January 01, 1970
The aim of this paper is to study \(m\)-partial isometries on Hilbert spaces, a natural extension of partial isometries and \(m\)-isometries. We establish structural and spectral results, characterize the \(m\)-partial isometric weighted shifts, and investigate similarity to \(m\)-isometries and \(m\)-partial isometries.
The class of \(m\)-isometric operators was introduced by Agler in 1990 [1]. He and Stankus investigated them further in the influential trilogy [2]–[4]. Since then \(m\)-isometries have been a subject of intense research. Much attention has been paid to their local properties [5], [6] and characterizing the \(m\)-isometric operators in certain classes of operators [7]–[10]. There have also been several results concerning \(m\)-isometric dilations, which are natural generalizations of the celebrated Sz.-Nagy dilation theorem for contractions [11]–[14]. Generalizations of \(m\)-isometries have also been proposed. For example, Gleason and Richter extended the concept to commuting tuples [15]. Sid Ahmed and Saddi introduced \(m\)-partial isometries [16], which serve as a simultaneous extension of \(m\)-isometries and partial isometries. The aim of this paper is to study \(m\)-partial isometries in more detail. In particular, we investigate their spectral properties and several similarity problems.
This paper is organized as follows. Section 2 introduces \(m\)-partial isometries and their basic properties. Weighted shifts are the focus of Section 3, in which Theorem 11 completely describes those that are \(m\)-partially isometric. Section 4 concerns the spectra of \(m\)-partial isometries. In particular, we prove a spectral-inclusion result (Theorem 18) and investigate \(m\)-isometric dilations of \(m\)-partial isometries (Theorem 26). Section 5 contains several results about similarity to \(m\)-isometries or \(m\)-partial isometries. We conclude this paper in Section 6, in which we pose several open problems that we hope will motivate further research on these topics.
In what follows, \(\mathcal{H}\) denotes a complex Hilbert space endowed with inner product \(\langle \cdot \, , \, \cdot\rangle\) and associated norm \(\|\cdot\|\). We denote by \(\mathcal{B}(\mathcal{H})\) the algebra of all bounded linear operators on \(\mathcal{H}\). For each \(T \in \mathcal{B}(\mathcal{H})\), we denote by \(\sigma(T)\), \(\mathcal{R}(T)\), \(\mathcal{N}(T)\), and \(T^{*}\) the spectrum, range, kernel, and adjoint of \(T\), respectively. Moreover, \(\sigma(T), \sigma_{\mathrm{ap}}(T)\), and \(\sigma_{\mathrm{p}}(T)\) represent the spectrum, approximate point spectrum and point spectrum of \(T\), respectively. A (closed) subspace \(\mathcal{M}\subseteq \mathcal{H}\) is invariant for \(T\) (or \(T\)-invariant) if \(T\mathcal{M}\subseteq \mathcal{M}\). The orthogonal complement of \(\mathcal{M}\) is denoted by \(\mathcal{M}^\perp\) and the orthogonal projection onto \(\mathcal{M}\) by \(P_{\mathcal{M}}\). We let \(\mathbb{D}\), \(\mathbb{D}^{-}\), and \(\mathbb{T}\) denote the open unit disk, the closed unit disk, and the unit circle in the complex plane \(\mathbb{C}\), respectively. We write \(\sim\) for similarity and \(\sim_{+}\) for similarity via a positive intertwiner, and we let \(\mathsf{M}_n\) denote the set of \(n \times n\) complex matrices.
Recall that \(T \in \mathcal{B}(\mathcal{H})\) is an isometry if \(T^{*} T=I\), in which \(I\) is the identity operator. Equivalently, \(T\) is a contraction with a contractive left inverse \(S\); that is, there exists an \(S \in \mathcal{B}(\mathcal{H})\) with \(S T=I\) and \(\| S \| \leq 1\). The Wold–von Neumann theorem decomposes an isometry as a direct sum of a unitary operator and a unilateral shift. This algebraic characterization extends up to similarity if one replaces “contraction” with “bounded powers” [17], [18]. Thus, \(T\) is similar to an isometry if and only if it is power bounded and has a left inverse with bounded powers. In finite-dimensional spaces, isometries are unitaries. Consequently, \(A \in \mathsf{M}_{n}\) is similar to an isometry if and only if it is similar to a unitary. This occurs if and only if \(A\) is diagonalizable and \(\sigma(A) \subseteq \mathbb{T}\).
Recall that \(T \in \mathcal{B}(\mathcal{H})\) is a partial isometry if \(T T^{*} T=T\); that is, \(T^{*}\) is a generalized inverse of \(T\). Mbekhta proved that \(T \in \mathcal{B}(\mathcal{H})\) is a partial isometry if and only if it is a contraction with a contractive generalized inverse; that is, an \(S\) such that \(STS = S\), \(TST = T\), and \(\| S \| \leq 1\) [19]; see also [20].
Acknowledgments. SRG was partially supported by NSF Grant DMS-2452084. MB was supported by the subsidy granted to AGH University of Krakow by Polish Ministry of Science and Higher Education.
For \(T \in \mathcal{B}(\mathcal{H})\) and integer \(m \geq 1\), define the selfadjoint operator \[\label{eq:Beta} \beta_m(T) := \sum_{k=0}^m (-1)^{m-k} \binom{m}{k} \, T^{*k} T^k ,\tag{1}\] in which \(\textstyle\binom{m}{k}\) is a binomial coefficient. If \(\beta_m(T) = 0\), then \(T\) is an \(m\)-isometry. The definition \(\beta_m(T) = 0\) is equivalent to \[\label{eq:NormCondition} \sum_{k=0}^m (-1)^{m-k} \binom{m}{k} \| T^k \mathbf{x} \|^2 = 0\tag{2}\] for all \(\mathbf{x} \in \mathcal{H}\). Whenever we speak of an \(m\)-isometry it is with the understanding that \(m \geq 1\). A \(1\)-isometry is an isometry in the usual sense: \(T^*T = I\). An \(m\)-isometry is a strict \(m\)-isometry if it is not an \((m-1)\)-isometry. Since \[\label{eq:BetaInduction} \beta_{m+1}(T) = T^* \beta_m(T) T - \beta_m(T),\tag{3}\] it follows that an \(m\)-isometry is an \(n\)-isometry for all \(n \geq m\). If \(T\) is an invertible \(m\)-isometry, then so is \(T^{-1}\) because \[\label{eq:InverseAlso} (-1)^m\beta_m(T^{-1}) = (T^*)^{-m}\beta_m(T) T^{-m} =0.\tag{4}\]
Recall that \(T \in \mathcal{B}(\mathcal{H})\) is a partial isometry if \(TT^*T = T\); that is, if \(T\beta_1(T) = 0\). More generally, \(T\) is an \(m\)-partial isometry if \(T \beta_m(T) = 0\); it is strict if it is not an \((m-1)\)-partial isometry. These definitions originate in [16], although its consequences appear to be little pursued in the literature.
There is another characterization of partial isometries: \(T\in \mathcal{B}(\mathcal{H})\) is a partial isometry if and only if \(T^*T\) is an orthogonal projection. In fact, \(T^*T\) is the orthogonal projection onto \(\mathcal{N}(T)^{\perp}\). This occurs precisely when \(I - T^*T = -\beta_1(T)\) is the orthogonal projection onto \(\mathcal{N}(T)\). The next lemma generalizes this observation.
Lemma 1. Let \(T\in \mathcal{B}(\mathcal{H})\). The following are equivalent.
\(T\) is an \(m\)-partial isometry.
\((-1)^m\beta_m(T)\) is the orthogonal projection onto \(\mathcal{N}(T)\).
Proof. (a) \(\Rightarrow\) (b). Suppose that \(T\) is an \(m\)-partial isometry. Then \(\mathcal{R}(\beta_{m}(T)) \subseteq \mathcal{N}(T)\) by definition. Moreover, \((-1)^m \beta_m(T) \mathbf{h} = \mathbf{h}\) for any \(\mathbf{h} \in \mathcal{N}(T)\). Therefore, \(\mathcal{R}(\beta_m(T)) = \mathcal{N}(T)\). Since \(\beta_m(T)\) is selfadjoint, the definition \(T \beta_m(T) = 0\) ensures that \(\beta_{m}(T)T^*\mathbf{h} = \mathbf{0}\) for every \(\mathbf{h}\in \mathcal{H}\). With respect to the orthogonal decomposition \(\mathcal{H}= \mathcal{N}(T)\oplus \mathcal{R}(T^*)^-\), we have \((-1)^m\beta_{m}(T) = \big[ \begin{smallmatrix} I & {\color{gray}0} \\ {\color{gray}0} & {\color{gray}0} \end{smallmatrix}\big]\), so \((-1)^m\beta_{m}(T) = P_{\mathcal{N}(T)}\).
