Schmidt decomposable products of projections

TL;DR

This paper characterizes when the product of two orthogonal projections on a Hilbert space admits a singular value decomposition, linking it to orthonormal bases, diagonalizability of their difference, and geometric properties of the Grassmann manifold.

Contribution

It provides a new characterization of operators as products of projections with SVD, connecting algebraic, geometric, and operator-theoretic perspectives.

Findings

Operators $T=PQ$ have SVD iff certain orthonormal bases exist.

$A=P-Q$ is diagonalizable if and only if $T$ has SVD.

Connections to Toeplitz, Hankel, Wiener-Hopf operators and Grassmannian geometry.

Abstract

We characterize operators $T=PQ$ ($P,Q$ orthogonal projections in a Hilbert space $H$) which have a singular value decomposition. A spatial characterizations is given: this condition occurs if and only if there exist orthonormal bases $\{\psi_n\}$ of $R(P)$ and $\{\xi_n\}$ of $R(Q)$ such that $\langle\xi_n,\psi_m\rangle=0$ if $n\ne m$. Also it is shown that this is equivalent to $A=P-Q$ being diagonalizable. Several examples are studied, relating Toeplitz, Hankel and Wiener-Hopf operators to this condition. We also examine the relationship with the differential geometry of the Grassmann manifold of underlying the Hilbert space: if $T=PQ$ has a singular value decomposition, then the generic parts of $P$ and $Q$ are joined by a minimal geodesic with diagonalizable exponent.

Equations

$$T=\sum _{n\ge 1}{s}_n⟨,{\xi }_n⟩{\psi }_n=\sum _{n\ge 1}{s}_n{\psi }_n\otimes {\xi }_n,$$

$$\overline{R(T)}\subset \mathcal{S},N(T{)}^{\perp }\subset \mathcal{T}and(\mathcal{S}\ominus \overline{R(T)})\oplus (\mathcal{T}\ominus N(T{)}^{\perp })\subset R(T{)}^{\perp }\cap N(T).$$

$$⟨{\psi }_n,{\xi }_m⟩=0forn\ne m.$$

$$P_{\mathcal{S}}{P}_{\mathcal{T}}=(\sum _{n\ge 1}⟨,{\psi }_n⟩{\psi }_n)(\sum _{m\ge 1}⟨,{\xi }_m⟩{\xi }_m)=\sum _{n\ge 1}⟨{\psi }_n,{\xi }_n⟩{\psi }_n\otimes {\xi }_n.$$

$$⟨{\psi }_n^{\prime },{\xi }_n⟩=|⟨{\psi }_n,{\xi }_n⟩|=s_n$$

$$P_{\mathcal{S}}{P}_{\mathcal{T}}=\sum _{n\ge 1}|⟨{\psi }_n,{\xi }_n⟩|{\psi }_n^{\prime }\otimes {\xi }_n.$$

$$T=\sum _{n\ge 1}{s}_n{\psi }_n\otimes {\xi }_n$$

$$T^{\ast }=\sum _{n\ge 1}{s}_n⟨,{\psi }_n⟩{\xi }_n,T{T}^{\ast }T=\sum _{n\ge 1}{s}_n^3⟨,{\xi }_n⟩{\psi }_n,$$

$$T^2=\sum _{n,m\ge 1}{s}_n{s}_m⟨{\psi }_n,{\xi }_n⟩⟨,{\xi }_n⟩{\psi }_n=\sum _{n\ge 1}{s}_n\left(\sum _{m\ge 1}⟨,s_m⟨{\xi }_n,{\psi }_m⟩{\xi }_m\right){\psi }_n.$$

$$\sum _{m\ge 1}{s}_n{s}_m⟨{\xi }_n,{\psi }_m⟩{\xi }_m=s_n^3{\xi }_n.$$

$$⟨\psi ,\xi ⟩=0,$$

$$(\mathcal{S}\ominus \overline{R(T)})\oplus (\mathcal{T}\ominus N(T{)}^{\perp })\subset R(T{)}^{\perp }\cap N(T).$$

$$A{\xi }_k=P{\xi }_k-Q{\xi }_k=PQ{\xi }_k-{\xi }_k=T{\xi }_k-{\xi }_k=s_k{\psi }_k-{\xi }_k$$

$$A{\psi }_k=P{\psi }_k-Q{\psi }_k=P{\psi }_k-QP{\psi }_k={\psi }_k-T^{\ast }{\psi }_k={\psi }_k-s_k{\xi }_k.$$

$$A^2{\xi }_k=(1-s_k^2){\xi }_k.A^2{\psi }_k=(1-s_k{)}^2{\psi }_k.$$

$${\nu }_k=((1-s_k^2{)}^{1/2}-1){\xi }_k+s_k{\psi }_{k}and{\omega }_k=(-(1-s_k^2{)}^{1/2}-1){\xi }_k+s_k{\psi }_k$$

$$N(A-1)=R(P)\cap N(Q),N(A+1)=N(P)\cap R(Q),$$

$${\mathcal{H}}_0=\{N(A)\oplus N(A-1)\oplus N(A+1){\}}^{\perp },$$

$$A_0=A{|}_{{\mathcal{H}}_0}=P_0-Q_0,$$

$$P_0=P_V=\frac{1}{2}\{1+A_0+V(1-A_0^2{)}^{1/2}\},Q_0=Q_V=\frac{1}{2}\{1-A_0+V(1-A_0^2{)}^{1/2}\}.$$

$$1\oplus 0.$$

$$P_0{Q}_0=P_V{Q}_V=\frac{V}{2}\{V(1-A_0^2{)}^{1/2}+1-A_0\}(1-A_0^2{)}^{1/2}.$$

$$A_0=\sum _{n\ge 1}{\lambda }_n(E_n-F_n),$$

$$P_0{Q}_0{\nu }_k^n=\frac{1}{2}(1-{\lambda }_n^2){\nu }_k^n+\frac{1}{2}(1-{\lambda }_n)(1-{\lambda }_n^2{)}^{1/2}{\omega }_k^n$$

$$P_0{Q}_0{\omega }_k^n=\frac{1}{2}(1-{\lambda }_n^2){\omega }_k^n+\frac{1}{2}(1+{\lambda }_n)(1-{\lambda }_n^2{)}^{1/2}{\nu }_k^n.$$

$$\frac{1}{2}\left(\begin{array}[]{cc}(1-\lambda_{n}^{2})&(1+\lambda_{n})(1-\lambda_{n}^{2})^{1/2}\\ (1-\lambda_{n})(1-\lambda_{n}^{2})^{1/2}&(1-\lambda_{n}^{2})\end{array}\right),$$

$$1\le \Vert p-q\Vert \le \Vert P-Q\Vert \le 1.$$

$$1-PQP=1-P\oplus P{Q}^{\perp }P$$

$$P_{I}f={\chi }_{I}fand{Q}_{J}f=\left({\chi }_J\hat{f}\right)\stackrel{ˇ}{},$$

$$P_I=M_{{\chi }_I}and{Q}_J=U_{\mathcal{F}}^{\ast }{M}_{{\chi }_J}{U}_{\mathcal{F}}.$$

Full text

2010 MSC: 47A05, 47A68, 47B35, 47B75.

Keywords: Projections, products of projections, differences of projections.

1 Introduction

Let ${\cal H}$ be a Hilbert space, ${\cal B}({\cal H})$ the space of bounded linear operators, ${\cal P}({\cal H})\subset{\cal B}({\cal H})$ the set of orthogonal projections. In what follows $R(T)$ denotes the the range of $T\in{\cal B}({\cal H})$ and $N(T)$ its nullspace. Given a closed subspace ${\cal S}\subset{\cal H}$, the orthogonal projection onto ${\cal S}$ is denoted by $P_{\cal S}$. In this paper we study part of the set ${\cal P}\cdot{\cal P}=\{PQ:P,Q\in{\cal P}({\cal H})\}$, namely, the subset of all $T=PQ$ such that $T^{*}T=PQP$ is diagonalizable. Operators in ${\cal P}\cdot{\cal P}$ are special cases of generalized Toeplitz operators as well as of Wiener-Hopf operators. As we shall see in a section of examples, they give rise to classical Toeplitz and Wiener-Hopf operators. Therefore this paper can be regarded as the study of operators in these classes, having a diagonal structure.

