and \(T|_{{\mathcal{M}}^\perp}\!\) are direct summands, that is, parts, of \(T\kern-1pt.\)

A nonzero operator \(T\kern-1pt\) on a linear space is nilpotent of index \(j\) if \({T^j\kern-1pt=O}\) for some integer \({j>1}\) and \({T^i\!\ne O}\) for every positive integer \({i<j}\), where \(O\) denotes the null operator.A nilpotent operator without a specified index will be supposed to be of index 2.An involution is an invertible operator on a linear space for which \({T^{-1}\!=T}\); equivalently, for which \({T^2\!=I}\), where \(I\) stands for the identity operator.

An operator \(T\kern-1pt\) acting on a Hilbert space is normal if it commutes with \(T^*\kern-3pt\) (i.e., \(T^*T\kern-1pt=T\kern1pt T^*\kern-1pt).\) It is quasinormal if it commutes with \({T^*T\kern-1pt}\) (i.e., if \({(T^*T\kern-1pt-T\kern1pt T^*)\kern1pt T\!=O}).\) So normal operators are quasinormal.A self-adjoint is a normal operator such that \({T=T^*\!}\), and a unitary is an invertible normal operator such that \({T^{-1}\!=T^*}\!.\) The following assertions are pairwise equivalent.

(For the equivalence among items (i) to (iv), see, e.g., [5]; (iv) and (v) are equivalent because an invertible quasinormal is normal.)An operator satisfying any of the above equivalent conditions is called a symmetry.

3 Completely Normaloid Operators↩︎

A Hilbert-space operator \(T\kern-1pt\) is hyponormal if \({T\kern1pt T^*\kern-2pt\le T^*T}\!.\) (Quasinormal operators are hyponormal.) An operator \(T\kern-1pt\) acting on a normed space \({\mathcal{X}}\) is paranormal if \(\|Tx\|^2\!\le{\|T^2x\|\kern.5pt\|x\|}\) and \(k\)-paranormal if \(\|Tx\|^{k+1}\!\le{\|T^{k+1}x\|\kern.5pt\|x\|^k}\kern-1pt\) for some positive integer \(k\), for every \({x\in{\mathcal{X}}}.\) A Banach-space operator \(T\kern-1pt\) is normaloid if its spectral radius coincides with its norm (i.e., \({r(T)=\|T\|}\)), which is equivalent to \({\|T^k\|\kern-1pt=\kern-.5pt\|T\|^k}\!\) for every integer \({k\kern-1pt\ge\kern-1pt1}.\) These classes are related by proper inclusion: \[\nsc Normal \subset Hyponormal \subset Paranormal \subset k-Paranormal \subset Normaloid. \eqno{(\dagger)}\] In fact, paranormal operators are normaloid [6] (Example 4.1 below shows that the inclusion is proper).It is known that the inverse of an invertible paranormal operator is paranormal [6], that integer powers of paranormal operators are paranormal [7] and [8], and that the restriction of a paranormal operator to an invariant subspace is paranormal [9].It is also known that \(k\)-paranormal operators are normaloid and that restrictions of \(k\)-paranormal operators to invariant subspaces are \(k\)-paranormal (see, e.g., [10]\(\kern.5pt\)).Properties of powers and inverses of \(k\)-paranormal operators have been investigated in, for instance, [10].

A Banach-space operator is hereditarily normaloid if every restriction of it to an invariant subspace is normaloid.Let HN denote the class of all hereditarily normaloid operators.The class HN was investigated in [11] (see also [12], [13]\(\kern.5pt\)).This is a large class of operators.It lies properly between the \(k\)-paranormal and the normaloid operators (see, e.g., [10]\(\kern.5pt\)).We will be dealing with a still larger class.

Definition. A Hilbert-space operator is completely normaloid if every restriction of it to a reducing subspace is normaloid (i.e., if every part of it is normaloid).

Let CN stand for the class of all completely normaloid Hilbert-space operators.This class is trivially included in the class of normaloid operators and includes the class of all irreducible normaloid operators. Then we get a refinement of chain \((\dagger)\): \[\nsc k-Paranormal \subset HN \subset CN \subset Normaloid. \eqno{(\ddag)}\] To verify that \({\sc HN\kern-1.5pt\subseteq\kern-1.5pt\sc CN}\), take a Hilbert space \({\mathcal{H}}\) and an operator \(T\kern-1.5pt\) on \({\mathcal{H}}.\) Let \(\it Lat\,({\mathcal{H}})\) be the lattice of all subspaces of \({\mathcal{H}}.\) Let \(\sc NLD\) denote the collection of all normaloid operators on any \({\!{\mathcal{M}}\in\it Lat\,({\mathcal{H}})}.\) Let \({\it Lat\,(T)\subseteq\it Lat\,({\mathcal{H}})}\) be the lattice of all \(T\)-invariant subspaces for \(T\kern-1pt.\) Let \(\it Red\,(T)\) be the lattice of all reducible subspaces of \(T\kern-1pt.\) \[T\in\sc HN \iff T|_{\mathcal{M}}\in\sc NLD\; \rm for all \;{\mathcal{M}}\in\it Lat\,(T)\] \[\Longrightarrow\quad T|_{\mathcal{M}}\in\sc NLD\; \rm for all \;{\mathcal{M}}\in\it Red\,(T) \subseteq\it Lat\,(T) \iff T\in\sc CN.\]

Remark 3.1. In [9], a Hilbert-space operator was said to be invariant normaloid if the restriction of it to every invariant subspace is normaloid. (See also [14]\(\kern.5pt\)).The notion of a normaloid operator given in [9] was in terms of the numerical radius (i.e., \({w(T)=\|T\|}\)) rather than in terms of the spectral radius (i.e., \({r(T)=\|T\|}\)).These definitions of normaloidness, however, are equivalent (see, e.g., [5]\(\kern.5pt\)). Moreover, it was shown in [9] that if an invariant normaloid (i.e., a hereditarily normaloid) Hilbert-space operator is compact, then it is normal.This can be extended to completely normaloid. \[\it If\/ {T\in\sc CN} is compact, then it is normal\/.\] In fact, the proof in [9] actually shows that if a Hilbert-space operator is compact and all its direct summands are normaloid, then it is normal.Therefore, all subclasses of normaloid operators considered here, when acting on a finite-dimensional space, are normal. But there are nonnormal normaloid operators on finite-dimensional spaces:for instance, \({T\kern-2pt=\kern-1pt A\oplus I}\kern-.5pt\) with \({A\kern-1pt=\kern-1pt\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}0 & 1 \cr 1/2 & 0 \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}.\) \(\blacksquare\)

4 The Root Problem↩︎

Let \({\kern 1pt\mathcal{C}}\) be a class of operators on a Hilbert space of dimension greater than one containing normal operators.Recall that powers and parts of a normal operator are again normal.Let \(n\) be a positive integer.The \(n\)th root problem (or the root problem, for short) asks for classes \(\kern-.5pt{\kern 1pt\mathcal{C}}\kern-1pt\) of operators satisfying the following condition.