(b) \(\Rightarrow\) (a). Suppose that \((-1)^m\beta_m(T) = P_{\mathcal{N}(T)}\). If \(\mathbf{h} = \mathbf{h}_1 + \mathbf{h}_2\), in which \(\mathbf{h}_{1}\in \mathcal{N}(T)\) and \(\mathbf{h}_{2}\in \mathcal{R}(T^*)^-\), then \(T\beta_{m}(T)\mathbf{h} = (-1)^m T P_{\mathcal{N}(T)} \mathbf{h} = (-1)^m T\mathbf{h}_{1} = \mathbf{0}\). Thus, \(T\) is an \(m\)-partial isometry. ◻
Remark 2. Suppose that \(T \in \mathcal{B}(\mathcal{H})\) is an \(m\)-partial isometry with \(\mathcal{N}(T) = \{\mathbf{0}\}\). Then Lemma 1 ensures that \((-1)^m \beta_m(T) = 0\), so \(T\) is an \(m\)-isometry.
Example 3. For each integer \(m\geq 1\), we claim that \(T=\big[ \begin{smallmatrix} 0 & 0 \\ m^{-1/2} & 0 \end{smallmatrix}\big]\) is an \(m'\)-partial isometry if and only if \(m' = m\). Indeed, \[(-1)^{m'}\beta_{m'}(T) = \begin{bmatrix} 1-\frac{m'}{m} & 0\\ 0 & 0 \end{bmatrix},\] so \((-1)^{m'}\beta_{m'}(T) = P_{\mathcal{N}(T)}\) if and only if \(m' = m\). In particular, \(T\) is a strict \(m\)-partial isometry that is not an \((m+1)\)-partial isometry.
Example 4. Suppose that \(T \in \mathcal{B}(\mathcal{H})\) has the block decomposition \[T= \begin{bmatrix} 0 & A\\ 0 & B \end{bmatrix}\] with respect to \(\mathcal{H}= \mathcal{N}(T) \oplus \mathcal{R}(T^*)^-\). Let \(M=A^*A+B^*B\) and observe that \[T^*T= \begin{bmatrix} 0 & 0\\ 0 & M \end{bmatrix} \quad \text{and} \quad T^{*2}T^2= \begin{bmatrix} 0 & 0\\ 0 & B^*MB \end{bmatrix},\] so \[\label{eq:TTB} T^{*2}T^2-2T^*T+I = \begin{bmatrix} I & 0\\ 0 & B^*MB-2M+I \end{bmatrix}.\tag{5}\] Thus, \(T\) is a \(2\)-partial isometry if and only if \(T(T^{*2}T^2-2T^*T+I)=0\); that is, \[A(B^*MB-2M+I)=0 \quad \text{and} \quad B(B^*MB-2M+I)=0.\] The condition \(B^*MB-2M+I=0\) ensures that \(T\) is a \(2\)-partial isometry. If this occurs, 5 is not \(0\), so \(T\) is not a \(2\)-isometry, whereas \(T(T^{*2}T^2-2T^*T+I)=0\), so \(T\) is a \(2\)-partial isometry. Such block operator matrices appear in [21], where the authors studied conditions which may guarantee similarity to a partial isometry.
It is well-known that the adjoint of any partial isometry is again a partial isometry. However, this property does not hold in general for \(m\)-partial isometries.
Example 5. A computation confirms that \[T = \begin{bmatrix} 0 & 0 & 0 \\ \sqrt{ \frac{2}{3}} & 0 & 0 \\ 0 & \frac{\sqrt{2}}{2} & 0 \end{bmatrix} \in \mathcal{B}(\mathbb{C}^3)\] is a \(2\)-partial isometry and \(T^*\) is not.
However, there are certain conditions that ensure that the adjoint of an \(m\)-partial isometry is an \(m\)-partial isometry.
Remark 6. We claim that if \(T\in \mathcal{B}(\mathbb{C}^2)\) is a \(2\)-partial isometry, then \(T^*\) is a \(2\)-partial isometry. If \(\operatorname{rank}T = 0\), the assertion is trivial. If \(\operatorname{rank}T = 2\), then \(T\) is invertible and hence \(T\) is a \(2\)-isometry. Theorem 2.7 in [9] ensures that \(T\) is unitary, so \(T^*\) is a \(2\)-partial isometry. Now suppose that \(\operatorname{rank}T = 1\). Then there are unit vectors \(\mathbf{u},\mathbf{v}\in \mathbb{C}^2\) and a scalar \(\alpha > 0\) such that \(T\mathbf{x} = \alpha \langle \mathbf{x}, \mathbf{v} \rangle \mathbf{u}\) for all \(\mathbf{x}\in \mathbb{C}^2\). Moreover, \(\mathcal{N}(T) = \operatorname{span}\{ \mathbf{v} \}^{\perp}\), \(T\mathbf{v}=\alpha \mathbf{u}\), and \(T^2 \mathbf{v}=\alpha^2\langle \mathbf{u}, \mathbf{v} \rangle \mathbf{u}\), so the \(2\)-partial isometry condition is \(\alpha^4 |\langle u,v\rangle|^2-2\alpha^2+1=0\). Similarly, \(T^*\mathbf{x}=\alpha \langle \mathbf{x},\mathbf{u} \rangle\mathbf{v}\) for all \(\mathbf{x} \in \mathbb{C}^2\) and \(\mathcal{N}(T^*)^{\perp}=\operatorname{span}\{u\}\). Then \(T^*\mathbf{u}=\alpha \mathbf{v}\) and \((T^*)^2 \mathbf{u} =\alpha^2\langle \mathbf{v},\mathbf{u} \rangle \mathbf{v}\). Since \(\alpha^4 |\langle v,u\rangle|^2-2\alpha^2+1=0\), we conclude that \(T^*\) is a \(2\)-partial isometry.
Remark 7. Recall that \(T \in \mathcal{B}(\mathcal{H})\) is complex symmetric if there is a conjugation \(C\) (a conjugate-linear, isometric involution) such that \(T = CT^*C\) [22]–[25]. If \(T\) is a complex symmetric \(m\)-partial isometry, then \(T^*\) is too since \(C (T^{*k}T^k)C = T^k T^{*k}\).
If \(T\in \mathcal{B}(\mathcal{H})\) is a partial isometry, then \(TT^*T = T\); that is, \(T^*\) is a generalized inverse of \(T\). The lemma below generalizes this to \(m\)-partial isometries.
Lemma 8. Let \(T\in \mathcal{B}(\mathcal{H})\) be an \(m\)-partial isometry. Then for \[S = \sum_{k=1}^{m} (-1)^{k-1} \binom{m}{k} T^{\ast k}T^{k-1}\] we have \(T = TST\) and \(S = STS\).
Proof. First, note that \[\begin{align} ST &= \sum_{k=1}^{m} (-1)^{-k+1} \binom{m}{k} T^{\ast k}T^{k}\\ &= (-1)^{m+1}\sum_{k=1}^{m} (-1)^{m-k} \binom{m}{k} T^{\ast k}T^{k}\\ &= (-1)^{m+1}\left( \beta_m(T)-(-1)^mI\right)\\ &= I-(-1)^m\beta_m(T). \end{align}\] Then Lemma 1 ensures that \(ST\) is the orthogonal projection onto \(\mathcal{R}(T^*)^-\). Then \[T = T(P_{\mathcal{N}(T)}+P_{\mathcal{R}(T^*)^-}) = TP_{\mathcal{R}(T^*)^-} = TST.\] In turn, since \(\mathcal{R}(S) \subset \mathcal{R}(T^*)\) it follows that \(STS = S\). ◻
Bermúdez, Martinón, and Negrín characterized the \(m\)-isometric weighted shifts in 2010 [8]. A polynomial description of their weight sequences was later obtained by Abdullah and Le [7]. In this section we study the corresponding problems for \(m\)-partial isometries. We completely describe \(m\)-partially isometric weighted shifts and illustrate the results with several examples. We begin with a computation that translates the equation \(T\beta_m(T)=0\) into a recurrence for the weights.