Also this paper is a kind of sequel to [8], [3] and [4], the first concerned with the whole set ${\cal P}\cdot{\cal P}$, the other two with ${\cal P}\cdot{\cal P}\cap{\cal K}({\cal H})$, where ${\cal K}({\cal H})$ denotes the ideal of compact operators acting in ${\cal H}$. Compact operators $T$ satisfy that $T^{*}T$ is diagonalizable.

We shall say that $T$ is S-decomposable if it has a singular value (or Schmidt) decomposition [24],

[TABLE]

where $\{\xi_{n}:n\geq 1\}$ and $\{\psi_{n}:n\geq 1\}$ are orthonormal systems, and $s_{n}>0$. In this case, $\{\psi_{n}\}$, $\{xi_{n}\}$ are orthonormal bases of $\overline{R(T)}$, $N(T)^{\perp}$, respectively and $T\xi_{n}=s_{n}\psi_{n}$, $T^{*}\psi_{n}=s_{n}\xi_{n}$, $T^{*}T\xi_{n}=s_{n}^{2}\xi_{n}$, $TT^{*}\psi_{n}=s_{n}^{2}\psi_{n}$ for all $n\geq 1$.

Clearly, $T$ is S-decomposable if and only if $T^{*}T$ (equivalently $TT^{*}$) is diagonalizable, if and only if $T^{*}$ is S-decomposable. Also it is clear that if $U,V$ are unitary operators, $T$ is S-decomposable if and only if $UTV$ is S-decomposable.

This paper is devoted to the study of the operators $T\in{\cal P}\cdot{\cal P}$ which are S-decomposable.

Let us describe the contents of the paper. In Section 2 we prove that $T=PQ$ is S-decomposable if and only if there exist orthonormal bases $\{\xi_{n}\}$ of $R(P)$ and $\{\psi_{n}\}$ of $R(Q)$ such that $\langle\xi_{n},\psi_{m}\rangle\neq 0$ if $n\neq m$. We also prove that $T=PQ$ is S-decomposable if and only if $A=P-Q$ is diagonalizable. This result is based on a Theorem by Chandler Davis ([10], Theorem 6.1), which characterizes operators which are the difference of two projections. A recent treatment of these operators can be found in [2]. The S-decomposability of $PQ$ is equivalent to that of $P(1-Q)$, $(1-P)Q$ and $(1-P)(1-Q)$. As a corollary we prove that $P-Q$ is diagonalizable if and only if $P+Q$ is, the eigenvalues $\pm\lambda_{n}$ of $P-Q$ which are different from $0,1$ correspond with the eigenvalues $1\pm(1-\lambda_{n})^{2}$, with the same multiplicity. Section 3 contains several interesting classes of examples of S-decomposable operators in ${\cal P}\cdot{\cal P}$. If ${\cal H}=L^{2}(\mathbb{R}^{n})$ and $I,J\subset\mathbb{R}^{n}$ are Lebesgue measurable set with finite positive measure, define $P_{I}f=\chi_{I}f$ and $Q_{J}f=(P_{J}\hat{f})\check{}$, for $f\in{\cal H}$. Here $\chi_{A}$ denotes the characteristic function of $A\subset\mathbb{R}^{n}$ and $\hat{}$, $\check{}$ denote the Fourier-Plancherel transform and its inverse. The product $P_{I}Q_{J}$ is a proper Wiener-Hopf operator, is also known as a concentration operator, and its study is related to mathematical formulations of the Heisenberg uncertainty principle. The reader is referred to [25], [19], [11], [12] for results concerning these products. Under the conditions described above, $P_{I}Q_{J}$ is a Hilbert-Schmidt operator, thus S-decomposable. This implies that also $P_{I^{\prime}}Q_{J^{\prime}}$ is S-decomposable (but non compact) when $I^{\prime}$ or $J^{\prime}$ have co-finite measure. It should be mentioned that the spectral description of the $P_{I}Q_{J}$ is no easy task (see [25] for the case when $I$, $J$ are intervals in $\mathbb{R}$). Another interesting family of examples is obtained if ${\cal H}=L^{2}(\mathbb{T})$ and ${\cal H}$ is decomposed as ${\cal H}={\cal H}_{+}\oplus{\cal H}_{-}$, where ${\cal H}_{+}=H^{2}(\mathbb{T})$. If $\varphi$, $\psi$ are continuous functions with modulus one, put $P=1-P_{\varphi{\cal H}_{+}}$ and $Q=P_{\psi{\cal H}_{+}}$. Then $PQ$ is a unitary operator times a Hankel operator with continuous symbol, and therefore a compact operator by a Theorem by Hartman [16]. Then $(1-P)Q$ is a unitary operator times a Toeplitz operator, and a non compact S-decomposable operator. On the other hand, using a result by Howland ([18], Theorem 9.2), one can find convenient non-continuous $\varphi$, $\psi$ such that $PQ$ is not S-decomposable.

In Section 4 we prove that, for two closed subspaces ${\cal S},{\cal T}$ of ${\cal H}$, the operator $T=P_{\cal S}P_{\cal T}$ is S-decomposable if and only if there exist isometries $X,Y:\ell^{2}\to{\cal H}$ with $R(X)={\cal S}$, $R(Y)={\cal T}$ such that $X^{*}Y\in{\cal B}(\ell^{2})$ is a diagonal matrix.

In Section 5 we characterize S-decomposability in terms of what we call Davis’ symmetry $V$: given two projections $P,Q$, the decomposition ${\cal H}=N(P+Q-1)\oplus N(P+Q-1)^{\perp}$ reduces simultaneously $P$ and $Q$. They act non trivially on the second subspace ${\cal H}^{\prime}=N(P+Q-1)^{\perp}$. Denote by $P^{\prime}$ and $Q^{\prime}$ the restrictions of $P$ and $Q$ to this subspace. Then the isometric part in the polar decompostion of $P^{\prime}+Q^{\prime}-1$ is a selfadjoint unitary operator $V$ which satisfies $VP^{\prime}V=Q^{\prime}$, $VQ^{\prime}V=P^{\prime}$. We relate these operator with the differential geometry of the space ${\cal P}({\cal H}^{\prime})$ of projections in ${\cal H}^{\prime}$ (or Grassmann manifold of ${\cal H}^{\prime}$). Specifically, with the unique short geodesic curve joining $P^{\prime}$ and $Q^{\prime}$ in ${\cal P}({\cal H}^{\prime})$. For instance, it is shown that $PQ$ is S-decomposable if and only if the velocity operator of the unique geodesic joining $P^{\prime}$ and $Q^{\prime}$ is diagonalizable.

In Section 5 it is shown that any contraction $\Gamma\in{\cal B}({\cal H})$ is the $1,1$ entry of a unitary operator times a product of projections acting in ${\cal H}\times{\cal H}$.

2 Products and differences of projections

If $T\in{\cal P}\cdot{\cal P}$, then $T=P_{\overline{R(T)}}P_{N(T)^{\perp}}$. This is a result of T. Crimmins (unpublished; there is a proof in [23] Theorem 8). Moreover, Crimmins proved that $T\in{\cal B}({\cal H})$ belongs to ${\cal P}\cdot{\cal P}$ if and only if $TT^{*}T=T^{2}$ [23]. However, the factorization $T=P_{\overline{R(T)}}P_{N(T)^{\perp}}$ is one among among many others. In [8], Theorem 3.7, it is proved that if $T\in{\cal P}\cdot{\cal P}$, then $T=P_{\cal S}P_{\cal T}$ if and only if

[TABLE]

In [8], for any $T\in{\cal P}\cdot{\cal P}$ the set of all pairs $({\cal S},{\cal T})$ of closed subspaces such that $T=P_{\cal S}P_{\cal T}$ is denoted by ${\cal X}_{T}$. Our first result is a characterization of ${\cal X}_{T}$ for S-decomposable $T$. The proof is essentially that of Theorem 4.1 in [4], where $T$ is supposed to be a compact element of ${\cal P}\cdot{\cal P}$. We include a proof for the reader’s convenience.