If \(T\kern-1pt\) lies in \({\kern 1pt\mathcal{C}}\) and \(T^n\kern-1pt\) is normal, then \(T\kern-1pt\) is normal.

Since \(T\kern-1pt\) belongs to \({\kern 1pt\mathcal{C}}\), and since it is expected to be normal, the class \({\kern 1pt\mathcal{C}}\) is supposed to contain normal operators; otherwise, the root problem is vacuous.

The functional calculi for operators acting on a complex Hilbert space ensure that a normal operator \(N\) has a normal \(n\)th root.In fact, if \({\psi\!:{\mathbb{C}\kern.5pt}\to{\mathbb{C}\kern.5pt}}\) is the function assigning to each \({\zeta\kern-1pt\in{\mathbb{C}\kern.5pt}}\) its principal \(n\)th root, that is, \({\psi(\zeta)=|\zeta|^{\frac{1}{\scriptstyle n}} exp \big(i\,Arg(\zeta)\kern1pt\frac{1}{\scriptstyle n}\kern-1pt\big)}\), then \({\psi(\zeta)^n\kern-1pt=\zeta}\) for every \({\zeta\kern-1pt\in{\mathbb{C}\kern.5pt}}\) and \(\psi(N)\) is an \(n\)th root of \(N\), which is again a normal operator.For more aspects of the characterisation of the \(n\)th root problem, see, e.g., [15], [16].However, there are two crucial features here.First, there is no uniqueness:a normal operator may have several normal square roots. (Trivial example:\(I\) and \(-I\) are normal square roots of the identity operator.)More important, there are square roots for normal operators that are not normal.(Trivial example: a nonzero nilpotent operator, which is never normal, is a square root of the null operator.)

Example 4.1. An example next to trivial emphasises some critical questions.Take \(T\kern-1pt={L\oplus I}\), where \(L\) is a nonzero nilpotent contraction.The contraction \(T\kern-1pt\) is not normal because \(L\) is not normal; \(L\) is not even normaloid — the only quasinilpotent normaloid operator is the null operator.But \({T^2\!=O\kern-.5pt\oplus I}\) is a nonzero normal (thus not nilpotent) with \({r(T^2)\kern-1pt=\kern-1pt1}\), and so \(T\kern-1pt\) is normaloid with \({\|T\|\kern-1pt=\kern-1ptr(T)\kern-1pt=\kern-1pt1}.\) A defect of this example is that, although \(T\kern-1pt\) is a reducible normaloid operator, it has a nonnormaloid part \(L\), thus dismissing any chance of \(T\) being normal. \(\blacksquare\)

For comparison, consider a well-known class satisfying the root problem:

If \(T\kern-1pt\) is hyponormal and \(T^n\kern-1.5pt\) is normal for some \(n\), then \(T\) is normal

[1], which was extended to paranormal operators in [2] and to \(k\)-paranormal operators in [3].(The root problem has been extended to classes of normaloid — and nonnormaloid — operators that properly include the paranormal operators; see, e.g., [3], [4], [17].For variants of the problem, for instance, involving subnormal and quasinormal operators together, as well as multivariable versions of it, see, e.g., [18], [19] and the references therein.)

Since the classes of hyponormal and paranormal operators (which are normaloid) satisfy the root problem, they do not present the defect described in Example 4.1.If a hyponormal (paranormal) is reducible, then its parts are hyponormal (paranormal), and so its parts are normaloid.The same property also applies to \(k\)-paranormal operators (which are normaloid):parts of \(k\)-paranormal are \(k\)-paranormal, thus normaloid — cf.chains \((\dagger)\) and \((\ddag).\) Recalling that CN stands for the class of normaloid operators with normaloid parts, consider the following question.

Question 4.2. Take \(T\kern-1pt\) in CN.Is \(T\kern-1pt\) normal whenever \(T^n\!\) is normal for some \(n\)?

5 Multiples of Unitary Operators↩︎

From now on all Banach and Hilbert spaces are complex.The aim of this section is to prove Theorem 5.2, which gives an ultimate condition for a normaloid operator to be a multiple of a unitary operator, and to disclose some useful corollaries of it.To begin with, we encapsulate some necessary basic results on invertibility.

Remark 5.1. Recall that the inverse of an invertible Banach-space operator is an operator (i.e., it is bounded) by the Inverse Mapping Theorem.

(a) If \(T\kern-1pt\) is an operator on a Banach space, then \({\sigma(T)^n=\sigma(T^n)}\) for every positive integer \(n\), and so

\(T^n\kern-1pt\) is invertible if and only if \(T\kern-1pt\) is invertible.

(b) Suppose \(T\kern-1pt\) is invertible.Recall that (since \({I=T\kern1pt T^{-1}}\)) \[1\le\|T\|\,\|T^{-1}\|.\] (c) So \({\|T^k\|^{-1/k}\!\le\|T^{-k}\|^{1/k}}\) for every \({k\kern-1pt\ge\kern-1pt1}.\) Thus, by the Gelfand–Beurling formula, which says that \({r(T)=\lim_k\|T^k\|^{1/k}}\) for every operator \(A\), we get \[r(T)^{-1}\le r(T^{-1}) \qquad\big(equivalently, \quad r(T^{-1})^{-1}\le r(T)\big).\] Normaloidness for both \(T\kern-1pt\) and \(T^{-1}\kern-1pt\) does not imply that the above inequalities become identities (as we will see in Proposition 5.6(c) below.) \(\blacksquare\)

Let \(T\kern-1pt\) be an operator on a Hilbert space.

Theorem 5.2. If \(T\kern-1pt\) is normaloid and \(T^n\kern-1pt\) is invertible with \({\|T^n\|^{-1}\!=\kern-1pt\|T^{-n}\|}\) for some positive integer \(n\), then \(T\kern-1pt\) is a multiple of a unitary operator.