Example 9. Let \((e_{n})_{n \geq 1}\) be an orthonormal basis of \(\mathcal{H}\). For a numerical sequence \((\omega_{n})_{n \geq 1}\), the associated weighted shift operator \(T\) on \(\mathcal{H}\) is defined by \(T e_{n}=\omega_{n} e_{n+1}\) for \(n \geq 1\). It is bounded if and only if the weight sequence is bounded; we assume this is the case in what follows. One can show that \[T^{k} e_{n} = \bigg( \prod_{j=n}^{n+k-1} \omega_{j} \bigg) e_{n+k}\] and \[T^{* k} e_n = \begin{cases} 0, & n\leq k,\\[3pt] \bigg(\displaystyle\prod_{j=n-k}^{n-1}\overline{\omega_j}\bigg)e_{n-k}, & n>k. \end{cases}\] Thus, \[T^{* k} T^{k} e_{n} = \bigg( \prod_{j=n}^{n+k-1} | \omega_{j} |^2 \bigg) e_{n}\] and hence \(T\) is an \(m\)-partial isometry if and only if the following holds for all \(n \geq 1\):\[\label{eq:WeightedCondition} \omega_{n} \bigg[ (-1)^{m} + \sum_{1 \leq k \leq m} (-1)^{m-k} \binom{m}{k} \bigg( \prod_{j=n}^{n+k-1} | \omega_{j} |^2 \bigg) \bigg] = 0.\tag{6}\]
Remark 10. Let \(a_n=|\omega_n|^2\) for \(n \geq 1\). If \(a_n,a_{n+1},\ldots,a_{n+m-2}\neq 0\), the relation \[(-1)^m + \sum_{k=1}^{m} (-1)^{m-k}\binom{m}{k} \prod_{j=n}^{n+k-1}a_j =0\] ensures that \[a_{n+m-1} = \frac{ (-1)^{m+1} - \displaystyle\sum_{k=1}^{m-1} (-1)^{m-k}\binom{m}{k} \prod_{j=n}^{n+k-1}a_j }{ \displaystyle\prod_{j=n}^{n+m-2}a_j }.\] Thus, admissible blocks of nonzero weights may be constructed recursively, so long as the right side is positive. If the right side is \(0\), then \(a_{n+m-1}=0\), so the string of nonzero weights terminates. Negative values are not admissible, since \(a_j=|\omega_j|^2\geq 0\), although the recursion can produce such values. If \(m=2\), then \[a_{n+1} = \frac{-1-(-1)\binom{2}{1}a_n }{ a_n } = \frac{-1+2a_n}{a_n} = 2-\frac{1}{a_n}.\] Thus, \(a_{n+1}\geq 0\) if and only if \(a_n \geq \frac{1}{2}\). For example, if \(a_n = \frac{1}{4}\), then \(a_{n+1} = -2<0\), which is inadmissible.
The next theorem characterizes \(m\)-partially isometric weighted shifts.
Theorem 11. Let \(T\in \mathcal{B}(\mathcal{H})\) be an \(m\)-partially isometric weighted shift with weights \((\omega_n)_{n\geq 1}\). Let \(Z=\{n\geq 1:\omega_n=0\}\).
If \(Z=\varnothing\), then \(T\) is an \(m\)-isometry.
If \(Z\neq\varnothing\), then write \(Z=\{n_r:r\in I\}\), in which \(I\) is an index set and \(n_1 < n_2 < \cdots\) is an increasing sequence (finite or infinite). Let \(n_0=0\) and \[\mathcal{H}_r = \operatorname{span}\{e_{n_{r-1}+1},e_{n_{r-1}+2},\ldots,e_{n_r}\}\] for each \(r \in I\). If \(I=\{1,2,\ldots,s\}\) is finite, let \(\mathcal{H}_{\infty} = \overline{\operatorname{span}\{e_{n_s+1},e_{n_s+2},\ldots\}}\). If \(I=\mathbb{N}\), then omit \(\mathcal{H}_{\infty}\) in what follows. Then \[\label{eq:OrthogonalDecomposition} \mathcal{H}= \bigg(\bigoplus_{r\in I}\mathcal{H}_r\bigg) \oplus \mathcal{H}_{\infty}\qquad{(1)}\] is an orthogonal decomposition into reducing subspaces for \(T\). With respect to this decomposition, \[T = \bigg(\bigoplus_{r\in I}N_r\bigg)\oplus S,\] in which each \(N_r = T|_{\mathcal{H}_r}\) is nilpotent of order \(d_r=n_r-n_{r-1}\). Moreover, \(S=T|_{\mathcal{H}_{\infty}}\) is an \(m\)-isometric weighted shift unless \(I = \mathbb{N}\), in which case it is omitted.
Proof. Suppose that \(Z=\varnothing\). Then \(\omega_n\neq 0\) for all \(n \geq 1\), so 6 ensures that \[(-1)^m + \sum_{k=1}^{m} (-1)^{m-k}\binom{m}{k} \prod_{j=n}^{n+k-1}a_j =0\] for all \(n\geq 1\). Equivalently, \(\beta_m(T)\mathbf{e}_n=0\) for all \(n \geq 1\), so \(T\) is an \(m\)-isometry.
Suppose that \(Z \neq \varnothing\). For each \(r\in I\), let \(\mathcal{K}_r =\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_{n_r}\}\). Since \(\omega_{n_r}=0\), it follows that \(T\mathbf{e}_j=\omega_j \mathbf{e}_{j+1}\in \mathcal{K}_r\) for all \(1\leq j\leq n_r\). Thus, \(\mathcal{K}_r\) is \(T\)-invariant. Since \[T^*\mathbf{e}_j= \begin{cases} 0 & \text{if j=1},\\[3pt] \overline{\omega_{j-1}} \mathbf{e}_{j-1} & \text{if j\geq 2}, \end{cases}\] we see that \(T^*\mathbf{e}_j\in \mathcal{K}_r\) for all \(1\leq j\leq n_r\). Thus, each \(\mathcal{K}_r\) reduces \(T\) and, since they are nested, their orthogonal differences \(\mathcal{H}_r=\mathcal{K}_r\ominus \mathcal{K}_{r-1} = \operatorname{span}\{\mathbf{e}_{n_{r-1}+1},\ldots,\mathbf{e}_{n_r}\}\) also reduce \(T\). If \(I=\{1,2,\ldots,s\}\), then \(\mathcal{H}_{\infty}=\mathcal{K}_s^\perp\) reduces \(T\). This yields ?? .
Since \(n_{r-1}\) and \(n_r\) are consecutive zero positions in the weight sequence, it follows that \(\omega_{n_{r-1}+1},\ldots,\omega_{n_r-1}\neq 0\). On \(\mathcal{H}_r\), the restriction \(N_r =T|_{\mathcal{H}_r}\) acts as follows: \[\mathbf{e}_{n_{r-1}+1} \mapsto \omega_{n_{r-1}+1} \mathbf{e}_{n_{r-1}+2} \mapsto \cdots \mapsto \omega_{n_r-1} \mathbf{e}_{n_r} \mapsto 0.\] Thus, \(N_r^{d_r}=0\), in which \(d_r=n_r-n_{r-1}\). Since an empty product equals \(1\), \[N_r^{d_r-1}e_{n_{r-1}+1} = \bigg( \prod_{j=n_{r-1}+1}^{n_r-1}\omega_j \bigg)e_{n_r} \neq 0,\] so \(N_r\) is nilpotent of order \(d_r\).
If \(I=\mathbb{N}\), there is nothing to prove. Suppose that \(I=\{1,2,\ldots,s\}\). By construction, \(\omega_n\neq 0\) for \(n>n_s\) and hence the factor \(\omega_n\) in 6 is nonzero, so \[\label{eq:WeightTail} (-1)^m + \sum_{k=1}^{m} (-1)^{m-k}\binom{m}{k} \prod_{j=n}^{n+k-1}a_j =0.\tag{7}\] With respect to the orthonormal basis \(\mathbf{e}_{n_s+1},\mathbf{e}_{n_s+2},\ldots\) of \(\mathcal{H}_{\infty}\), the restriction \(S=T|_{\mathcal{H}_{\infty}}\) is a weighted shift with nonzero weights. The identity 6 ensures that \(S\) is an \(m\)-isometry. ◻
Example 12. For each \(m\geq 1\), we construct an \(m\)-partially isometric weighted shift \(T\) that is similar to \(S \otimes S^*\), the tensor product of the unilateral shift \(S\) with its adjoint. First recall that \(S \otimes S^*\) is unitarily equivalent to \(\bigoplus_{n=1}^{\infty} J_n\), in which \(J_n\) denotes the \(n \times n\) nilpotent Jordan block [26]. Fix \(m \geq 1\) and place a \(0\) weight at each triangular number; that is, at positions \(1,3,6,10,15,\ldots\). This leaves room for separate stretches of \(\ell = 1,2,3,4,\ldots\) positive weights between zeros. To fill a stretch of \(\ell\geq 1\) positive weights use the sequence \[\sqrt{\frac{\ell}{\ell+m-1}},\, \sqrt{\frac{\ell-1}{\ell+m-2}},\, \sqrt{\frac{\ell-2}{\ell+m-3}},\, \ldots,\, \sqrt{\frac{2}{m+1}},\, \sqrt{\frac{1}{m}},\] each element of which belongs to \([2^{1-m},1)\); in particular, the weights are bounded above and below. Thus, the weight sequence is \[0,\, \sqrt{\frac{1}{m}},\, 0,\, \sqrt{\frac{2}{m+1}},\, \sqrt{\frac{1}{m}},\, 0,\, \sqrt{\frac{3}{m+2}},\, \sqrt{\frac{2}{m+1}},\, \sqrt{\frac{1}{m}},\, 0,\ldots .\] The weighted shift \(T\) so constructed is a bounded \(m\)-partial isometry and, moreover, it is similar to \(\bigoplus_{n=1}^{\infty} J_n\), in which \(J_n\) denotes the \(n \times n\) nilpotent Jordan block.
Example 13. Let \(\omega_2 = 0\) and \(\omega_n = 1\) for all \(n \neq 2\). Then, the associated weighted shift \(T\) is a \(1\)-partial isometry since the condition \(T\beta_1(T)=0\) demands that \(\omega_n(|\omega_n|^2-1)=0\) for all \(n \geq 1\). However, \(T\) is not a \(1\)-isometry (that is, an isometry) since it has nontrivial kernel. More generally, for every \(m\geq 1\), \[\omega_1=\frac{1}{\sqrt m},\qquad \omega_2=0,\quad \text{and} \quad \omega_n=1\quad (n\geq 3)\] defines an \(m\)-partial isometry that is neither nilpotent nor an \(m\)-isometry.