Theorem 2.1**.**

Let ${\cal S},{\cal T}\subset{\cal H}$ be closed subspaces of ${\cal H}$. Then $T=P_{\cal S}P_{\cal T}$ is S-decomposable if and only if there exist orthonormal bases $\{\psi_{n}:n\geq 1\}$ of ${\cal S}$, $\{\xi_{n}:n\geq 1\}$ of ${\cal T}$ such that $\langle\xi_{n},\psi_{m}\rangle=0$ if $n\neq m$. In such case, the numbers $|\langle\xi_{n},\psi_{n}\rangle|$ are the singular values of $T$.

Proof.

Suppose that $\{\psi_{n}\}$, $\{\xi_{n}\}$ are orthonormal bases of ${\cal S}$, ${\cal T}$ respectively, such that

[TABLE]

Therefore

[TABLE]

In order to get the Schmidt decomposition of $P_{\cal S}P_{\cal T}$, we only need to replace $\langle\psi_{n},\xi_{n}\rangle$ by the appropriate sequence of positive numbers: write $\langle\psi_{n},\xi_{n}\rangle=e^{i\theta_{n}}|\langle\psi_{n},\xi_{n}\rangle|$ and replace $\psi_{n}$ by $\psi^{\prime}_{n}=e^{-i\theta_{n}}\psi_{n}$. Then $\{\psi^{\prime}_{n}\}$ is still an orthonormal basis of ${\cal S}$, and

[TABLE]

are the singular values in the decomposition

[TABLE]

This shows that $P_{\cal S}P_{\cal T}$ is S-decomposable.

Conversely, if $T=P_{\cal S}P_{\cal T}$ is S-decomposable it has a singular value decomposition

[TABLE]

and it holds that $T^{2}=TT^{*}T$. Then

[TABLE]

and

[TABLE]

Using $TT^{*}T=T^{2}$ we get, for each $n\geq 1$

[TABLE]

Then $\langle\xi_{n},\psi_{m}\rangle=0$ if $n\neq m$ and $s_{n}=\langle\xi_{n},\psi_{n}\rangle$. Finally, we can extend the orthonormal bases $\{\psi_{n}\}$ of $\overline{R(T)}$ and $\{\xi_{n}\}$ of $N(T)^{\perp}$ to orthonormal bases of ${\cal S}$ and ${\cal T}$. In fact, if $\psi\in{\cal S}\ominus\overline{R(T)}$ and $\xi\in{\cal T}\ominus N(T)^{\perp}$, then

[TABLE]

because

[TABLE]

∎

Next, we show that $T=PQ$ is S-decomposable if and only if $A=P-Q$ is diagonalizable, and establish the relation between the singular values of $T$ and the eigenvalues of $A$. We present this equivalence as two separate theorems, to avoid too long a statement.

Theorem 2.2**.**

Suppose that $T=PQ$ is S-decomposable with singular values $s_{n}$. Then $A=P-Q$ is diagonalizable, with eigenvalues $\pm(1-s_{n}^{2})^{1/2}$, $n\geq 1$, plus, eventually, $0,-1$ and $1$.

Proof.

Put as above $T=\sum_{n\geq 1}s_{n}\psi_{n}\otimes\xi_{n}$, with $\xi_{n}\in R(Q)$ and $\psi_{n}\in R(P)$. First note that $s_{n}\leq s_{1}=\|T\|\leq\|P\|\|Q\|\leq 1$. Moreover, $s_{1}=1$ means that $T\xi_{1}=\eta_{1}$ and thus $\|P(Q\xi_{1})\|=1=\|\xi_{1}\|\geq\|Q\xi_{1}\|\geq\|P(Q\xi_{1})\|$, i.e., $\xi_{1}\in R(Q)$and $Q\xi_{1}=\xi_{1}\in R(P)$. Then $\xi_{1}=\psi_{1}$. The same happens for all $n$ such that $s_{n}=1$: the associated vectors $\xi_{n}=\psi_{n}$ generate $R(P)\cap R(Q)$. Note that $A=P-Q$ is trivial in this subspace.

Suppose that $s_{k}<1$. Apparently,

[TABLE]

and

[TABLE]

Then

[TABLE]

Since $s_{k}=|\langle\xi_{k},\psi_{k}\rangle|<1$ and $\|\xi_{k}\|=\|\psi_{k}\|=1$, it follows that $\xi_{k},\psi_{k}$ span a two-dimensional eigenspace for $A^{2}$, with eigenvalue $1-s_{k}^{2}$. Then

[TABLE]

are orthogonal eigenvectors for $A$, with eigenvalues $(1-s_{k}^{2})^{1/2}$ and $-(1-s_{k}^{2})^{1/2}$, respectively.

The orthogonal systems $\xi_{k}$ and $\psi_{k}$ can be extended to orthonormal bases of $R(P)$ and $R(Q)$, respectively (as in the proof of Theorem 2.1). On the extension of the system $\xi_{k}$, i.e., $R(P)\ominus R(T)$, $A=P-Q$ equals $1$. On the extension of $\psi_{k}$, $R(Q)\ominus N(T)^{\perp}$, $A$ equals $-1$. Together, these extended systems span $R(P)+R(Q)$, and here $A$ is diagonalizable. On the orthogonal complement of this subspace, namely $N(P)^{\perp}\cap N(Q)^{\perp}$, $A$ is trivial. ∎

Remark 2.3**.**

Note that, except for $1$ and $-1$, the eigenvalues $(1-s_{k}^{2})^{1/2}$ and $-(1-s_{k}^{2})^{1/2}$ of $A$ have the same multiplicity. Also note that

[TABLE]

and $N(A)=R(P)\cap R(Q)\oplus N(P)\cap N(Q)$.

The above result has a converse. In [10] Chandler Davis proved that operators $A=P-Q$ are characterized as follows: in the generic part of $A$, namely

[TABLE]

which reduces $P,Q$ and $A$, if we denote $P_{0}=P|_{{\cal H}_{0}}$, $Q_{0}=Q|_{{\cal H}_{0}}$ and

[TABLE]

there exists a symmetry $V$ ($V^{*}=V^{-1}=V$) such that $VA=-AV$ and

[TABLE]

$V$ is characterized by these properties. With these notations we have:

Theorem 2.4**.**

If $A=P-Q$ is diagonalizable with (non nil) eigenvalues $\pm\lambda_{n}$ ($0<|\lambda_{n}|<1$) and $\pm 1$ , then $T=PQ$ is S-decomposable with singular values $(1-\lambda_{n}^{2})^{1/2}$ and $1$.

Proof.