Since \(T\kern-1pt\) is invertible, \({\|T\|\ne0}.\) (a) Set \({S=\kern-1ptT/\|T\|}.\) As \(T\kern-1pt\) is normaloid, \(S\) is a normaloid contraction with \({\|S^k\|=1}\) for every positive integer \(k\) (because \({S^k\kern-1pt=T^k/\|T\|^k\kern-1pt=T^k/\|T^k\|}\kern.5pt\)). (b) Since \(T^n\) is invertible, \(S^n\) is invertible, and so is \(S.\) (c) Since \({\|T^n\|^{-1}\!=\kern-1pt\|T^{-n}\|}\) and \(T\) is normaloid, it follows that \({\|S^{-n}\|=1}.\;\) Indeed, \[\|S^{-n}\|=\|(T/\|T\|)^{-n}\|=\|T^{-n}\|/\|T\|^{-n} =\|T^{-n}\|\,\|T\|^n=\|T^n\|^{-1}\,\|T^n\|=1.\] (d) Thus, according to Remark 5.1(b), we get \({\|S^{-1}\|=1}.\;\) In fact, \[1=\|S\|^{-1}\le\|S^{-1}\|=\|S^{n-1}S^{-n}\|\le\|S^{n-1}\|\,\|S^{-n}\| =\|S^{-n}\|=1.\] (e) Since \({\|S\|=\|S^{-1}\|=1}\), \(S\) is unitary. Actually, on a Hilbert space, an invertible contraction whose inverse is a contraction is unitary (see, e.g., [5]\(\kern.5pt\)), which is a particular case of a classical result that says that an invertible power-bounded operator with a power-bounded inverse is similar to a unitary operator — cf. [20]\(\kern.5pt\). (f) And so \({T=\|T\| S}\) is a multiple of a unitary operator. 0◻

Immediate corollaries of Theorem 5.2 read as follows.

Corollary 5.3. If \(T\kern-.5pt\kern-1pt\) is normaloid and \(\kern-.5ptT^n\kern-1.5pt\) is a multiple of a unitary operator for some positive integer \(\kern-.5ptn\), then \(\kern-.5ptT\kern-1.5pt\) is a multiple of a unitary operator \(\kern-.5pt(\kern-.5pt\)and thus normal\()\).

Corollary 5.4. If \(T\kern-1pt\) is normaloid and \(T^n\kern-1pt\) is a scalar operator for some positive integer \(n\), then \(T\kern-1pt\) is a multiple of a unitary operator \((\)and thus normal\()\).

By a scalar operator we mean a multiple of the identity.The particular case in Corollary 5.4 has been considered in [4].Note that if \(T^n\kern-1pt\) is a zero multiple of a unitary or of the identity, the above corollaries hold trivially with \(T\kern-1pt\) being the null operator because the only normaloid nilpotent operator is the null operator. Otherwise, \(T^n\kern-1pt\) is invertible (and so is \(T\kern-1pt\)), as required in Theorem 5.2.It is also worth noting that in case of Corollary 5.4, the resulting operator \(T\kern-1pt\) is not necessarily a multiple of the identity (i.e., a scalar).For instance, \({T\kern-1pt=\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}0 & 1 \cr 1 & 0 \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}\) and \({T^2\kern-1pt=\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}1 & 0 \cr 0 & 1 \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}.\)

A version of the root problem asks for normaloid operators that become normal when raised to an integer power.We consider below three properties of normaloid operators that are required in the sequel, leading to two classes of operators where normaloidness implies normality.

Remark 5.5. Let \(T\kern-1pt\) be an operator on a Banach space. (a) If \(T\kern-1pt\) is normaloid, then \(T^k\kern-1pt\) is normaloid for every integer \({k\kern-1pt\ge\kern-1pt1}.\) (In fact, \({r(T)^k\kern-1.5pt=\kern-.5pt r(T^k)\kern-1pt\le\kern-1pt\|T^k\| \kern-1pt\le\kern-1pt\|T\|^k\kern-1.5pt=\kern-.5ptr(T)^k\!}\), so \({r(T^k)\kern-1pt=\kern-.5ptr(T)^k\kern-1.5pt=\kern-.5pt\|T\|^k \kern-1.5pt=\kern-.5pt\|T^k\|}.\)) (b) On a \(2\)-dimensional space, normaloidness coincides with normality.
(See, e.g., [4].) (c) For an involution, normaloidness coincides with normality. Theorem 5.2 can be used to supply a proof of this assertion.In fact, if \(T\kern-1pt\) is an involution (i.e., \({T^2\kern-1pt=I}\); equivalently, \({T^{-1}\kern-1pt=T}\)), then \({r(T)^2=r(T^2)=r(I)=1}.\) Thus, \[1=r(T)\le\|T\|=\|T^{-1}\|.\] If, in addition, \(T\kern-1pt\) is normaloid, then \({r(T)=\|T\|}\) so that \[\|T\|^{-1}=\|T^{-1}\|.\] Since \(T\) is an invertible normaloid for which the above identity holds, Theorem 5.2 with \({n=1}\) says that \(T\kern-1pt\) is a multiple of a unitary operator \(U.\) As \({\|T\|=1}\), we get \(T=\gamma\,U\) with \(\gamma\) in the unit circle.\(\kern-1pt\)Then \({T^*\!=\overline{\gamma}\,U^*\!=\overline{\gamma}\,U^{-1} \!=\overline{\gamma}\,\gamma^{-1}T^{-1}\!=T^{-1}\!=T}.\) \(\kern-.5pt\)So \(T\kern-1pt\) is a self-adjoint involution (i.e., a normal involution).The converse is trivial\(.\) Thus, in addition to the equivalent conditions in Section 2, \(T\kern-1pt\) is a symmetry if and only if

Consider the statement in Theorem 5.2.Since \(T\kern-1pt\) is an invertible normaloid, the condition \({\|T^n\|^{-1}\!=\kern-1pt\|T^{-n}\|}\) for some \(n\) is equivalent to \({\|T\|^{-n}\!=\kern-1pt\|T^{-n}\|}\) for such an \(n.\) The conclusion of Theorem 5.2 says that \(T\kern-1pt\) is normal.If \(T\kern-1pt\) is an invertible normal, then \(T^{-1}\!\) is normal; so \(T\kern-1pt\) and \(T^{-1}\!\) are normaloid, which implies by Remark 5.5(a) that \(\|T^k\|\kern-1pt=\kern-1pt\|T\|^k\!\) and \(\|T^{-k}\|\kern-1pt=\kern-1pt\|T^{-1}\|^k\!\) for every \(k.\) In this case, \[\|T^n\|^{-1}\!=\kern-1pt\|T^{-n}\| \kern-1pt\iff\kern-1pt \|T\|^{-n}\!=\kern-1pt\|T^{-1}\|^n \kern-2pt\kern-1pt \iff \|T\|^{-1}\!=\kern-1pt\|T^{-1}\| \kern-2pt\iff\kern-1pt 1\kern-1pt=\kern-1pt\|T\|\kern1pt\|T^{-1}\|.\] Proposition 5.6 below completes such an argument.

Proposition 5.6. Let \(T\kern-1pt\) be an invertible operator on a Banach space and consider the following assertions.