Example 14. A nilpotent weighted shift need not be an \(m\)-partial isometry. For example, let \(\omega_1=2\) and \(\omega_n=0\) for \(n\geq 2\). Then \(T^2=0\) and \(T\) is not an \(m\)-partial isometry for any \(m\geq 1\). Indeed, apply 6 with \(n=1\) and use the fact that all products of length at least \(2\) contain \(\omega_2=0\) to obtain \[\omega_1 \bigg[ (-1)^m+(-1)^{m-1}\binom{m}{1}|\omega_1|^2 \bigg]=0.\] Since \(\omega_1=2\neq 0\), this is equivalent to \((-1)^m+(-1)^{m-1}m|\omega_1|^2=0\), or, equivalently, \(4= |\omega_1|^2=\frac{1}{m}\), which is impossible.
Remark 15. To obtain an \(m\)-partial isometry that is not an \(m\)-isometry, it is necessary that at least one weight vanish. If no weight vanishes, then \(a_n \to 1\) since a weighted shift is \(m\)-isometric if and only if \(a_n = p(n+1)/p(n)\) for some polynomial \(p\) of degree at most \(m-1\) [7].
Example 16. Let \(m=2\). Then 6 says that \(\omega_n(1-2a_n+a_na_{n+1})=0\) for all \(n \geq 1\). Thus, \(a_{n+1} = 2 - 1/a_n\) whenever \(\omega_n\neq 0\); in particular, observe that \(a_n = 1\) if and only if \(a_{n+1}=1\). If \(a_1 = 1\), then \(a_n = 1\) for all \(n \geq 1\), in which case \(T\) is an isometry. Now suppose that \(a_1 \neq 1\). Then the sequence \(b_n = 1/(a_n-1)\) satisfies \[b_{n+1} = \frac{1}{a_{n+1}-1} = \frac{1}{1-\frac{1}{a_n}} = \frac{a_n}{a_n-1} = 1+\frac{1}{a_n-1} = b_n+1,\] so \(b_n=b_1+n-1\) and hence \[a_n = 1+\frac{1}{n-1+\frac{1}{a_1-1}}\] is nonnegative and tends to \(1\); in particular, \(\omega_n\) is a bounded sequence. Moreover, \(a_n = 0\) if and only if \(n = -1/(a_1-1)\); for example, \(a_1 = \frac{3}{4}\) yields \(a_2 = \frac{2}{3}\), \(a_3 = \frac{1}{2}\), and \(a_4 = 0\). Note that \(T\) is not a \(2\)-isometry since \(1-2a_4+a_4a_5=1\neq 0\). Thus, \(T\) is a \(2\)-partial isometry that is not a \(2\)-isometry. Note that \(T\) is a \(2\)-isometry if \(a_1 \neq 1 - \frac{1}{n}\) for any integer \(n \geq 1\).
In this section we study the spectrum of \(m\)-partial isometries. In particular, we prove a sharp spectral-inclusion theorem. We begin with the following lemma.
Lemma 17. Let \(K \subseteq \mathbb{C}\) be compact and \(r>0\). If \(\partial K\subseteq r\mathbb{D}^-\), then \(K\subseteq r\mathbb{D}^-\).
Proof. Since \(K\) is compact, the continuous function \(z\mapsto |z|\) attains its maximum on \(K\). Thus, there exists a \(w\in K\) such that \(|w|=\max_{z\in K}|z|\). If \(w \in \operatorname{int} K\), then \(K\) contains points of modulus greater than \(|w|\), which is impossible. Thus, \(w \in \partial K \subseteq r \mathbb{D}^-\) and hence each \(z \in K\) satisfies \(|z| \leq |w| \leq r\), so \(K \subseteq r\mathbb{D}^-\). ◻
The computation below owes much to [2] and [27].
Theorem 18. Let \(T\in \mathcal{B}(\mathcal{H})\) be an \(m\)-partial isometry.
For \(m\) odd, \(\sigma(T) \subseteq \mathbb{D}^-\). In particular, \(r(T) \leq 1.\)
For \(m\) even, \(\sigma(T)\subseteq (\sqrt{2}\,\mathbb{D})^-\) and \(\sigma_{\mathrm{p}}(T) \subseteq \sqrt{2}\,\mathbb{D}\). In particular, \(r(T) \leq \sqrt{2}.\)
Proof. Since \(\sigma(T)\) is compact and \(\partial\sigma(T)\subseteq \sigma_{\mathrm{ap}}(T)\), Lemma 17 says that it suffices to establish the desired inclusion for the approximate point spectrum.
Let \(\lambda \in \sigma_{\mathrm{ap}}(T)\) and let \(\mathbf{x}_n\) be unit vectors such that \(T\mathbf{x}_n = \lambda \mathbf{x}_n + \mathbf{o}_n(1)\), in which \(\mathbf{o}_n(1)\) is a sequence of vectors that tends to \(\mathbf{0}\) as \(n \to \infty\). The boundedness of \(T\) ensures that \(T^k \mathbf{x}_n = \lambda^k \mathbf{x}_n + \mathbf{o}_n(1)\) for each integer \(k\geq 1\). Therefore, \[\begin{align} 1 &= \| \mathbf{x}_n \|^2 \geq \| P_{\mathcal{N}(T)} \mathbf{x}_n \|^2 = \langle (-1)^m \beta_{m}(T)\mathbf{x}_n,\mathbf{x}_n \rangle \label{eq:NonNegP}\\ &= \sum_{k=0}^m (-1)^k \binom{m}{k} \langle T^{*k}T^k\mathbf{x}_n, \mathbf{x}_n \rangle = \sum_{k=0}^m (-1)^k \binom{m}{k} \| T^k \mathbf{x}_n \|^2 \nonumber \\ &= \sum_{k=0}^m (-1)^k \binom{m}{k} |\lambda|^{2k} + o_n(1) = (1 - |\lambda|^2)^m + o_n(1), \nonumber \end{align}\tag{8}\] in which \(o_n(1)\) is a scalar sequence tending to \(0\). We conclude that \[0 \leq (1 - |\lambda|^2)^m \leq 1,\] the lower bound coming from the nonnegativity of \(\| P_{\mathcal{N}(t)}\mathbf{x}_n \|\) in 8 . If \(m\) is odd, then \(|\lambda| \leq 1\). If \(m\) is even, then \((1 - |\lambda|^2) \in [-1,1]\) and hence \(|\lambda|\leq \sqrt{2}\). Since \(\sigma_{\mathrm{p}}(T) \subseteq \sigma_{\mathrm{ap}}(T)\), it suffices to show that \(T\) has no eigenvalues of modulus \(\sqrt{2}\). Suppose toward a contradiction that \(T \mathbf{x} = \lambda \mathbf{x}\), in which \(|\lambda|=\sqrt{2}\) and \(\mathbf{x}\) is a unit vector. Then 8 ensures that \(1 = \| P_{\mathcal{N}(T)}\mathbf{x} \|^2\), so \(\mathbf{x} \in \mathcal{N}(T)\), a contradiction. ◻
Example 19. Let \(m\) be odd and \(0 < a < 1\). Then \(T = \big[ \begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}\big]\), in which \[b= a\sqrt{ \frac{(1-a^2)^m}{1 - (1-a^2)^m } }\] is a strict \(m\)-partial isometry with \(\sigma_{\mathrm{p}}(T) = \{0,a\}\). Thus, Theorem 18.a is sharp for all odd \(m\).
Example 20. Let \(m\) be even and \(0 < a < \sqrt{2}\). Then \(T = \big[ \begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}\big]\), in which \[b = \frac{a (a^2-1)^{m/2}}{ \sqrt{ 1 - (a^2-1)^m } },\] is a strict \(m\)-partial isometry with \(\sigma_{\mathrm{p}}(T) = \{0,a\}\). Thus, Theorem 18.b is sharp for all even \(m\).
Remark 21. Any compact subset of \(\mathbb{D}^-\) (respectively, \((\sqrt{2}\,\mathbb{D})^-\)) that contains \(0\) can be the spectrum of an \(m\)-partial isometry with \(m\) odd (respectively, even). Simply take direct sums of the matrices in the previous examples.
Theorem 18 gives spectral bounds for \(m\)-partial isometries. We next record two boundary consequences.
Corollary 22. Let \(T\in\mathcal{B}(\mathcal{H})\) be an \(m\)-partial isometry.
If \(m\) is odd, then \(\sigma(T)\cap\mathbb{T}=\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\).
If \(m\) is even, then \(\sigma(T)\cap\sqrt{2}\,\mathbb{T}=\sigma_{\mathrm{ap}}(T)\cap\sqrt{2}\,\mathbb{T}\) and \(\sigma_{\mathrm{p}}(T)\cap\sqrt{2}\,\mathbb{T}=\varnothing\).
Proof. We use the standard inclusion \(\partial\sigma(T)\subseteq\sigma_{\mathrm{ap}}(T)\).