On the non generic parts $N(A-1)\oplus N(A+1)$, $T$ equals zero. In $N(A)=R(P)\cap R(Q)\oplus N(P)\cap N(Q)$, $T$ is

[TABLE]

Thus $PQ$ is diagonal (thus S-decomposable) in ${\cal H}_{0}^{\perp}$. In ${\cal H}_{0}$, after straightforward computations (note that $V$ commutes with $A_{0}^{2}$) one has

[TABLE]

Since $A_{0}$ is diagonalizable, and there exists the symmetry $V$ associated to $P_{0}$ and $Q_{0}$, which intertwines $A_{0}$ with $-A_{0}$, it follows that $A_{0}$ is of the form

[TABLE]

where $E_{n},F_{n}$ $(n\geq 1$) are pairwise orthogonal projections with $\dim R(E_{n})=\dim R(F_{n})=m_{n}\leq\infty$. The eigenvalues $\lambda_{n}$ of $A_{0}$ are different from $\pm 1$, because $N(A_{0}\pm 1)=\{0\}$. Fix an orthonormal basis $\{\nu^{n}_{k}:1\leq k\leq m_{n}\}$ for $R(E_{n})$. The fact that $VA=-AV$ implies that $V$ maps (the $\lambda_{n}$-eigenspace) $R(E_{n})$ onto (the $-\lambda_{n}$-eigenspace) $R(F_{n})$, and back. Then we can consider for $R(F_{n})$ the orthonormal basis given by $\omega^{n}_{k}=V\nu^{n}_{k}$. Thus also $V\omega_{k}^{n}=\nu_{k}^{n}$. Then

[TABLE]

and

[TABLE]

It follows that the $2$-dimensional subspace generated by (the orthonormal vectors) $\nu_{k}^{n}$ and $\omega_{k}^{n}$ is invariant for $P_{0}Q_{0}$. The matrix of $P_{0}Q_{0}$ restricted to this subspace (in this basis) is

[TABLE]

whose singular values are [math] and $(1-\lambda_{n}^{2})^{1/2}$. In the orthonormal basis $\{\nu_{k}^{n},\omega_{k}^{n}\}$ of ${\cal H}_{0}$ (paired in this fashion), the operator $P_{0}Q_{0}$ is block-diagonal, with $2\times 2$ blocks. It follows that $PQ$ is S-decomposable with singular values $(1-\lambda_{n}^{2})^{1/2}$ and, eventually, $1$. The singular value $1$ occurs only if $R(P)\cap R(Q)\neq\{0\}$. ∎

Remark 2.5**.**

The multiplicity of $(1-\lambda^{2}_{n})^{1/2}$ as a singular value of $PQ$ is $m_{n}$.

Remark 2.6**.**

From the above results, which relate eigenvalues of $P-Q$ and singular values of $PQ$, it follows that if $PQ$ is compact, and either $P$ or $Q$ have infinite rank, then $\|P-Q\|=1$. Indeed, if $PQ$ is compact, the singular values accumulate eventually at [math], and therefore the eigenvalues of $A$ accumulate at $1$. However, this result holds with more generality. It is a simple exercise that if $p\neq q$ are non nil projections in a C∗-algebra such that $pq=0$, then $\|p-q\|=1$. Our case consists in reasoning in the Calkin algebra: $p=\pi(P)$, $q=\pi(Q)$, where $\pi:{\cal B}({\cal H})\to{\cal B}({\cal H})/{\cal K}({\cal H})$ is the quotient homomorphism. Then

[TABLE]

The following result will be useful to provide further examples. In a special case (see Example 3.1 in Section 3), it was proven by M. Smith ([26], Th. 3.1)

Proposition 2.7**.**

$PQ$* is S-decomposable if and only if $P(1-Q)$ is S-decomposable (and therefore if and only if $(1-P)Q$ or $(1-P)(1-Q)$ are S-decomposable).*

Proof.

$P(1-Q)$ is S-decomposable if and only if $P(1-Q)P=P-PQP$ is diagonalizable. This operator acts non trivially only in $R(P)$. Thus, it is diagonalizable if and only if it is diagonalizable in $R(P)$. Adding $1-P$ (equal to the identity in $N(P)$), one obtains that this latter fact is equivalent to $1-PQP=1-P\oplus P-PQP$ being diagonalizable in ${\cal H}=N(P)\oplus R(P)$. Clearly $1-PQP$ is diagonalizable if and only if $PQP$ also is, i.e., if and only if $PQ$ is S-decomposable. ∎

As a direct consequence of this fact, one obtains the following corollary

Corollary 2.8**.**

Let $P,Q$ be projections. Then $P-Q$ is diagonalizable if and only if $P+Q$ is diagonalizable. In that case, $\lambda_{n}$ is an eigenvalue of $P-Q$ with $0<|\lambda_{n}|<1$ if and only if $1\pm(1-\lambda_{n})^{2}$ is an eigenvalue of $P+Q$, with the same multiplicity.

Proof.

By the above results, any eigenvalue $\lambda_{n}=\pm(1-s_{n}^{2})^{1/2}$, where $s_{n}$ is a singular value of $PQ$, or equivalently, $s_{n}^{2}$ is an eigenvalue of $PQP$. On the other hand, from the proof of Proposition 2.7, the eigenvalues of

[TABLE]

are $1$, and $1-s_{n}^{2}$. Then again by Theorem 2.2, the eigenvalues of $P-Q^{\perp}=P+Q-1$ are $\pm s_{n}$, and thus the eigenvalues of $P+Q$ are $1\pm s_{n}=1\pm(1-\lambda_{n}^{2})^{1/2}$. Since $P-Q$ is a difference of projections, the eigenvalues $+\lambda$ and $-\lambda$ (when $0<|\lambda|<1$) have the same multiplicity (see [2]), and by the above results, these add up to the multiplicity of $s=(1-\lambda^{2})^{1/2}$ as a singular value of $PQ$. This number clearly equals the multiplicity of $(1-s^{2})^{1/2}$ as a singular value of $PQ^{\perp}$. Note that $P+Q-1=P-Q^{\perp}$ is also a difference of projections, therefore the multiplicities of $\pm s=\pm(1-\lambda^{2})^{1/2}$ coincide ($0<s<1$). ∎

Remark 2.9**.**

The multiplicity of $1$ as an (eventual) eigenvalue of $P-Q$ is the dimension of $R(P)\cap N(Q)$, the multiplicity of $-1$ is the dimension of $N(P)\cap R(Q)$, the sum of these multiplicities is the multiplicity of [math] in $P-Q^{\perp}$, or the multiplicity of $1$ in $P+Q$. Similarly, the multiplicity of [math] in $P-Q$ equals the sum of the multiplicities of [math] and $2$ in $P+Q$.

Remark 2.10**.**

To study the examples in the next section, it will also be useful to note that if $P$ has infinite rank and $PQ$ is compact, then $P(1-Q)$ is S-decomposable but non compact.

3 Examples

Example 3.1**.**

Let $I,J\subset\mathbb{R}^{n}$ be Lebesgue-measurable sets of finite measure. Let $P_{I},Q_{J}$ be the projections in $L^{2}(\mathbb{R}^{n},dx)$ given by

[TABLE]

where $\chi_{L}$ denotes the characteristic function of the set $L$. Equivalently, denoting by $U_{\cal F}$ the Fourier transform regarded as a unitary operator acting in $L^{2}(\mathbb{R}^{n},dx)$, then

[TABLE]

In [11] (Lemma 2) it is proven that $P_{I}Q_{J}$ is a Hilbert-Schmidt operator. See also [12]. Then $T=P_{I}Q_{J}$ is S-decomposable (with square summable singular values) These products play a relevant role in operator theoretic formulations of the uncertainty principle [11], [12].

In this case one has the spectral picture of $A=P_{I}-Q_{J}$. It is known [19], [12] that

[TABLE]

and $R(P_{I})\cap R(Q_{J})$ is infinite dimensional. Thus $N(A)=R(P_{I})\cap R(Q_{J})$ is infinite dimensional, $N(A\pm 1)=\{0\}$, and the eigenvalues of $A$ are of the form $\pm(1-s_{k}^{2})^{1/2}$, where the sequence $s_{k}$ belongs to $\ell^{2}(\mathbb{Z})$. In special cases, e.g. $I=[0,T],J=[-\Omega,\Omega]$ intervals in $\mathbb{R}$, the eigenfunctions are known and the eigenvalues have multiplicity one [17].

If one relaxes the condition that the sets be of finite measure, $P_{I}Q_{J}$ ceases to be compact. Using Proposition 2.7, one obtains non compact examples: replacing the above conditions by $|\mathbb{R}^{n}\setminus I|<\infty$ or $|\mathbb{R}^{n}\setminus J|<\infty$ (see also [26], one obtains non-compact, S-decomposable products of projections.