(a) If \(\kern-.5pt T\kern-1.5pt\) is normaloid, then (i)\(\kern-1pt\) implies (ii),\(\kern-1pt\) and (i)\(\kern-1pt\) also implies that \(\kern-.5pt T^{-1}\kern-2pt\) is normaloid. (b) If both \(T\kern-1pt\) and \(T^{-1}\kern-1.5pt\) are normaloid, then assertions (i) to (v) are equivalent. (c) \(T\kern-1pt\) and \(T^{-1}\kern-1.5pt\) both being normaloid does not imply any of the assertions (i) to (v).

(a) If (i) holds, then \({\|T^{-k}\kern-1pt\|\le\|T^{-1}\|^k\kern-2pt=\|T\|^{-k}}\kern-2pt\), and hence \({\|T\|^k\|T^{-k}\|\le1}\), for every \({k\ge1}.\) Thus, by Remark 5.1(b), if \(T\kern-1pt\) is normaloid, then \({1\le\|T^k\|\,\|T^{-k}\|}={\|T\|^k\|T^{-k}\|\le1}.\) Hence, \({\|T^k\|\,\|T^{-k}\|=1}\); that is, \({\|T^{-k}\|=\|T^k\|^{-1}}\!.\) So (i) \(\Rightarrow\) (ii). Also, if \(T\kern-1pt\) is normaloid and (i) holds, then (ii) holds as we saw above, and so (v) holds by the Gelfand–Beurling formula.Thus, since \({\|T\|^{-1}\!=r(T)^{-1}}\!\) (for \(T\kern-1pt\) is normaloid), \[\|T^{-1}\| \;{{{\buildrel_{\scriptstyle\rm(i)}\over=}}}\; \|T\|^{-1}=r(T)^{-1} \;{{{\buildrel_{\scriptstyle\rm(v)}\over=}}}\; r(T^{-1}),\] and so (i) implies that \(T^{-1}\kern-1pt\) is normaloid whenever \(T\kern-1pt\) is. (b) In general, for every invertible operator, (ii) \(\Rightarrow\) \(\big\{\)(iii) and (iv)\(\big\}\) trivially, and (iv) \(\Rightarrow\) (v) by the Gelfand–Beurling formula.If \(T\kern-1pt\) is normaloid, then (i) \(\Rightarrow\) (ii) by item (a).Moreover, if \(T\kern-1pt\) and \(T^{-1}\kern-1pt\) are normaloid, then (iii) \(\Rightarrow\) (i).Indeed, \[\|T\|^{-n}=\|T^n\|^{-1} \;{{{\buildrel_{\scriptstyle\rm(iii)}\over=}}}\; \|T^{-n}\|=\|(T^{-1})^n\|=\|T^{-1}\|^n,\] so that \({\|T\|^{-1}=\|T^{-1}\|}.\) Finally, if \(T\kern-1pt\) and \(T^{-1}\kern-1pt\) are normaloid, then (v) \(\Rightarrow\) (i).

(c) For instance, if \({T\!=\kern-.5pt\frac{1}{2}I\oplus 2I}\), then each of (i) up to (v) fails. 0◻

Corollary 5.7. Let \(T\kern-1pt\) be an invertible normaloid operator on a Hilbert space.If \({\|T\|^{-1}\kern-1pt=\|T^{-1}\|}\), then \(T\kern-1pt\) is a multiple of a unitary operator, thus normal.

\(\!\)Take any positive integer \(n.\) If \(T\kern-1pt\) is invertible, normaloid and \({\|T\|^{-1}\!=\kern-1pt\|T^{-1}\|}\), then Proposition 5.6(a) ensures that \({\|T^n\|^{-1}\kern-1pt=\|T^{-n}\|}.\) So \(T\kern-1pt\) is a multiple of a unitary operator by Theorem 5.2. 0◻

Corollary 5.7, or Proposition 5.6(a), ensures that \(T^{-1}\) is normaloid as well.

Corollary 5.8. Suppose \(T\kern-1pt\) is an invertible normaloid operator on a Hilbert space with a normaloid inverse.The following assertions are equivalent.

If \(T\) and \(T^{-1}\) are normaloid, then Proposition 5.6(b) ensures that (a) to (c) are equivalent, (d) trivially implies (c), and (c) implies (d) by Theorem 5.2. 0◻

Although being an invertible normaloid with a normaloid inverse does not imply any of the above assertions, in this case the assertions are equivalent:they all hold or they all fail together.Thus, if we add an extra assumption to Corollary 5.7, namely, the inverse of \(T\kern-1pt\) is normaloid, then the central condition, “\({\|T\|^{-1}=\|T^{-1}\|}\)”, can be replaced by any of the equivalent conditions in Corollary 5.8.

6 The Root Problem for Completely Normaloid Operators↩︎

Let \(T\kern-1pt\) be a normaloid operator on a Hilbert space.Suppose all parts of \(T\kern-1pt\) are normaloid.If \(T^n\!\) is normal for some \(n\), then \(T\kern-1pt\) is normal.In other words:

Theorem 6.1. On an arbitrary Hilbert space, if an operator \(T\) is completely normaloid and \(T^n\!\) is normal for some positive integer \(n\), then \(T\kern-1pt\) is normal.

Theorem 6.1 gives an affirmative answer to Question 4.2.

First we establish in Lemma 6.3 a version of the above result for a separable Hilbert space.But before proving Lemma 6.3, we need a few results for direct integrals on separable Hilbert spaces, which are summarised below and reflect the minimum required in the proof of Lemma 6.3.For a comprehensive treatment on direct integrals, including all those properties, see [21] (for a brief review focusing on normal operators, see [22]; see also [23]\(\kern.5pt).\)

(a) A separable Hilbert space \({\mathcal{H}}\) can be represented as a direct integral (see, e.g., [24]) of a family \({\{{\mathcal{H}}_\lambda\}\!=\!\{{\mathcal{H}}_\lambda\}_{\lambda\in\Lambda}}\!\) of Hilbert spaces \({\mathcal{H}}_\lambda\), referred to as a field of fibre spaces, with respect to a measure space \({(\Lambda,\mu)}\), where \(\mu\) is a positive Borel measure on the \(\sigma\)-algebra of Borel subsets of a locally compact complete metric space \(\Lambda.\) In the present case, \(\Lambda\) will be a compact set in \({\mathbb{C}\kern.5pt}\); thus, \(\mu\) is finite.A direct integral representation of \({\mathcal{H}}\) is usually denoted by \[{\mathcal{H}}=\!\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu.\] An element \(x\) in \(\kern-1pt{\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu}\) is a function \({x\!:\Lambda\!\to\!\bigcup_\lambda\!{\mathcal{H}}_\lambda}\) such that \({x_\lambda\in{\mathcal{H}}_\lambda}\) for each \(\lambda\) in \(\Lambda\), and \({\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu}\) is the collection of all these functions \({x=\{x_\lambda\}=\{x_\lambda\}_{\lambda\in\Lambda}}\) for which \({\int_\Lambda\|x_\lambda\|^2d\mu<\infty}\), satisfying some additional measurability conditions including the separability of each fibre space \({\mathcal{H}}_\lambda\) (see, e.g., [21]). Direct integrals extend the notion of direct sums.