(a) If \(m\) is odd, then Theorem 18 gives \(\sigma(T)\subseteq\mathbb{D}^-\). Hence every point of \(\sigma(T)\cap\mathbb{T}\) belongs to \(\partial\sigma(T)\), and therefore to \(\sigma_{\mathrm{ap}}(T)\). The reverse inclusion is automatic.
(b) If \(m\) is even, then Theorem 18 gives \(\sigma(T)\subseteq\sqrt{2}\,\mathbb{D}^-\). Thus, every point of \(\sigma(T)\cap\sqrt{2}\,\mathbb{T}\) belongs to \(\partial\sigma(T)\), and hence to \(\sigma_{\mathrm{ap}}(T)\). The reverse inclusion is automatic. Finally, Theorem 18 also gives \(\sigma_{\mathrm{p}}(T)\subseteq\sqrt{2}\,\mathbb{D}\), so \(\sigma_{\mathrm{p}}(T)\cap\sqrt{2}\,\mathbb{T}=\varnothing\). ◻
Proposition 23. Let \(T\in\mathcal{B}(\mathcal{H})\) be an \(m\)-partial isometry.
If \(\lambda\in\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\), then \(\overline{\lambda}\in\sigma_{\mathrm{ap}}(T^*)\).
If \(\lambda\in\sigma_{\mathrm{p}}(T)\cap\mathbb{T}\), then \(\overline{\lambda}\in\sigma_{\mathrm{p}}(T^*)\).
If \(T\mathbf{x}=\lambda \mathbf{x}\) and \(T\mathbf{y}=\mu \mathbf{y}\) with \(\lambda\neq\mu\), and \(|\lambda|=1\) or \(|\mu|=1\), then \(\mathbf{x}\perp \mathbf{y}\).
Let \(\lambda\in\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\) and \(\mu\in\sigma_{\mathrm{ap}}(T)\) with \(\lambda\neq\mu\). If \(\mathbf{x}_n\) and \(\mathbf{y}_n\) are unit approximate eigensequences for \(\lambda\) and \(\mu\), respectively, then \(\langle{\mathbf{x}_n},{\mathbf{y}_n}\rangle\to0\).
Proof. (a) Let \(\lambda\in\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\) and choose unit vectors \(\mathbf{x}_n\) such that \((T-\lambda I)\mathbf{x}_n\to0\). The proof of Theorem 18 shows that \[\| P_{\mathcal{N}(T)}\mathbf{x}_n \|^2 =(1-|\lambda|^2)^m+\mathbf{o}_n(1)=\mathbf{o}_n(1).\] Consequently, \(\beta_m(T)\mathbf{x}_n=(-1)^m P_{\mathcal{N}(T)}\mathbf{x}_n\to0\). On the other hand, since \(T^kx_n=\lambda^kx_n+\mathbf{o}_n(1)\) for each fixed \(k\), we have \[\begin{align} \beta_m(T)\mathbf{x}_n &=\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}T^{*k}T^k \mathbf{x}_n\\ &=\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}\lambda^kT^{*k} \mathbf{x}_n+\mathbf{o}_n(1) \\ &=(\lambda T^*-I)^m \mathbf{x}_n+\mathbf{o}_n(1). \end{align}\] Thus, \((\lambda T^*-I)^m \mathbf{x}_n\to0\). Since \(|\lambda|=1\), we have \(\lambda T^*-I=\lambda(T^*-\overline{\lambda} I)\), so \[(T^*-\overline{\lambda} I)^m\mathbf{x}_n\to0.\] If \(T^*-\overline{\lambda} I\) were bounded below, then so would its \(m\)th power be, which is impossible because \(\| \mathbf{x}_n \|=1\). Therefore, \(\overline{\lambda}\in\sigma_{\mathrm{ap}}(T^*)\).
(b) Suppose that \(\lambda\in\sigma_{\mathrm{p}}(T)\cap\mathbb{T}\) and \(T\mathbf{x}=\lambda \mathbf{x}\) with \(\mathbf{x}\neq \mathbf{0}\). Then \[\| P_{\mathcal{N}(T)}\mathbf{x} \|^2=(1-|\lambda|^2)^m\| \mathbf{x} \|^2=0,\] so \(P_{\mathcal{N}(T)}\mathbf{x}=0\) and hence \(\beta_m(T)\mathbf{x}=0\). As above, but with an exact eigenvector, \[\beta_m(T)\mathbf{x}=(\lambda T^*-I)^m\mathbf{x}.\] Thus, \((T^*-\overline{\lambda} I)^m \mathbf{x}=\mathbf{0}\). Let \(q\geq1\) be minimal such that \((T^*-\overline{\lambda} I)^q \mathbf{x}=\mathbf{0}\). Then \(\mathbf{y}=(T^*-\overline{\lambda} I)^{q-1} \mathbf{x} \neq \mathbf{0}\) and \((T^*-\overline{\lambda} I)\mathbf{y}=0\), so \(\overline{\lambda}\in\sigma_{\mathrm{p}}(T^*)\).
(c) Without loss of generality, suppose that \(|\lambda|=1\). Then \(P_{\mathcal{N}(T)}\mathbf{x}= \mathbf{0}\) and hence Lemma 1 yields \[\begin{align} 0 &=\langle P_{\mathcal{N}(T)}\mathbf{x}, \mathbf{y}\rangle =\sum_{k=0}^{m}(-1)^k\binom{m}{k}\langle T^k \mathbf{x} , T^k\mathbf{y} \rangle\\ &=\sum_{k=0}^{m}(-1)^k\binom{m}{k}(\lambda\overline{\mu})^k\langle \mathbf{x},\mathbf{y} \rangle =(1-\lambda\overline{\mu})^m\langle \mathbf{x},\mathbf{y}\rangle. \end{align}\] Since \(|\lambda|=1\) and \(\lambda\neq\mu\), we get \(\langle \mathbf{x}, \mathbf{y} \rangle=0\).
(d) Let \(\lambda\in\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\) and \(\mu\in\sigma_{\mathrm{ap}}(T)\) with \(\lambda\neq\mu\). Let \(\mathbf{x}_n\) and \(\mathbf{y}_n\) be unit approximate eigensequences for \(\lambda\) and \(\mu\). As above, \(\| P_{\mathcal{N}(T)}\mathbf{x}_n \|\to 0\) and \[\langle {P_{\mathcal{N}(T)}\mathbf{x}_n, \mathbf{y}_n}\rangle =(1-\lambda\overline{\mu})^m\langle \mathbf{x}_n,\mathbf{y}_n \rangle+\mathbf{o}_n(1).\] The left side tends to \(0\) and \(1-\lambda\overline{\mu}\neq0\) since \(|\lambda|=1\) and \(\lambda\neq\mu\), so \(\langle \mathbf{x}_n,\mathbf{y}_n\rangle\to0\). ◻
Corollary 24. Suppose that \(T\) and \(T^*\) are \(m\)-partial isometries. Then \[\lambda\in\sigma_{\mathrm{ap}}(T)\cap\mathbb{T} \quad\Longleftrightarrow\quad \overline{\lambda}\in\sigma_{\mathrm{ap}}(T^*)\cap\mathbb{T}\] and \[\lambda\in\sigma_{\mathrm{p}}(T)\cap\mathbb{T} \quad\Longleftrightarrow\quad \overline{\lambda}\in\sigma_{\mathrm{p}}(T^*)\cap\mathbb{T}.\] Consequently, \[\sigma(T)\cap\mathbb{T}=\sigma_{\mathrm{ap}}(T)\cap\mathbb{T} \quad\text{and}\quad \sigma(T^*)\cap\mathbb{T}=\sigma_{\mathrm{ap}}(T^*)\cap\mathbb{T}.\]
Proof. The equivalences follow by applying Proposition 23 to \(T\) and \(T^*\). We prove the spectral equality for \(T\); the proof for \(T^*\) is identical. We use the decomposition \[\label{eq:spectrum-decomposition} \sigma(T)=\sigma_{\mathrm{ap}}(T)\cup\{\lambda\in\mathbb{C}:\overline{\lambda}\in\sigma_{\mathrm{p}}(T^*)\}.\tag{9}\] Let \(\lambda\in\sigma(T)\cap\mathbb{T}\). If \(\lambda\notin\sigma_{\mathrm{ap}}(T)\), then 9 implies that \(\overline{\lambda}\in\sigma_{\mathrm{p}}(T^*)\cap\mathbb{T}\). By Proposition 23, \(\lambda\in\sigma_{\mathrm{p}}(T)\subseteq\sigma_{\mathrm{ap}}(T)\), a contradiction. Hence \(\sigma(T)\cap\mathbb{T}\subseteq\sigma_{\mathrm{ap}}(T)\cap\mathbb{T}\). The reverse inclusion is automatic. ◻
Remark 25. Proposition 23 and Corollary 24 are reduction-free analogues of [16]. Their results assume that the relevant kernels are reducing subspaces. Here, no reducing hypothesis is imposed. Under the reducing hypothesis, the nonzero approximate spectral values are forced onto \(\mathbb{T}\), so Proposition 23 and Corollary 24 recover the corresponding conclusions of [16].