Note also that, due to Theorem 2.1, in the above cases (i.e. both $I$ and $J$ have finite or co-finite measure), the subspaces $R(P_{I})=\{f\in L^{2}(\mathbb{R}^{n}):f|_{\mathbb{R}^{n}\setminus I}=0\}$ and $R(Q_{J})=\{g\in L^{2}(\mathbb{R}^{n}):\hat{g}|_{\mathbb{R}^{n}\setminus J}=0\}$ have orthonormal bases $\{f_{n}\}$ and $\{g_{n}\}$, repectively, which satisfy $\langle f_{n},g_{m}\rangle=0$ if $n\neq m$.

We study more carefully the case

[TABLE]

not covered above. Straightforward computations (see [19]) show that the operator $P_{I}Q_{J}$, acting in $L^{2}(0,+\infty)$ is given by

[TABLE]

Let us prove that $P_{I}Q_{J}P_{I}$ is non compact. For $n\in\mathbb{N}$, let

[TABLE]

Apparently $e_{n}\in L^{2}(\mathbb{R})$ and $\|e_{n}\|_{2}^{2}=\frac{n}{2}$. Note that

[TABLE]

Changing variables $v=x-t$ in the first integral and $u=t-x$ in the second, one obtains

[TABLE]

The second integral, which we shall denote $\lambda_{n}$, can be computed. Denote by $\mathbb{L}$ the usual Laplace transform. Then

[TABLE]

[TABLE]

Let us denote by $F_{n}(x)$ the left hand integral,

[TABLE]

Lemma 3.2**.**

With the current notations,

[TABLE]

Proof.

Compute

[TABLE]

Integrating by parts, and using that (by means of the L’Hospital rule !), we get

[TABLE]

and that $F_{n}(0)=0$. Then

[TABLE]

[TABLE]

which, by computations similar as above involving the Laplace transform, equals

[TABLE]

Then

[TABLE]

[TABLE]

Then

[TABLE]

[TABLE]

Therefore

[TABLE]

∎

Proposition 3.3**.**

If $I=[0,+\infty)$ and $J=[-1,1]$, then $P_{I}Q_{J}P_{I}$ is non compact, with

[TABLE]

Proof.

If $P_{I}Q_{J}P_{I}$ were compact, there would exist a subsequence $f_{k}=\frac{1}{\|e_{n_{k}}\|_{2}}e_{n_{k}}$ such that $P_{I}Q_{J}P_{I}f_{k}$ is convergent. By the above lemma, this would imply that the sequence $f_{k}$ is convergent. This is clearly not the case. For instance,

[TABLE]

which is less than $\frac{1}{2}$ by the geometric-arithmetic inequality. This clearly implies that the sequence of the unit vectors $f_{k}$ cannot be convergent.

The last assertions follow from the above lemma. ∎

Remark 3.4**.**

Note that Example 3.1 above shows, in particular, that the Volterra-like integral operator

[TABLE]

is unbounded in $L^{2}(0,+\infty)$ (though it is a Volterra operator on any finite interval $[0,r]$, thus compact with trivial spectrum in $L^{2}(0,r)$, for $r<\infty$). Indeed, if it were bounded, then $T=P_{I}Q_{J}P_{I}-B$,

[TABLE]

would be bounded. But the computations above show that the functions $e_{n}(x)=e^{(-\frac{1}{n}+i)x}$ are eigenfunctions for $T$, with unbounded eigenvalues $\lambda_{n}$.

Example 3.5**.**

Let ${\cal H}=L^{2}(\mathbb{T},dt)$ where $\mathbb{T}$ is the $1$-torus, and consider the decomposition

[TABLE]

where ${\cal H}_{+}$ is the Hardy space. Let $\varphi,\psi$ be continuous functions in $\mathbb{T}$ with $|\varphi(e^{it})|=|\psi(e^{it})|=1$ for all $t$, and

[TABLE]

Since $\varphi$ and $\psi$ are unimodular, the multiplication operators $M_{\varphi}$, $M_{\psi}$ are unitary in ${\cal H}$ and thus

[TABLE]

Note that $P_{-}M_{\bar{\varphi}\psi}|_{{\cal H}_{+}}=H(\bar{\varphi}\psi)$ is the Hankel operator with symbol $\bar{\varphi}\psi$, which is compact by Hartman’s theorem [16] (see also Theorem 5.5 in [20]). Thus $T=PQ$ is compact, and therefore S-decomposable.

Again using Proposition 2.7, one obtains non compact S-decomposable examples. For instance, put now

[TABLE]

In this case

[TABLE]

is decomposable Thus the operator $P_{+}M_{\bar{\varphi}\psi}P_{+}$ is non-compact and S-decomposable in $L^{2}(\mathbb{T})$. Since it acts non trivially in ${\cal H}_{+}$, it follows that the Toeplitz operator $T_{\bar{\varphi}\psi}$ is S-decomposable in ${\cal H}_{+}$.

On the other hand, using Theorem 2.2, it follows that

[TABLE]

diagonalizable. In [5] it was shown that $\pm 1$ are eigenvalues of $A$ only if the winding numbers of $\varphi$ and $\psi$ do not coincide. The other eigenvalues of $A$ are $\pm(1-s_{n}^{2})^{1/2}$, where $s_{n}$ are the singular values of $T_{\bar{\varphi}\psi}$, and [math]. Since this operator has closed range (being a Fredholm operator), the eigenvalues do not accumulate at $\pm 1$. The nullspace of $A$ is infinite dimensional, it contains the subspace $\varphi\psi{\cal H}_{+}$.

Again, using Theorem 2.1, one obtains that, with the above hypothesis on $\varphi$ and $\psi$, there exist orthonormal bases $\{f_{n}\}$ and $\{g_{n}\}$ of ${\cal H}_{+}$ such that $\langle\varphi f_{n},\psi g_{m}\rangle=0$ if $n\neq m$.

In [18] (Theorem 5.2) J.S. Howland proved that if the function $f$ on $\mathbb{T}$ is $C^{2}$ on the complement of a finite set $\{z_{1},\dots,z_{n}\}$ at which the lateral limits $f(z_{i}^{\pm})$ and $f^{\prime}(z_{i}^{\pm})$ exist, and one defines the jump of $f$ at $z$ to be

[TABLE]

then the absolutely continuous part of the Hankel operator $H(f)$ is unitarily equivalent to

[TABLE]

where $M_{i,z}$ denotes the operator of multiplication by the variable $z$ in $L^{2}(-\frac{1}{2}j(z_{i}),\frac{1}{2}j(z_{i}))$. In particular, this implies that if $\bar{\varphi}\psi$ is piecewise $C^{2}$ with jumps as $f$ above, then $PQ$ can be decomposed as a finite direct sum of operators, some of which are multiplication by the variable in $L^{2}$ of an interval. Clearly these operators are not S-decomposable. Then $PQ$ is not S-decomposable.

Example 3.6**.**

Let ${\cal H}={\cal L}\times{\cal S}$, $B:{\cal S}\to{\cal L}$ a bounded operator, and $E=E_{B}$ the idempotent operator given by the matrix

[TABLE]

Any idempotent in ${\cal B}({\cal H})$ can be expanded in this form. In [1] the reader can find a study of the properties of $E$ in terms of those of $B$. Consider $P=P_{R(E)}=P_{\cal L}$ and $Q=P_{N(E)}$ and $T=PQ$. Straightforward computations show that $R(E)={\cal L}$ and that

[TABLE]

Then

[TABLE]

Apparently, $T$ is S-decomposable if and only if $BB^{*}(1+BB^{*})^{-1}$ is diagonalizable, which is equivalent to $BB^{*}$ being diagonalizable, or $B$ S-decomposable. Note also that $T$ is compact if and only if $B$ is compact.