(b) A decomposable operator \(T\kern-1pt\) on \({{\mathcal{H}}\kern-1pt=\!\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu}\) is one such that \({Tx_\lambda\kern-1pt=T_\lambda x_\lambda\kern-1pt}\) in \({\mathcal{H}}_\lambda\) (\(\mu\)-a.e.)for a collection \(\{T_\lambda\}\) of operators \({T_\lambda\!:{\mathcal{H}}_\lambda\!\to\kern-1pt{\mathcal{H}}_\lambda}\) with \({\mu-{\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda}\|T_\lambda\|\kern-1pt<\!\infty}\), so that \({T\kern-1ptx\kern-1pt=\kern-1pt\{T_\lambda x_\lambda\kern-.5pt\}}\) lies in \({\!\int_\Lambda^{_\oplus}\kern-1pt{\mathcal{H}}_\lambda d\mu}\) for every \({x=\{x_\lambda\kern-.5pt\}}\) in \({\!\int_\Lambda^{_\oplus}\kern-1pt{\mathcal{H}}_\lambda d\mu}\) (satisfying additional measurability conditions).A decomposable operator \({T\kern-1pt\!:{\mathcal{H}}\kern-1pt\to\kern-1pt{\mathcal{H}}}\) is denoted by \[T\kern-1pt=\!\int_\Lambda^{_\oplus}T_\lambda d\mu,\] which is referred to as the direct integral of \(T\kern-1pt.\) The induced uniform norm of \(T\) is (b\(_1\)) \({\|T\|=\mu-{\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda}\|T_\lambda\|}.\) If \(S\) on \({{\mathcal{H}}=\!\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu}\) is decomposable with \({S=\int_\Lambda^{_\oplus} S_\lambda d\mu}\), then (b\(_2\)) \({S=T\kern-1pt}\) if and only if \({S_\lambda=T_\lambda}\) \(\mu\)-a.e., (b\(_3\)) \(TS\) is decomposable and \({TS=\!\int_\Lambda^{_\oplus} T_\lambda S_\lambda d\mu}\), (b\(_4\)) \(S^*\) is decomposable, and \({S^*\!=\!\int_\Lambda^{_\oplus} S_\lambda^* d\mu}.\) From (b\(_2\)), (b\(_3\)) and (b\(_4\)) we may infer the following further properties. (b\(_5\)) For every positive integer \(n\), \(T^n\kern-1pt\) is decomposable, and \({T^n\kern-1pt=\!\int_\Lambda^{_\oplus} T_\lambda^n d\mu}.\) (b\(_6\)) A decomposable \(T\kern-1pt\) is normal if and only if the operators \(T_\lambda\kern-1pt\) are normal \(\mu\)-a.e. (c) If \(N\) is a normal operator on \({\mathcal{H}}\), then it is a special kind of decomposable operator, namely, a diagonalisable operator, where every \(N_\lambda\) is a scalar operator, \[N=\!\int_\Lambda^{_\oplus}\lambda\,I_\lambda d\mu,\] with \({\Lambda=\sigma(N)}\), the spectrum of the normal operator \(N\), where \(\lambda\) lies in \(\sigma(N)\) and each \(I_\lambda\) is the identity operator on \({\mathcal{H}}_\lambda.\) An important property of the von Neumann algebra generated by a normal operator \(N\) on \({\mathcal{H}}\) is that every operator in its commutant is decomposable on \({{\mathcal{H}}=\int_{\sigma(N)}^{_\oplus}{\mathcal{H}}_\lambda\,d\mu}.\) In fact, an operator is decomposable if and only if it commutes with a diagonalisable operator (see, e.g., [21]). Hence, (c\(_1\)) if \(T\kern-1pt\) commutes with \(N\), then \({T\kern-1pt=\!\int_{\sigma(N)}^{_\oplus}T_\lambda\,d\mu}\) with each \(T_\lambda\) acting on \({\mathcal{H}}_\lambda.\)

That having been said, assertions about properties of the operators \(T_\lambda\) of a decomposable operator \({T\kern-1pt=\!\int_\Lambda^{_\oplus} T_\lambda d\mu}\) are to be understood \(\mu\)-almost everywhere.In particular, we say that the operators \(T_\lambda\) are normaloid (\(\mu\)-a.e\(.\)) if \({\|T_\lambda^n\|=\|T_\lambda\|^n}\) for every positive integer \(n\) or, equivalently, \({\sigma(T_\lambda)=\|T_\lambda\|}\) or, still equivalently, if \({w(T_\lambda)=\|T_\lambda\|}\), where these identities are understood \(\mu\)-almost everywhere over \(\Lambda\).

The following auxiliary result is required in the proof of Lemma 6.3.

Proposition 6.2. Let \({{\mathcal{H}}=\!\int_\Lambda^{_\oplus}{\mathcal{H}}_\lambda d\mu}\) be a direct integral of a separable Hilbert space and let \({T\kern-1pt=\!\int_\Lambda^{_\oplus}T_\lambda d\mu}\) be a decomposable operator on \({\mathcal{H}}.\) If \(T\kern-1pt\) is completely normaloid, then \(T_\lambda\) is normaloid \(\mu\)-a.e.

Recall that (i) an operator \(A\) is normaloid if and only if \({\|A\|=w(A)}\), where \(w(A)\) stands for the numerical radius of \(A\) (cf. Remark 3.1), (ii) if \({A=\!\int_\Lambda^{_\oplus}T_\lambda d\mu}\) is a decomposable operator, then \({\|A\|=\mu-{\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda}\|A_\lambda\|}\) (cf. (b\(_1)\kern.5pt)\), and (iii) if \(\Lambda\) is a locally compact Hausdorff space and, in particular, if \(\Lambda\) is a compact set in \({\mathbb{C}\kern.5pt}\) (as we have assumed above), then \({w(A)={\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda}\,w(A_\lambda)}\) [25].