Since partial isometries have isometric dilations, it is natural to ask whether \(m\)-partial isometries possess \(m\)-isometric dilations; see Question 4 in Section 6.
Theorem 26. Let \(T\in \mathcal{B}(\mathcal{H})\) be an \(m\)-partial isometry.
If \(m\) is even and \(r(T) > 1\), then \(T\) has no \(m'\)-isometric dilation for any \(m'\).
If \(m = 3\), then \(T\) has a \(3\)-isometric dilation.
Proof. (a) If \(T\) has an \(m'\)-isometric dilation, then \(\|T^n\|^2/ n^{m'-1}\) is bounded as \(n\to \infty\) [2]. However, the spectral-radius formula ensures that \(1<r(T) \leq \| T^n\|^{1/n}\) for all \(n \geq 1\), so the powers of \(T\) grow exponentially.
(b) Lemma 1 implies that \(\beta_3(T) \leq 0\) and [13] ensures that \(T\) has a \(3\)-isometric dilation. ◻
We turn our attention to the problem of similarity to an \(m\)-isometry or \(m\)-partial isometry. The next result proves that the challenge of similarity to an \(m\)-isometry (or an \(m\)-partial isometry) can be interpreted as a renormalization problem.
Theorem 27. An operator \(T \in \mathcal{B}(\mathcal{H})\) is similar to an \(m\)-isometry (respectively, \(m\)-partial isometry) if and only if there exists an equivalent Hilbert norm such that \(T\) is an \(m\)-isometry (respectively, \(m\)-partial isometry) with respect to this norm.
Proof. Suppose that \(T=P^{-1} S P\), in which \(P \in \mathcal{B}(\mathcal{H})\) is invertible and \(S\) is an \(m\)-isometry (respectively, \(m\)-partial isometry). Then \(|\!|\!| \mathbf{x} |\!|\!| = \langle P^{*} P x, x \rangle^{1 / 2}\) is an equivalent Hilbert norm and \(T\) is an \(m\)-isometry (respectively, \(m\)-partial isometry) with respect to this norm. For the converse, suppose that \(|\!|\!| \,\cdot\, |\!|\!|\) is a Hilbert norm with inner product \(\langle\!\langle \, \cdot\, ,\, \cdot\, \rangle\!\rangle\). Then there is a positive, invertible \(A \in \mathcal{B}(\mathcal{H})\) such that \(\langle\!\langle \mathbf{x}, \mathbf{y} \rangle\!\rangle = \langle A \mathbf{x}, \mathbf{y} \rangle\). Let \(P = A^{1/2}\). Then for any \(\mathbf{x} \in \mathcal{H}\) (respectively, for any \(\mathbf{x} \in \mathcal{N}(T)^{\perp_{\textrm{New}}}=P^{-1}((P\mathcal{N}(T))^\perp)=P^{-1}(\mathcal{N}(PTP^{-1})^\perp))\), \[\sum_{k=0}^{m} (-1)^{k} \binom{m}{k} \, |\!|\!| T^{m-k} \mathbf{x} |\!|\!|^2 = 0 \iff \sum_{k=0}^{m} (-1)^{k} \binom{m}{k} \, \|P T^{m-k} \mathbf{x}\|^2 = 0 .\] Set \(\mathbf{x} = P^{-1} \mathbf{y}\) in the last equality and we conclude that \(T\) is similar to an \(m\)-isometry (respectively, \(m\)-partial isometry). ◻
The spectrum of an invertible \(m\)-isometry is contained in the unit circle [2], hence so is the spectrum of any operator similar to an invertible \(m\)-isometry. Below we show that, in a sense, the converse result is also true in finite-dimensional spaces.
Theorem 28. Let \(\mathcal{H}\) be a finite-dimensional Hilbert space and let \(T\in \mathcal{B}(\mathcal{H})\). Let \(m \geq 1\) be odd. The following are equivalent.
\(T\) is similar to an \(m\)-isometry.
There is an equivalent Hilbert norm with respect to which \(T\) is an \(m\)-isometry.
\(\sup_{n\in \mathbb{Z}_{\neq 0}} \frac{\| T^{n}\|^{2}}{|n|^{m-1}} < \infty\).
\(\sigma(T) \subset \mathbb{T}\) and the maximum Jordan block size of \(T\) is at most \(\frac{m+1}{2}\).
Proof. (a) \(\Leftrightarrow\) (b) This equivalence follows from two facts. First, every equivalent Hilbert norm on \(\mathcal{H}\) is of the form \(|\!|\!| \mathbf{x} |\!|\!|=\| A\mathbf{x} \|\) for a positive invertible \(A \in \mathcal{B}(\mathcal{H})\). Second, if \(T = Q^{-1}SQ\) and \(Q = UA\), in which \(U\) is unitary and \(A\) is positive and invertible, then \(T = P^{-1}(U^*SU)P\). Since \(m\)-isometries are preserved by unitary equivalence, the desired result follows from a straightforward use of 2 .
(a) \(\Rightarrow\) (c) Suppose that \(T\) is similar to an \(m\)-isometry. Then \(\sigma(T) \subset \mathbb{T}\) [2], so \(T\) is invertible. The proof of [2] ensures that \(\| T^n \|^2/n^{m-1}\) is bounded as \(n \to \infty\). Since 4 ensures that \(T^{-1}\) is similar to an \(m\)-isometry, we obtain (c).
(c) \(\Rightarrow\) (d) Since \(r(T) = \lim_{n\to \infty} \sqrt[n]{\| T^{n}\|}\), we infer from (c) that \(r(T) = r(T^{-1}) = 1\). Using the fact that \(\sigma(T^{-1}) = \{ 1/ \lambda : \lambda \in \sigma(T) \}\), it follows that \(\sigma(T) \subseteq \mathbb{T}\). Suppose that the largest Jordan block of \(T\) is \(J_k(\lambda)\) and \(\mathbf{v}\in \mathcal{H}\) is a generalized eigenvector of type \(k\) [28]. Then [28] ensure that for every \(n\geq 1\), \[T^n \mathbf{v} = \sum_{j=0}^{k-1} \binom{n}{j}\lambda^{n-j}(T-\lambda I)^j \mathbf{v}.\] Thus, \(\| T^{n}\mathbf{v} \|^{2}\) is a polynomial in \(n\) of degree \(2k-2\). Condition (c) ensures that \(2k-2 \leq m-1\); that is, \(k \leq \frac{m+1}{2}\).
(d) \(\Rightarrow\) (a) Without loss of generality, suppose that \(\mathcal{H}= \mathbb{C}^{n}\) endowed with the standard inner product. Then \(T = P^{-1}JP\), in which \(P \in \mathcal{M}_n\) is invertible and \(J = J_1 \oplus J_2 \oplus \cdots \oplus J_p\) with each \(J_i\) a Jordan block for some \(\lambda \in \sigma(T)\). Condition (d) ensures that \(T\) is similar to \(V + N\), in which \(V\) is unitary, \(N\) is nilpotent of order at most \(\frac{m+1}{2}\), and \(VN = NV\). Thus, \(T\) is \(m\)-isometric by [9]. ◻
The previous result does not hold if \(\mathcal{H}\) is infinite dimensional.
Example 29. Let \(m\geq 2\) and define \[\omega_{n} = \begin{cases} 1 & \text{if n\leq 1},\\ \frac{n^{m-1}}{(n-1)^{m-1}} & \text{if n \geq 2}. \end{cases}\] Let \(T \in \mathcal{B}(\ell^{2}(\mathbb{Z}))\) be the bilateral weighted shift with weights \(\omega_n\); that is, \(T\mathbf{e}_{n} = \omega_{n+1}\mathbf{e}_{n+1}\) for \(n\in \mathbb{Z}\). Since \(\omega_n \to 1\) as \(n \to \pm \infty\), it follows that \(\sigma(T) = \mathbb{T}\) [29]. In particular, \(T\) is invertible. For \(p \geq 1\), observe that \[T^{p} \mathbf{e}_{n} = \begin{cases} \frac{(n+p)^{m-1}}{n^{m-1}}\mathbf{e}_{n+1} & \text{for n\geq 1},\\ (n+p)^{m-1}\mathbf{e}_{n+1} & \text{for -p+1 < n < 1},\\ \mathbf{e}_{n+1} & \text{for n\leq -p+1}. \end{cases}\] Thus, \[\label{ExFormPolyPositivePowers} \| T^{p}\|^{2} = (1+p)^{m-1}.\tag{10}\] Since \(\omega_{n} \geq 1\) for each \(n\in\mathbb{Z}\), we have \(\| T\mathbf{x} \| \geq \| \mathbf{x} \|\) for \(\mathbf{x}\in \ell^{2}(\mathbb{Z})\). Hence \(\| T^{-p} \| \leq 1\) for \(p\geq 1\). Combining this with 10 shows that condition (c) of Theorem 28 holds.