If one applies Theorem 2.1 to this example, one obtains that $B$ is S-decomposable if and only if there exist orthonormal bases $\{(0,v_{n})\}$ of $\{0\}\times{\cal L}$ and $\{(w_{n},Bw_{n})\}$ of the graph of $B$, such that $\langle(0,v_{n}),(w_{m},Bw_{m})\rangle=\langle v_{n},Bw_{m}\rangle=0$ if $n\neq m$. This fact can be proved straightforwardly.

4 Moore-Penrose pseudoinverses

Penrose [21] and Greville [13] proved that , for $n\times n$ square matrices, the Moore-Penrose inverse of an idempotent matrix $E$ is a product of orthogonal projections. More precisely, it holds that

[TABLE]

Since for matrices $(A^{\dagger})^{\dagger}=A$, Penrose-Greville theorem can be stated as follows: an $n\times n$ matrix $E$ is idempotent if and only if $E^{\dagger}$ is a product of two orthogonal projections. This result was extended to infinite dimensional Hilbert space operators in [7] , provided that $PQ$ is supposed to have closed range. In the case that $R(PQ)$ is not closed, there is still a similar characterization, but one needs to define the Moore-Penrose inverse for certain unbounded operators. The reader is referred to [8]. As in example 3.6, if $E$ is an idempotent operator, in terms of the decomposition ${\cal H}=R(E)\oplus R(E)^{\perp}$, one has

[TABLE]

where $B:R(E)^{\perp}\to R(E)$.

Combining the above facts and previous results we obtain the following:

Corollary 4.1**.**

Let $E\in{\cal B}({\cal H})$ be an idempotent operator. Then the following are equivalent:

$E$* is S-decomposable* 2. 2.

$B$* is S-decomposable* 3. 3.

$P_{N(E)^{\perp}}P_{R(E)}$* is S-decomposable.* 4. 4.

$P_{N(E)}P_{R(E)}$* is S-decomposable.* 5. 5.

$P_{R(E)}-P_{N(E)}$* is diagonalizable.* 6. 6.

$P_{R(E)}+P_{N(E)}$* is diagonalizable.* 7. 7.

There exist orthonormal bases $\{\eta_{n}\}$ of $R(E)$ and $\{\nu_{n}\}$ of $N(E)$ such that $\langle\eta_{n},\nu_{m}\rangle=0$ if $n\neq m$.

Some of these conditions were proven in [1].

Remark 4.2**.**

By a theorem by Buckholtz ([6], Theorem 1), since ${\cal H}$ is the direct sum of $R(E)$ and $N(E)$, it follows that $P_{R(E)}-P_{N(E)}$ is invertible for every idempotent $E$, which in turn implies that $P_{R(E)}+P_{N(E)}$ is invertible. In fact, for any $P,Q\in{\cal P}({\cal H})$, $P-Q$ is invertible if and only if $\|PQ\|<1$ and $\|(1-P)(1-Q)\|<1$, while $P+Q$ is invertible if and only if $\|(1-P)(1-Q)\|<1$. In geometric terms, $\|PQ\|$ is the cosine of the (Dixmier) angle between $R(P)$ and $(Q)$, and $\|(1-P)(1-Q)\|$ is the cosine of the angle between $N(P)$ and $N(Q)$. If ${\cal H}$ is the direct sum of $R(P)$ and $RQ)$, these angles coincide and are not zero.

Finally, note that if $T$ is S-decomposable with expansion $T=\sum_{n\geq 1}{\cal S}_{n}\langle\ \ \ ,\xi_{n}\rangle\psi_{n}$, then

[TABLE]

5 Isometries

Given a subspace ${\cal S}\subset{\cal H}$ with a given orthonormal basis ${\bf B}_{\cal S}=\{\xi_{n}:b\geq 1\}$, an isometry is defined,

[TABLE]

whose range is ${\cal S}$. Observe that, by definition, the set of all S-decomposable operators in ${\cal H}$ can be described as

[TABLE]

The condition of bi-orthogonality of Theorem 2.1 can be written in terms of the corresponding isometries.

Proposition 5.1**.**

Let ${\cal S},{\cal T}$ be closed subspaces of ${\cal H}$. Then $T=P_{\cal S}P_{\cal T}$ is S-decomposable if and only if there exist isometries $X,Y:\ell^{2}\to{\cal H}$, with range ${\cal S}$ and ${\cal T}$, respectively, such that

[TABLE]

is a diagonal matrix.

Proof.

Suppose that $T$ is decomposable, then by Theorem (2.1), there exist orthonormal bases ${\bf B}_{\cal S}=\{\xi_{k}:k\geq 1\}$ and ${\bf B}_{\cal T}=\{\psi_{n}:n\geq 1\}$ of ${\cal S}$ and ${\cal T}$ such that $\langle\xi_{n},\psi_{k}\rangle=0$ if $n\neq k$. Consider the isometries

[TABLE]

Then

[TABLE]

i.e. $X^{*}Y$ is a diagonal matrix whose entries are $\langle\psi_{n},\xi_{n}\rangle$.

Conversely, suppose that $X,Y:\ell^{2}\to{\cal H}$ are isometries with $R(X)={\cal S}$ and $R(Y)={\cal T}$, such that $X^{*}Y$ is a diagonal matrix. Denote by $\{e_{n}:n\geq 1\}$ the canonical basis of $\ell^{2}$. Then $\xi_{n}=X(e_{n})$ and $\psi_{k}=Y(e_{k})$ form orthonormal bases of ${\cal S}$ and ${\cal T}$. Moreover

[TABLE]

∎

6 Davis’ symmetry

Let $P,Q$ be projections, and consider

[TABLE]

This subspace reduces $P$ and $Q$, denote by $P^{\prime}=P|_{{\cal H}^{\prime}}$ and $Q^{\prime}=Q|_{{\cal H}^{\prime}}$, as operators acting in ${\cal H}^{\prime}$. Note that

[TABLE]

and thus $S^{\prime}=P^{\prime}+Q^{\prime}-1$ is a selfadjoint operator with trivial kernel (and thus dense range) in ${\cal H}^{\prime}$. Let

[TABLE]

be the polar decomposition. It follows that $V$ is a selfadjoint unitary operator, i.e., a symmetry. The fact that

[TABLE]

implies that the symmetry $V$ intertwines $P^{\prime}$ and $Q^{\prime}$:

[TABLE]

Also one recovers $P^{\prime}$ and $Q^{\prime}$ in terms of $V$ and the difference $A=P^{\prime}-Q^{\prime}$, by means of the formulas of the previous section:

[TABLE]

These facts were proved by Chandler Davis in [10]. Then $T=PQ$, in the decomposition ${\cal H}={\cal H}^{\prime\perp}\oplus{\cal H}^{\prime}$ is given by

[TABLE]

The following result is a straightforward consequence of the results in the previous section:

Proposition 6.1**.**

$T=PQ$* is S-decomposable if and only $Q^{\prime}VQ^{\prime}$ is diagonalizable (equivalently: $P^{\prime}VP^{\prime}$ is diagonalizable). If $\{\xi_{n}\}$ is an orthonormal system of eigenvectors for $Q^{\prime}VQ^{\prime}$, then $\langle V\xi_{n},\xi_{k}\rangle=0$ if $n\neq k$.*

Proof.

If $Q^{\prime}VQ^{\prime}=\sum_{n\geq 1}\lambda_{n}\langle\ \ \ ,\xi_{n}\rangle\xi_{n}$, then

[TABLE]

and thus the orthonormal systems $\{\xi_{n}\}$ and $\{V\xi_{n}\}$ are bi-orthogonal. ∎

Remark 6.2**.**

Suppose that

[TABLE]

Then

[TABLE]

In particular, if all the singular values have multiplicity $1$, then $V\psi_{n}=\pm\xi_{n}$.