Let \(\Lambda'\) be the set of all \({\lambda\in\Lambda}\) for which \(T_\lambda\) is not normaloid.That is, \[\Lambda'=\big\{\lambda\in\Lambda\!:\|T_\lambda\|>w(T_\lambda)\big\}.\] As the operator \(T\) is decomposable, the maps \({\lambda\mapsto\|T_\lambda\|}\) and \({\lambda\mapsto w(T_\lambda)}\) are measurable, and so is the map \({\lambda\mapsto\|T_\lambda\|\kern-1pt-\kern-1pt w(T_\lambda)}\), which ensures that \(\Lambda'\) is a measurable set.

Suppose \({\mu(\Lambda')>0}.\) We can express \(\Lambda'\) as the countable union \({\Lambda'=\bigcup_{k=1}^\infty\Lambda'_k}\), where \[\Lambda'_k =\big\{\lambda\in\Lambda\!:\|T_\lambda\|>w(T_\lambda)+{\textstyle{\frac{1}{k}}}\big\}\] It is clear that each \(\Lambda'_k\) is also measurable and \({\Lambda'\subset\Lambda'_k}.\) Since \({\mu(\Lambda')>0}\), there exists an integer \(k\) such that \({\mu(\Lambda'_k)>0}.\) Let \(T_k\) denote the restriction of \(T\kern-1pt\) to the subspace \({{\mathcal{H}}_k\kern-1pt=\!\int_{\Lambda'_k}^\oplus{\mathcal{H}}_\lambda d\mu}\) of \({\mathcal{H}}.\) As \({\mathcal{H}}_k\) reduces \(T\kern-1pt\) (invariant to \(T\kern-1pt\) and \(T^*\!\) — cf\(.\) (b\(_4)\kern.5pt)\), and \(T\kern-1pt\) is completely normaloid, the operator \(T_k\) is normaloid.Therefore, \[\|T_k\|=w(T_k). \eqno{(\S)}\] Moreover, as we also saw above \[\|T_k\|={\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda'_k}\|T_\lambda\| \quad \text{and}\quad w(T_k)={\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda'_k}w(T_\lambda).\] For every \({\lambda\in\Lambda'_k}\), we have the pointwise inequality \({\|T_\lambda\|>\omega(T_\lambda)+\frac{1}{k}}.\) Taking the essential supremum over \(\Lambda'_k\) yields \[\begin{align} \|T_k\| &={\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda'_k}\|T_\lambda\| \ge{\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda'_k}\big(w(T_\lambda)+{\textstyle{\frac{1}{k}}}\big) \\ &=\big({\kern 1pt{\rm ess}\kern.5pt\sup\kern 1.5pt}_{\lambda\in\Lambda'_k}\omega(T_\lambda\big)+{\textstyle{\frac{1}{k}}} =\omega(T_k)+{\textstyle{\frac{1}{k}}}. \end{align}\] Thus we get \({\|T_k\|\ge w(T_k)+\frac{1}{k}}\), which contradicts the identity \((\S).\) Such a contradiction ensures that the assumption \({\mu(\Lambda')>0}\) is false.Therefore, \({\mu(\Lambda')=0}\), which means that \({\|T_\lambda\|=w(T_\lambda)}\) for \(\mu\)-almost every \({\lambda\in\Lambda}.\) Thus, \(T_\lambda\) is normaloid \(\mu\)-almost everywhere. 0◻

Lemma 6.3. On a separable Hilbert space, if an operator \(T\kern-1pt\) is completely normaloid and \(T^n\!\) is normal for some positive integer \(n\), then \(T\kern-1pt\) is normal.

Let \(T\kern-1pt\) be an operator on a separable Hilbert space \({\mathcal{H}}.\) Take an arbitrary positive integer \(n.\) If \(T^n\kern-1pt\) is a normal operator on \({{\mathcal{H}}=\!\int_{\sigma(T^n)}^{_\oplus}\!{\mathcal{H}}_\lambda d\mu}\), then, as we saw in (c), it is a diagonalisable operator, \[T^n=\!\int_{\sigma(T^n)}^{_\oplus}\lambda\,I_\lambda d\mu.\] Since \(T\kern-1pt\) commutes with the normal operator \(T^n\!\), (c\(_1\)) says that \(T\kern-1pt\) is a decomposable operator on \({{\mathcal{H}}=\!\int_{\sigma(T^n)}^{_\oplus}\!{\mathcal{H}}_\lambda d\mu}\), which means that \[T\kern-1pt=\!\int_{\sigma(T^n)}^{_\oplus} T_\lambda d\mu,\] with each \(T_\lambda\) acting on each \({\mathcal{H}}_\lambda.\) Hence, by (b\(_5\)), \[T^n=\!\int_{\sigma(T^n)}^{_\oplus}T_\lambda^nd\mu,\] Thus, according to (b\(_2\)), \(\mu\)-a.e., \[T_\lambda^n=\lambda\,I_\lambda.\] Now suppose \(T\kern-1pt\) is completely normaloid so that \(T_\lambda\) is normaloid \(\mu\)-a.e.by Proposition 6.2.Then the above identity and Corollary 5.4 ensure that \(T_\lambda\) is normal \(\mu\)-a.e.Thus \({T\kern-1pt=\!\int_{\sigma(T^n)}^{_\oplus}T_\lambda d\mu}\) is normal on \({\mathcal{H}}\) according to (b\(_6\)). 0◻

To prove the general case where \({\mathcal{H}}\) is not necessarily separable, we borrow an approach introduced in [26].