Suppose toward a contradiction that there is an \(m\)-isometry \(V\in \mathcal{B}(\mathcal{H})\) such that \(V = STS^{-1}\) for some invertible \(S \in \mathcal{B}(\mathcal{H})\). Since \(T\) is invertible, so is \(V\) and hence \(V^{-1}\) is \(m\)-isometric; see 4 . For every \(\mathbf{u}\in \ell^{2}(\mathbb{Z})\) there is a polynomial \(p_{\mathbf{u}}\in \mathbb{R}[x]\) of degree at most \(m-1\) such that \(p_{\mathbf{u}}(n) = \| V^{-n}S\mathbf{u} \|^{2}\) for all \(n\geq 0\) [2]. Then for all \(\mathbf{u} \in \ell^2(\mathbb{Z})\) and \(n \geq 0\), \[p_{\mathbf{u}}(n) = \| V^{-n}S\mathbf{u} \|^{2} = \| ST^{-n}\mathbf{u} \|^{2} \leq \| S \|^{2} \| \mathbf{u} \|^{2}.\] The polynomial \(p_{\mathbf{u}}\) is bounded and hence constant. Thus, \(\| V^{-1}S\mathbf{u} \|^{2} = p_{\mathbf{u}}(1) = p_{\mathbf{u}}(0) = \| S\mathbf{u} \|^{2}\) for all \(\mathbf{u}\in \ell^{2}(\mathbb{Z})\), which means that \(V^{-1}\) isometric. Since \(V\) is also invertible, it is unitary. However, 10 ensures that \[(1+n)^{m-1} = \lVert T^{n}\rVert^{2}\leq \lVert S\rVert^{2}\lVert S^{-1}\rVert^{2}\lVert V^{n}\rVert^{2} = \lVert S\rVert^{2}\lVert S^{-1}\rVert^{2}\] for all \(n \geq 1\), which contradicts the assumption that \(m \geq 2\).
The main result of this section is the following theorem, which describes strict \(m\)-isometries on finite-dimensional Hilbert spaces.
Theorem 30. Let \(\mathcal{H}\) be an \(n\)-dimensional Hilbert space and let \(m\) be an odd integer such that \(3 \leq m \leq 2n-1\). Then \(T \in \mathcal{B}(\mathcal{H})\) is similar to a strict \(m\)-isometry if and only if \(T = U + Q\), in which \(U\) is diagonalizable with at most \(\frac{2n+1-m}{2}\) distinct unimodular eigenvalues, \(Q\) is nilpotent of order \(\frac{m+1}{2}\), and \(UQ = QU\).
Proof. \((\Leftarrow)\) This follows from [9].
\((\Rightarrow)\) Suppose that \(T \in \mathcal{B}(\mathcal{H})\) is similar to a strict \(m\)-isometry. On a finite-dimensional Hilbert space, strict \(m\)-isometries with \(m\) odd are precisely those operators of the form \(U+Q\), in which \(U\) is unitary, \(Q\) is nilpotent of order \(\frac{m+1}{2}\), and \(UQ = QU\) [9]. Thus, there exists an invertible \(P\), unitary \(U\), and nilpotent \(Q\) of order \(\frac{m+1}{2}\) such that \(T = PUP^{-1} + PQP^{-1}\) and \(UQ = QU\). Then \(PUP^{-1}\) is diagonalizable with unimodular eigenvalues, \(PQP^{-1}\) is nilpotent of order \(\frac{m+1}{2}\), and \(PUP^{-1}\) and \(PQP^{-1}\) commute. Suppose that \(T\) has \(r\) distinct eigenvalues. Aside from the largest Jordan block, the remaining \(r-1\) eigenvalues account for at least \(r-1\) dimensions. Therefore, the largest Jordan block has size at most \(n-(r-1)\). If \(r \geq \frac{2n+3-m}{2}\), then \(n-(r-1) \leq \frac{m-1}{2}<\frac{m+1}{2}\), which contradicts [9]. ◻
Corollary 31. Let \(\mathcal{H}\) be an \(n\)-dimensional Hilbert space. Then \(T \in \mathcal{B}(\mathcal{H})\) is similar to a strict \((2n-1)\)-isometry if and only if \(T = \alpha I + Q\), in which \(\alpha \in \mathbb{T}\) and \(Q\) is nilpotent of order \(n\).
Proof. Let \(m = 2n-1\) above and deduce that \(U\) has exactly one eigenvalue. ◻
For each \(A, B \in \mathcal{B}(\mathcal{H})\), let \(L_{A}, R_{B} \in \mathcal{B}(\mathcal{B}(\mathcal{H}))\) denote left multiplication by \(A\) and right multiplication by \(B\), respectively, defined for \(X \in \mathcal{B}(\mathcal{H})\) by \[L_{A}(X) = A X \qquad \text{and} \qquad R_{B}(X) = X B.\] For \(m \in \mathbb{Z}_{\geq 1}\), define \[\begin{align} \Delta_{A^{*}, B}^{m}(X) &= (L_{A^{*}} R_{B} - I)^{m}(X) \\ &= \bigg( \sum_{k=0}^{m} (-1)^{k} \binom{m}{k} (L_{A^{*}} R_{B})^{m-k} \bigg)(X) \\ &= \sum_{k=0}^{m} (-1)^{k} \binom{m}{k} \, A^{*(m-k)} X B^{m-k}. \end{align}\] Then \(\Delta^1_{T^*,T} = (-1)^0 T^*T + (-1)^1 I = T^*T - I\) and \[\beta_{m}(T) = (-1)^{m} \Delta_{T^{*}, T}^{m}(I).\]
The next theorem explores when an operator is similar to an \(m\)-partial isometry.
Theorem 32. Let \(T \in \mathcal{B}(\mathcal{H})\) and \(m \in \mathbb{N}\). Then \(T\) is similar to an \(m\)-partial isometry if and only if there exists an \(S \in \mathcal{B}(\mathcal{H})\) such that \(T\Delta_{S,T}^{m}(I)=0\) and \(T^* \sim_{+} S\).
Proof. \((\Rightarrow)\) Suppose that \(V = A^{-1}TA\) for some \(m\)-partial isometry \(V\). Then \[\begin{align} 0&= \sum_{k=0}^{ m }(-1)^{k}\binom{m}{k}V V^{* m-k} V^{m-k}\\ &= A^{-1}\Bigg (\sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} T A A^{*} T^{* m-k}(A^{-1})^{*} A^{-1}T^{m-k}\Bigg )A, \end{align}\] so, \[T\sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} (A A^{*}) T^{* m-k}(A A^*)^{-1}T^{m-k}=0.\] Then \(S=(AA^*) T^*(AA^*)^{-1}\) satisfies \(T^* \sim_+ S\) and \[\sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} TS^{ m-k} T^{m-k}=0.\]
\((\Leftarrow)\) Suppose that \(T^{*}=P S P^{-1}\) for some \(P>0\) and \[T\triangle_{S,T}^{m}(I) = T \sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} S^{ m-k} T^{m-k}=0.\] Then \[T\sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} P^{-1}T^{*{m-k}}P T^{m-k}=0.\] Let \(P^{1 / 2}\) be the positive square root of \(P\) and let \(V=P^{1 / 2} T P^{-1 / 2}\), so that \[\begin{align} & V\sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} V^{* m-k} V^{m-k}\\ &= P^{1 / 2}\Bigg( \sum_{k=0}^{ m }(-1)^{k}\binom{m}{k} T P^{-1 / 2} P^{-1 / 2} T^{* m-k}P^{1 / 2} P^{1 / 2}T^{m-k}\Bigg)P^{-1 / 2}\\ &= 0. \end{align}\] Thus, \(V\) is an \(m\)-partial isometry that is similar to \(T\). ◻
The following theorem generalizes [20] and [19].
Theorem 33. Let \(T\in\mathcal{B}(\mathcal{H})\) and let \(m\in\mathbb{N}\). Then \(T\) is similar to an \(m\)-partial isometry whose adjoint is also an \(m\)-partial isometry if and only if there exists an equivalent Hilbertian norm \(|\!|\!| \cdot |\!|\!|\) on \(\mathcal{H}\) such that if \(T^{\sharp}\) denotes the adjoint of \(T\) with respect to the inner product associated with this norm, then \[T\Delta^m_{T^{\sharp},T}(I)=0 \qquad\text{and}\qquad T^{\sharp}\Delta^m_{T,T^{\sharp}}(I)=0.\] If \(m\) is odd, then \(T\) and \(T^{\sharp}\) are \(m\)-concave with respect to \(|\!|\!| \cdot |\!|\!|\); that is, for all \(\mathbf{x} \in \mathcal{H}\), \[\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}|\!|\!| T^{k}\mathbf{x} |\!|\!|^{2}\leq 0 \quad \text{and}\quad \sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}|\!|\!| (T^{\sharp})^{k} \mathbf{x} |\!|\!|^{2}\leq 0.\]
Proof. \((\Rightarrow)\) Suppose that \(T\) is similar to an \(m\)-partial isometry \(V\) whose adjoint \(V^*\) is also an \(m\)-partial isometry. Then there exists an invertible \(X\in\mathcal{B}(\mathcal{H})\) such that \(V=XTX^{-1}\). Define an inner product on \(\mathcal{H}\) by \[\langle\!\langle \mathbf{u}, \mathbf{v} \rangle\!\rangle = \langle X\mathbf{u},X\mathbf{v}\rangle,\] for \(\mathbf{u},\mathbf{v}\in\mathcal{H}\); the associated norm is \(|\!|\!| \mathbf{u} |\!|\!| = \langle\!\langle \mathbf{u},\mathbf{u} \rangle\!\rangle^{1/2} = \| X \mathbf{u} \|\). Since \(X^*X\) is positive and invertible, \(|\!|\!| \cdot |\!|\!|\) is a Hilbertian norm equivalent to the original one. Let \(T^{\sharp}=X^{-1}V^*X\) denote the adjoint of \(T\) with respect to this new inner product.