Davis’ symmetry is related to the metric geometry of the set ${\cal P}({\cal H})$ of projections in ${\cal H}$ (also called Grassmannian manifold of ${\cal H}$). If one measures the length of a continuous piecewise smooth curve $p(t)\in{\cal P}({\cal H})$, $t\in I$, by means of

[TABLE]

it was shown ([22], [9]) that curves in ${\cal P}({\cal H})$ of the form

[TABLE]

for $X^{*}=X$ with $\|X\|\leq\pi/2$, such that $X$ is $P$-codiagonal (i.e $PXP=P^{\perp}XP^{\perp}=0$) have minimal length along their paths for $|t|\leq 1$. That is, any curve joining a pair of projections in this path cannot be shorter that the part of $P(t)$ which joins these projections. Given two projections $P,Q$, in [2] it was shown that there exists a unique $X$ ($X^{*}=X$, $\|X\|\leq\pi/2$, $X$ is $P$-codiagonal) such that $e^{iX}Pe^{-iX}=Q$ if and only if

[TABLE]

Let us denote $X=X_{P,Q}$ if such is the case. Also in [2] it was shown that $V$ and $X_{P,Q}$ are related by

[TABLE]

Note that since (always in the case $N(P+Q-1)=\{0\}$) $\|X_{P,Q}\|\leq\pi/2$, $X_{P,Q}$ is obtained from $V$ by means of the usual $\log$ function:

[TABLE]

Define the geodesic distance $d(P,Q)$ in ${\cal P}({\cal H})$ as

[TABLE]

Porta and Recht proved in [22] that

[TABLE]

Remark 6.3**.**

Formula (2) has a geometric interpretation. The fact that $X_{P,Q}$ is $P$-codiagonal, is equivalent to saying that $X_{P,Q}$ and $2P-1$ anti-commute, it follows that $e^{itX_{P,Q}}(2P-1)=(2P-1)e^{-itX_{P,Q}}$. Then, in particular $V=e^{\frac{i}{2}X_{P,Q}}(2P-1)e^{-\frac{i}{2}X_{P,Q}}$, or equivalently

[TABLE]

In other words, the projection $\frac{1}{2}(1+V)$ (onto the eigenspace where the symmetry $V$ acts as the identity) is the midpoint of the geodesic $P(t)$ joining $P$ and $Q$.

From the above facts, the following is apparent:

Corollary 6.4**.**

Let $P,Q$ be projections and, as above, $P^{\prime},Q^{\prime}$ the respective reductions to $N(P+Q-1)^{\perp}$, and let $V$ be Davis’ symmetry induced by these. Then

[TABLE]

Thus $PQ$ is S-decomposable if and only if $P^{\prime}e^{X_{P^{\prime},Q^{\prime}}}P^{\prime}$ is diagonalizable.

Proof.

Since $V=e^{iX_{P^{\prime},Q^{\prime}}}(2P^{\prime}-1)$, then

[TABLE]

Similarly, $V=P^{\prime}e^{iX_{P^{\prime},Q^{\prime}}}=e^{-iX_{P^{\prime},Q^{\prime}}}Q^{\prime}$, and so forth. ∎

Remark 6.5**.**

Since $Q^{\prime}=e^{iX_{P^{\prime},Q^{\prime}}}P^{\prime}e^{-iX_{P^{\prime},Q^{\prime}}}(2P^{\prime}-1)P^{\prime}$, it also follows that

[TABLE]

and

[TABLE]

Remark 6.6**.**

If the matrix of $X_{P^{\prime},Q^{\prime}}$ in terms of $P^{\prime}$ is given by

[TABLE]

then

[TABLE]

From this last remark, it follows that

Theorem 6.7**.**

$PQ$* is S-decomposable if and only if $Z$ is S-decomposable, if and only if $X_{P^{\prime},Q^{\prime}}$ is diagonalizable.*

Proof.

[TABLE]

Thus $X_{P^{\prime},Q^{\prime}}$ is diagonalizable if and only if $Z$ is S-decomposable. Indeed, if $Z$ is S-decomposable,

[TABLE]

Note that $\{v_{n}\}$ span $R(P^{\prime})$ and $\{w_{n}\}$ span $R(P^{\prime})^{\perp}$, therefore, they are pairwise orthogonal systems of vectors. Then

[TABLE]

For each fixed $n$, the two dimensional space generated by $v_{n}$ and $w_{n}$ reduces $X_{P^{\prime},Q^{\prime}}$. As in a previous argument, $X_{P^{\prime},Q^{\prime}}$ can be diagonalized in each of these spaces, providing a diagonalization of the whole operator $X_{P^{\prime},Q^{\prime}}$. The converse statement is apparent. ∎

Finally, let us further exploit formula (2).

Corollary 6.8**.**

If $A^{\prime}=P^{\prime}-Q^{\prime}$, then

[TABLE]

Proof.

In $N(P+Q-1)^{\perp}$, $P^{\prime}=P_{V}=\frac{1}{2}\{1+A^{\prime}+V(1-A^{\prime 2})^{1/2}\}$, thus

[TABLE]

∎

In particular, if $PQ$ is S-decomposable, with singular values of simple multiplicity, one has the following

Theorem 6.9**.**

Let $PQ$ be S-decomposable, $P^{\prime}Q^{\prime}=\sum_{n\geq 1}s_{n}\langle\ \ \ ,\xi_{n}\rangle\psi_{n}$, with $s_{n}$ of multiplicity $1$. Then $X_{P^{\prime},Q^{\prime}}$ is diagonalized as follows

[TABLE]

where

[TABLE]

and (as in the proof of Theorem 2.2)

[TABLE]

Proof.

If $PQ$ is S-decomposable, considering the decomposition of $PQ|_{N(P+Q-1)^{\perp}}=P^{\prime}Q^{\prime}$, in the proof of Theorem 2.2,

[TABLE]

for $\nu_{n},\omega_{n}$ described above. Then

[TABLE]

Recall that $VA=-VA$, or equivalently, $VAV=-A$ (see remarks before Theorem 2.4). Note that in $N(P+Q-1)^{\perp}$ we have erased the eigenvalues $\pm 1$ from $A$. Then, using Theorem 2.2, the fact that the singular values of $P^{\prime}Q^{\prime}$ have simple multiplicity implies that the (non nil) eigenvalues of $A^{\prime}$ have single multiplicity. These two assertions imply that

[TABLE]

Thus, in the diagonalization of $A^{\prime}$, we may replace $\xi_{n},\psi_{n}$ by scalar multiples (of modulus one) in order that

[TABLE]

Then

[TABLE]

Thus, by the formula in the above Corollary,

[TABLE]

Note that this is a block diagonal operator, with $2\times 2$ blocks, given by the subspaces generated by the (orthonormal) vectors $\nu_{n}$ and $\omega_{n}$ for each $n$. Each block, in this basis, is given by

[TABLE]

whose eigenvalues are $s_{n}+i(1-s_{n}^{2})^{1/2}$ and $s_{n}-i(1-s_{n}^{2})^{1/2}$, with (orthonormal) eigenvectors

[TABLE]

respectively, and the proof follows. ∎

Note that since $0<s_{n}$, the logarithms of these eigenvalues have modulus smaller than $\pi/2$, a fact predicted by the condition $\|X_{P^{\prime},Q^{\prime}}\|\leq\pi/2$.

Examples 6.10**.**

Let us review the examples in Section 3:

For $I,J\subset\mathbb{R}^{n}$ of finite Lebesgue measure, it is known (see [19], [12]) that

[TABLE]

Thus $P_{I}^{\prime}=P_{I}$ and $Q_{J}^{\prime}=Q_{J}$. It is also known (see for instance [17]) that in the particular case when $I$ and $J$ are intervals, the singular values of of $P_{I}Q_{J}$ have multiplicity one. Moreover the functions $\psi_{n}$ and $\xi_{n}$ are known to be the prolate spheroidal functions, for precise $I$ and $J$ (intervals in $\mathbb{R}$) [17]. It follows that one can compute explicitely the eigenvectors of $X_{P_{I},Q_{J}}$ for such intervals $I,J$. 2. 2.