Proof of Theorem 6.1. Suppose \(T\kern-1pt\) is a completely normaloid operator on a Hilbert space \({\mathcal{H}}\) such that \(T^n\) is normal for some positive integer \(n.\) Take an arbitrary \({x\in{\mathcal{H}}}.\) With \({\mathbb{N}\kern.5pt}\) denoting the set of all positive integers, \({{\mathbb{N}\kern.5pt}_0\kern-1pt={\mathbb{N}\kern.5pt}\cup\kern-1pt\{0\}}\), and with \(\bigvee\) standing for closure of span, set \[{\mathcal{H}}_x=\bigvee\big\{{T^*}^{i_k}T^{j_k}\cdots{T^*}^{i_1}T^{j_1}x\!:\, (i_1,\dots,i_k),(j_1,\dots,j_k)\in{\mathbb{N}\kern.5pt}_0^k,\, k\in{\mathbb{N}\kern.5pt}\big\}\;\;\subseteq\;\;{\mathcal{H}},\] which is a subspace of \({\mathcal{H}}.\) Since \({\mathcal{H}}_x\) is spanned by a countable set, it is separable, thus a separable Hilbert space.Also, as is readily verified, \({\mathcal{H}}_x\) is invariant to both \(T\kern-1pt\) and \(T^*\kern-1pt\), and so it reduces \(T\kern-1pt.\) Set \({T_x=T|_{{\mathcal{H}}_x}}\!\) on \({\mathcal{H}}_x.\) Since \(T\kern-1pt\) is completely normaloid and \({\mathcal{H}}_x\) reduces \(T\!\), it follows that \(T_x\) is completely normaloid as well.Since \(T^n\) is normal and \({\mathcal{H}}_x\) reduces \(T\kern-1pt\) (and so it reduces \(T^n\)), we get \[\begin{align} T_x^{*n}T_x^n &\kern-6pt=\kern-6pt& {T|_{{\mathcal{H}}_x}\!\!}^{*n}\,{T|_{{\mathcal{H}}_x}\!\!}^n =T^{*n}|_{{\mathcal{H}}_x}T^n|_{{\mathcal{H}}_x} =T^{*n}T^n|_{{\mathcal{H}}_x} \\ &\kern-6pt=\kern-6pt& T^nT^{*n}|_{{\mathcal{H}}_x} =T^n|_{{\mathcal{H}}_x}T^{*n}|_{{\mathcal{H}}_x} ={T|_{{\mathcal{H}}_x}\!\!}^n\,{T|_{{\mathcal{H}}_x}\!\!}^{*n} =T_x^nT_x^{*n}. \end{align}\] Then \(T_x^n\) is normal.Since \({\mathcal{H}}_x\) is separable, \(T_x\) on \({\mathcal{H}}_x\) is completely normaloid, and \(T_x^n\) is normal, it follows from Lemma 6.3 that \(T_x\) is normal.Thus, as \({\mathcal{H}}_x\) reduces \(T\kern-1pt\), \[T\kern1ptT^*x =T|_{{\mathcal{H}}_x}\!\!\,T|_{{\mathcal{H}}_x}^*x =T_xT^*_xx =T_x^*T_xx =T|_{{\mathcal{H}}_x}^*T|_{{\mathcal{H}}_x}\!\!\,x =T^*Tx.\] As \(x\) was arbitrarily taken from \({\mathcal{H}}\), the above identity ensures that \({T\kern1pt T^*\!=T^*T\kern-1pt}\), and therefore \(T\) is normal. 0◻

7 Some Consequences of Theorem 6.1↩︎

To avoid trivialities, we agree that all spaces have dimensions greater than \(1.\) The subspaces \({\mathcal{N}}(T)\) and \({\mathcal{R}}(T)^-\!\) of \({\mathcal{H}}\) stand for the kernel and closure of the range of an operator \(T\kern-1pt\) on a Hilbert space \({\mathcal{H}}.\) First we pose a few elementary auxiliary results.

Remark 7.1. Let \(T\kern-1pt\) be an operator on a Hilbert space. (i) If \({T\kern-1pt\in{\sc CN}}\), then

by the very definition of CN (we have already applied this property), and

Indeed, \({T\kern-1pt\in\sc CN\Longrightarrow\kern-1pt T^k\kern-1.5pt\in\sc CN}\) for every positive integer \(k\) since powers of normaloid operators are normaloid.The converse fails: \({T^2\!\in\sc CN{\,{{\Longrightarrow}\kern-11pt{\slash}}\;\;\;}\kern-.5pt T\kern-1pt\in\sc CN}.\) (In fact, if \({T\kern-1pt=L\oplus I}\) with \(L\) being a nilpotent with \({\|L\|\kern-1pt>\kern-1.5pt1}\), then \(T^2\) is normal, thus \({T^2\!\in\sc CN}\) trivially, but \(T\) is not normaloid.) (ii) If \(T^k\kern-1pt\) is normal for some positive integer \(k\), then the subspaces

by the Fuglede–Putnam Theorem:if an operator commutes with a normal operator \(N\), then it commutes with \(N^*\!.\) (As \(F\) commutes with \(T^k\kern-2pt\), \(\kern1pt F\kern1pt T^{*k}\kern-2pt=\kern-1pt T^{*k}F\kern-1pt\), so \(T^kF^*\kern-2pt=\kern-1pt F^*T^k\kern-2pt\), hence \({\mathcal{N}}(T^k)\) and \({\mathcal{R}}(T^k)^-\!\) are \(F^*\kern-1pt\)-invariant, and, trivially \(F\)-invariant.) \(\blacksquare\)

Corollary 7.2. Let \(T\kern-1pt\) be an operator on a Hilbert space \({\mathcal{H}}\), and take an arbitrary integer \({k\ge2}.\) Suppose \(T^k\!\) and all its parts are normaloid \((\)i.e., suppose \({T^k\kern-1.5pt\in\sc CN}).\) If \(T^n\kern-1.5pt\) is normal for some positive integer \(n\), then \[T=L\oplus S \quadon\quad {\mathcal{H}}={\mathcal{N}}(T^n)\oplus{\mathcal{N}}(T^n)^\perp,\] where \(L\) is a nilpotent operator on \({\mathcal{N}}(T^n)\) of index \({j\le\min\{k,n\}}\), \(S\) on \({\mathcal{N}}(T^n)^\perp\!\) is such that \(S^k\kern-1.5pt\) is normal \((\)as well as \(S^n)\), and any of these parts may be absent.

Since \(T^n\) is normal, and since \({\mathcal{N}}(T^n)\) reduces \(T\kern-1pt\) (cf. Remark 7.1.(c)\(\kern.5pt\)), \[T=L\oplus S \quadon\quad {\mathcal{H}}={\mathcal{N}}(T^n)\oplus{\mathcal{N}}(T^n)^\perp\!,\] where \({L=T|_{{\mathcal{N}}(T^n)}}\) and \({S=T|_{{\mathcal{N}}(T^n)^\perp\!}}\) are parts of \(T\kern-1pt.\) Thus, \({L^n\kern-1pt=(T|_{{\mathcal{N}}(T^n)})^n}={T^n|_{{\mathcal{N}}(T^n)}=O}\), so that \(L\) is nilpotent of index \({j\le n}\), and \({S^n\kern-1pt=(T|_{{\mathcal{N}}(T^n)^\perp})^n}={T^n|_{{\mathcal{N}}(T^n)^\perp}\!}\) is normal. Take an arbitrary \({k\ge1}\), and suppose \({T^k=L^k\oplus S^k}\) lies in CN so that \(L^k\) and \(S^k\) are again in CN (cf. Remark 7.1(a)\(\kern.5pt\)). (i) Since \(L^k\) is normaloid, and recalling that \(L^k\) is nilpotent because \(L\) is nilpotent, we get \({L^k=O}\), and so \(L\) is nilpotent of index \({j\le k}.\) (ii) Since \(S^k\) lies in CN and since \((S^k)^n=(S^n)^k\) is normal because \(S^n\) is normal, Theorem 6.1 ensures that \(S^k\) is normal. 0◻

For \({k=1}\), Corollary 7.2 collapses to Theorem 6.1, where \({L=O}.\)

There exist irreducible (thus not normal) involutions; sample: \(T\kern-1.5pt={\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}1 & 1 \cr 0 & \!-1 \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}.\) But involutions, in spite of having a normal square trivially, are normaloid if and only if they are normal.This and Theorem 6.1 suggest the following question.