Since \(V\) is an \(m\)-partial isometry, \(V\beta_m(V)=0\). Moreover, \(X\) is a unitary operator from \((\mathcal{H},\langle\!\langle \cdot,\cdot \rangle\!\rangle)\) onto \((\mathcal{H},\langle\cdot,\cdot\rangle)\), so \[\beta_m^{\sharp}(T)=X^{-1}\beta_m(V)X,\] in which \[\beta_m^{\sharp}(T) = \sum_{k=0}^m(-1)^{m-k}\binom{m}{k}(T^{\sharp})^kT^k.\] Therefore, \[T\beta_m^{\sharp}(T) = X^{-1}V\beta_m(V)X = 0.\] Since \[\beta_m^{\sharp}(T) = \sum_{k=0}^m(-1)^{m-k}\binom{m}{k}(T^{\sharp})^kT^k = \Delta^m_{T^{\sharp},T}(I),\] we obtain \[T\Delta^m_{T^{\sharp},T}(I)=0.\] Similarly, since \(V^*\) is an \(m\)-partial isometry, \(V^*\beta_m(V^*)=0\). Since \(T^{\sharp}=X^{-1}V^*X\), it follows that \(T^{\sharp}\beta_m^{\sharp}(T^{\sharp}) =X^{-1}V^*\beta_m(V^*)X =0\). Moreover, since \((T^{\sharp})^\sharp=T\), \[\beta_m^{\sharp}(T^{\sharp}) = \sum_{k=0}^m(-1)^{m-k}\binom{m}{k}T^k(T^{\sharp})^k = \Delta^m_{T,T^{\sharp}}(I).\] Consequently, \(T^{\sharp}\Delta^m_{T,T^{\sharp}}(I)=0\).
If \(m\) is odd, then the \(m\)-concavity of \(T\) follows from Lemma 1 applied to \(T\) on the Hilbert space \((\mathcal{H},\langle\!\langle \cdot,\cdot \rangle\!\rangle)\). Similarly, since \(T^{\sharp}\) is also an \(m\)-partial isometry with respect to \(|\!|\!| \cdot |\!|\!|\), Lemma 1 applied to \(T^{\sharp}\) gives the \(m\)-concavity of \(T^{\sharp}\).
(\(\Leftarrow)\) Suppose that there exists an equivalent Hilbertian norm \(|\!|\!| \cdot |\!|\!|\) on \(\mathcal{H}\) such that \[T\Delta^m_{T^{\sharp},T}(I)=0 \qquad\text{and}\qquad T^{\sharp}\Delta^m_{T,T^{\sharp}}(I)=0,\] in which \(T^{\sharp}\) denotes the adjoint of \(T\) with respect to this inner product associated with this norm. Since \[\Delta^m_{T^{\sharp},T}(I) = \sum_{k=0}^m(-1)^{m-k}\binom{m}{k}(T^{\sharp})^kT^k = \beta_m^{\sharp}(T),\] we get \(T\beta_m^{\sharp}(T)=0\). Thus, \(T\) is an \(m\)-partial isometry on \((\mathcal{H},|\!|\!| \cdot |\!|\!|)\). Since \((T^{\sharp})^\sharp=T\), \[\Delta^m_{T,T^{\sharp}}(I) =\sum_{k=0}^m(-1)^{m-k}\binom{m}{k}T^k(T^{\sharp})^k = \beta_m^{\sharp}(T^{\sharp}).\] The identities \[T^{\sharp}\Delta^m_{T,T^{\sharp}}(I)=0 \quad\text{and}\quad T^{\sharp}\beta_m^{\sharp}(T^{\sharp})=0\] are therefore equivalent. Thus, \(T^{\sharp}\) is also an \(m\)-partial isometry on \((\mathcal{H},|\!|\!| \cdot |\!|\!|)\).
Since \(|\!|\!| \cdot |\!|\!|\) is an equivalent Hilbertian norm, there exists a positive invertible \(G\in\mathcal{B}(\mathcal{H})\) such that \[\langle\!\langle \mathbf{x},\mathbf{y} \rangle\!\rangle = \langle G\mathbf{x},\mathbf{y}\rangle\] for all \(\mathbf{x},\mathbf{y}\in \mathcal{H}\). Let \(X=G^{1/2}\). Then \(X\) is a unitary operator from \((\mathcal{H},\langle\!\langle \cdot,\cdot \rangle\!\rangle)\) onto the original Hilbert space \((\mathcal{H},\langle\cdot,\cdot\rangle)\). Then \(V=XTX^{-1}\) is similar to \(T\). Since \(T\) is an \(m\)-partial isometry on \((\mathcal{H},|\!|\!| \cdot |\!|\!|)\), unitary equivalence ensures that \(V\) is an \(m\)-partial isometry on the original Hilbert space. Furthermore, \(V^*=XT^{\sharp} X^{-1}\). Since \(T^{\sharp}\) is an \(m\)-partial isometry on \((\mathcal{H},|\!|\!| \cdot |\!|\!|)\), unitary equivalence ensures that \(V^*\) is an \(m\)-partial isometry on the original Hilbert space. ◻
There are many potential avenues for additional investigation on the subject of \(m\)-partial isometries. We conclude this paper with several open questions that we hope will motivate further research.
Question 1. Let \(T\) be an \(m\)-partial isometry and let \[S=\sum_{k=1}^m (-1)^{k-1}\binom{m}{k}T^{*k}T^{k-1}\] be the generalized inverse of \(T\) appearing in Lemma 8. For \(m=1\), this reduces to \(S=T^*\), which is again a partial isometry whenever \(T\) is. For \(m\geq 2\), when is \(S\) necessarily an \(m\)-partial isometry? Similar to an \(m\)-partial isometry?
Question 2. In [19] Mbekhta proved that that \(T\) is a partial isometry if and only if it has a contractive generalized inverse. Does there exist a similar characterization of \(m\)-partial isometries?
In [20] the authors considered so-called regular operators; that is, operators \(T\in \mathcal{B}(\mathcal{H})\) with closed range such that \[\mathcal{N}(T) \subset \bigcap_{n\geq 1}\mathcal{R}(T^n).\] They showed that regular partial isometries are precisely direct sums of isometries, unitaries, and co-isometries.
Question 3. Do regular \(m\)-partial isometries have a decomposition as the direct sum of an \(m\)-isometry, invertible \(m\)-isometry, and the adjoint of \(m\)-isometry?
Question 4. Does every \(m\)-partial isometry \(T\) possess an \(m\)-isometric dilation if we assume \(r(T)\leq 1\)?
Garcia and Sherman proved that \(A \in \mathsf{M}_{n}\) is similar to a partial isometry if and only if (a) \(\sigma(A) \subseteq \mathbb{D}^{-}\); (b) if \(\zeta \in \sigma(A) \cap \mathbb{T}\), then its algebraic and geometric multiplicities are equal; and (c) \(\dim \mathcal{N}(A) \geq \dim \mathcal{N}(A-\lambda I)\) for each \(\lambda \in \sigma(A) \cap \mathbb{D}\) [30]. Does an analogous characterization exist for \(m\)-partial isometries?
Question 5. Can \(m\)-partial isometries on finite-dimensional Hilbert spaces be concretely classified up to similarity?
The condition \(T\beta_{2}(T) = 0\) can be written as \[T(2T^*-T^{*2}T)T = T,\] so \(S = 2T^*-T^{*2}T\) is a generalized inverse for \(T\) and \(ST\) is the orthogonal projection onto \(\mathcal{R}(T^*)\). Moreover, \(STS = S\), so \(S\) is almost the Moore-Penrose inverse of \(T\). However, \(TS\) need not be selfadjoint: if \[T = \begin{bmatrix} a & 0 \\ 1 & 0 \end{bmatrix}\] with \(|a|^2 = \frac{1+\sqrt{5}}{2}\), then \(T\) is a \(2\)-partial isometry and \[T S = \begin{bmatrix} -1 & 2a \\[2mm] \overline{a}\,\frac{1-\sqrt{5}}{2} & 2 \end{bmatrix}\] is not selfadjoint. All is not lost, however. The term “selfadjoint” above refers to the standard inner product on \(\mathbb{C}^2\). Since different inner products give rise to different adjoints, the next question is natural.
Question 6. If \(T\) is a \(2\)-partial isometry, is \(2T^*-T^{*2}T\) the Moore–Penrose inverse of \(T\) with respect to some inner product?