Question 7.3. Is there an irreducible normaloid operator with a normal square ?

Answer 7.4. Consider the following result from [27]. A Hilbert-space operator \(T\kern-1pt\) is the square root of a normal operator if and only if it is of the form \(T\kern-1pt={A\oplus\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}B & \kern 4pt C \cr O & \kern-1pt-B \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}\), where \(A\) and \(B\) are normal operators and \(C\) is positive \((\)i.e., nonnegative and injective\()\) and commutes with \(B\). Suppose \(T^2\kern-1pt\) is normal.So we can apply the above result.If \(T\kern-1pt\) is irreducible, then one of the above two parts is absent.If part \(\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}B & \kern 4pt C \cr O & \kern-1pt-B \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)\) is absent, then \({T\kern-1pt=A}\) is an irreducible normal operator, which is a contradiction.\(\kern-1pt\)This implies that part \(A\) is absent. \(\kern-1pt\)Thus, \({T\kern-1pt=\big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}B & \kern 4pt C \cr O & \kern-1pt-B \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)}\) is irreducible and \({T^2\!=\kern-1pt \big(\null\,\vcenter{ \baselineskip=8pt\mathsurround=0pt\ialign{ \hfil {\scriptstyle##} \hfil && \hfil {\scriptstyle##} \hfil \crcr \mathstrut \crcr \noalign{\kern-\baselineskip}B^2 & O \cr O & B^2\kern-1pt \cr \crcr \mathstrut \crcr \noalign{\kern-\baselineskip} \crcr }}\!\big)=B^2\oplus B^2}\kern-1pt.\) Hence, \(\|T^2\|=\|B^2\|=\|B\|^2\) because \(B\) is normal.If \(T\kern-1pt\) is normaloid, then \({\|T^2\|=\|T\|^2}\kern-1pt\), so that \(\|T\|=\|B\|\), which is again a contradiction since \({C\kern-1pt\ne\kern-1pt O}\) (\(C\) is injective)\(.\) Therefore, \[\it if\/ T^2\kern-1pt is normal and\/ T\kern-1pt is irreducible, then\/ T\kern-1pt is not normaloid\/.\]

We conclude with two remarks along the lines of Answer 7.4 and Theorem 6.1.

Remark 7.5. Let \(T\kern-1pt\) be an operator on an arbitrary Hilbert space.If it is in CN, then it is normaloid, but it may be irreducible.Since it is normaloid, all powers of it are normaloid, but if it is irreducible, none are normal, as we will see in item (b) below.First, we verify a generalisation of Answer 7.4. (a) If \(T\kern-1pt\) is normaloid and \(T^n\kern-1.5pt\) is normal for some integer \({n\kern-1pt\ge\kern-1pt1}\), then \(T\kern-1pt\) is reducible. Indeed, let \(T\kern1pt\) be a normaloid operator, take an arbitrary integer \({n\ge1}\), and suppose \(T^n\kern-1pt\) is normal.If \(T^n\kern-1pt\) is scalar, then Corollary 5.4 ensures that \(T\kern-1pt\) is normal and so reducible. Thus, suppose the normal operator \(T^n\kern-1pt\) is not scalar.Since \(T^n\kern-1.5pt\) commutes with \(T\kern-1pt\), the Fuglede–Putnam Theorem (as applied in Remark 7.1(c)\(\kern.5pt\)) ensures that \(T\kern-1pt\) commutes with \(T^{*n}\kern-1pt.\) So there exists a nonscalar operator (viz., \(T^n\kern-1pt\)) that commutes with \(T\kern-1pt\) and with \(T^*\kern-1pt\), and therefore \(T\kern-1pt\) is reducible (see, e.g., [28]\(\kern.5pt\)). Since \(T\kern-1pt\) is an irreducible operator in CN if and only if \(T\kern-1pt\) is an irreducible normaloid, a straightforward application of item (a) reads as follows (compare with Answer 7.4). (b) If \(T\kern-1pt\) is an irreducible operator in CN, then \(T^n\kern-1.5pt\) is not normal for all \({n\ge1}\), which also comes as a consequence of Theorem 6.1. \(\blacksquare\)

Remark 7.6. Let \(T\kern-1pt\) be again a normaloid operator acting on a separable Hilbert space \({\mathcal{H}}\), as in Lemma 6.3.If \(T^n\) is normal, then \(T\kern-1pt\) can be decomposed into a countable orthogonal direct sum \[T={\bigoplus}_{k=0}^\infty T_k \quad\;on\;\quad {\mathcal{H}}={\bigoplus}_{k=0}^\infty{\mathcal{H}}_k,\] where \({T_0=T|_{{\mathcal{H}}_0}}\) is nilpotent and the remaining \({T_k=T|_{{\mathcal{H}}_k}}\) are similar to normal operators \(N_k\) on \({\mathcal{H}}_k\) (cf. [29]\(\kern.5pt\)). Since \(T_0\) is normaloid, it is the null operator.Let \(G\) be an invertible (with a bounded inverse) operator on \({\mathcal{H}}_k\) for which \({T_k=G^{-1}N_kG}.\) Since \({\sigma(T_k)=\sigma(N_k)}\) for the normal operator \(N_k\), there is a scalar \(\lambda\) such that \({T_k-\lambda I}\) and \({N_k-\lambda I}\) are invertible, and \[T_{a,k}=\frac{(T_k-\lambda I)}{\|T_k-\lambda I\|} =\frac{G^{-1}(N_k-\lambda I)G}{\|T_k-\lambda I\|} =G^{-1}N_{a,k}G,\] is an invertible contraction similar to an invertible normal \(N_{a,k}.\) The fact that \(T_{a,k}\) is an invertible contraction implies that \[T_{a,k} is similar to a unitary operator,\] say \(T_{a,k}={W^{-1}UW}.\!\) Then \({N_{a,k}G\,W^{-1}=G\,W^{-1}U}\), which implies by the Fuglede–Putnam Theorem that \(N_{a,k}\) is unitary and the operator \(G\,T_{a,k}G^{-1}\) and its inverse are both power bounded.Hence \(T_{a,k}\) is a generalised scalar (cf. [30]\(\kern.5pt\)). Evidently, \({\sigma(T_{a,k})\subseteq{\mathbb{T}\kern.5pt}}\), the unit circle. \(\blacksquare\)

References↩︎