As in Example 3.5, consider ${\cal H}=L^{2}(\mathbb{T})$ and

[TABLE]

for $\varphi,\psi$ continuous functions in $\mathbb{T}$, of modulus $1$. It was shown in [5] that if $\varphi$ and $\psi$ have the same winding number, then

[TABLE] 3. 3.

As in example 3.6, let ${\cal H}={\cal L}\times{\cal S}$ and $B:{\cal S}\to{\cal L}$ a bounded operator, $P=P_{R(E)}=P_{\cal L}$ and $Q=P_{N(E)}$ and $T=PQ$. Elementary computations show that

[TABLE]

Thus this nullspace is trivial if and only if $B$ has trivial nullspace and dense range. Suppose that this is the case. Also it is straightforward to verify that

[TABLE]

and that

[TABLE]

Then

[TABLE]

Thus

[TABLE]

This computation is apparent if $B$ (and thus $|P+Q-1|$) is invertible, but also makes sense when $B$ has trivial nullspace and dense range. If $B=W|B|=|B^{*}|W$ are the polar decompositions of $B$, one has

[TABLE]

where $W(1+|B^{*}|^{2})^{-1/2}$ can be replaced by $(1+|B|^{2})^{-1/2}W^{*}$.

Therefore

[TABLE]

Suppose now that $B$ is S-decomposable, $B=\sum_{n\geq 1}s_{n}\langle\ \ ,e_{n}\rangle f_{n}$, where since $B$ has trivial nullspace and dense range, where $\{e_{n}\}$ and $\{f_{n}\}$ are orthonormal bases of ${\cal S}$ and ${\cal L}$, respectively. Then

[TABLE]

and $W$ is a unitary operators ($W:{\cal S}\to{\cal L}$), with $We_{n}=f_{n}$. Let $\xi_{n}=(e_{n},0)$, $\psi_{n}=(0,f_{n})$. Then $\{\xi_{n},\psi_{n}\}$ span a reducing subspace of $T=PQ$, $P=P_{R(E)}$, $Q=P_{N(E)}$, and in view of the above formulas, also reducing for $V$ and $X_{P,Q}$. Elementary computations show that the matrix of $e^{iX_{P,Q}}$ in the basis of this reducing subspace is

[TABLE]

Let $\theta_{n}$ be defined by $\cos(\theta_{n})=\frac{s_{n}}{(1+s_{n}^{2})^{1/2}}$ and $\sin(\theta_{n})=\frac{1}{(1+s_{n}^{2})^{1/2}}$ (or equivalently, since $s_{n}>0$: $\tan(\theta_{n})=\frac{1}{s_{n}}$), then the matrix of $X_{P,Q}$ in this reducing subspace is

[TABLE]

Recall [2] that if $P$ and $Q$ are projections such that $N(P+Q-1)=\{0\}$, there exists a unique exponent $X_{P,Q}$ with $d(P,Q)=\|X_{P,Q}\|$. In particular, one has the following consequence:

Corollary 6.11**.**

Let $B:{\cal S}\to{\cal L}$ with trivial nullspace and dense range, and $E$ as in Example 3.6.

1. (a)

If $B$ is invertible, then the geodesic dictance between $P_{R(E)}$ and $P_{N(E)}$ is

[TABLE] 2. (b)

If $B$ is non invertible (i.e. $B^{-1}$ is unbounded), then

[TABLE]

Proof.

Suppose that $B$ is S-decomposable. If $B$ is invertible, $s_{n}\in(\|B^{-1}\|^{-1},\|B\|)$, and if $B$ is non invertible there exists a decreasing subsequence $s_{n_{k}}$ of singular values of $B$, such that $s_{n_{k}}\to 0$. Thus the claims follow from the previous computations.

Suppose now $B$ arbitrary. Clearly $|B|$ can be approximated by positive invertible operators $A_{k}$ with finite spectrum, in particular, diagonalizable. If $B=W|B|$, then $B_{k}=WA_{k}$ approximate $B$ (as in 6.10.3). Since $B$ has trivial nullspace and dense range, $W$ is a unitary operator. Then $B_{k}$ are S-decomposable, with finite singular values (increasingly ordered)$s_{k,i}$, $1\leq i\leq n_{k}$. Note that $P=P_{R(E)}$ and $Q=P_{N(E)}$ are continuous functions of $B$. Denote by $E_{k}$, $P_{k}=P_{R(E_{k})}$ and $Q_{k}=P_{N(E_{k})}$ the operators acting in ${\cal L}\times{\cal S}$ which correspond to $B_{k}$. Then

[TABLE]

From the previous case, $d(P_{k},Q_{k})=\tan^{-1}(\frac{1}{s_{k,1}})$. If $B$ is invertible, $\frac{1}{s_{k,1}}\to\|B^{-1}\|$. Otherwise, $\frac{1}{s_{k,1}}\to\infty$. ∎

Remark 6.12**.**

As mentioned in the beginning of Section 2, if $T=PQ$, there may exist many factorizations, and that there exist a canonical factorization

[TABLE]

with the following minimality property: for any $\xi\in{\cal H}$, and any other factorization $T=PQ$, one has

[TABLE]

. In [3] it was shown that in example 3.1 the factorization $T=P_{I}Q_{J}$ is canonical.

In example 3.6 suppose that $B:{\cal S}\to{\cal L}$ has trivial nullspace and dense range. Elementary computations show that for $T=P_{R(E)}P_{N(E)}$,

[TABLE]

Then $\overline{R(T)}=R(E)$ and $N(T)=N(E))$, and this decomposition is also canonical.

Also in [3], it was shown that $R(P_{I})+R(Q_{J})$ is a closed proper direct sum, therefore $P_{I}Q_{J}$ is a different example from $P_{R(E)}P_{N(E)}$, for which $R(E)+N(E)$ is the whole space.

7 Dilations of contractions

Let $\Gamma$ be a contraction in a Hilbert space ${\cal H}_{0}$. P.R. Halmos showed in [14], that $\Gamma$ is the $1,1$ corner of a unitary operator $U$ acting in ${\cal H}_{0}\times{\cal H}_{0}$, namely

[TABLE]

If

[TABLE]

then

[TABLE]

i.e. $\Gamma$ factors as a unitary operator times a product of projections, on a bigger space. Apparently, $\Gamma$ is S-decomposable in ${\cal H}$ if and only if $QP$ is decomposable in ${\cal H}\times{\cal H}$

Moreover, if

[TABLE]

then

[TABLE]

Lemma 7.1**.**

On ${\cal H}\times{\cal H}$, one has that

[TABLE]

and

[TABLE]

Proof.

A vector in $R(P)$ is of the form \xi=\left(\begin{array}[]{c}\xi_{1}\\ 0\end{array}\right). $Q\xi=\xi$ if and only if

[TABLE]

Note that $(1-\Gamma\Gamma^{*})^{1/2}\Gamma=\Gamma(1-\Gamma^{*}\Gamma)^{1/2}$. Thus $\Gamma^{*}\Gamma\xi_{1}=\xi_{1}$ implies that $(1-\Gamma^{*}\Gamma)^{1/2}\xi_{1}=0$.

A vector $\xi\in R(P)$ belongs to $N(Q)$ if and only if $\Gamma^{*}\Gamma\xi_{1}=0$, i.e., $\xi_{1}\in N(\Gamma)$.

The other two statements are similar. ∎

Remark 7.2**.**

Straightforward computations show that

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Suppose that $N(\Gamma)=N(\Gamma^{*})=\{0\}$ (i.e., $P+Q-1$ has trivial nullspace and dense range). If $\Gamma=W|\Gamma|=|\Gamma^{*}|W$ are the polar decompositions (with $W$ a unitary operator), then

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With similar computations as above, one sees that if $\Gamma$ is S-decomposable with singular values $0<s_{n}\leq 1$, then the spectrum of $X_{P,Q}$ is $\{\pm\theta_{n}:cos(\theta_{n})=s_{n}\}$. With an argument as in Corollary 6.11, one has:

Corollary 7.3**.**

Let $\Gamma$ be a contraction in ${\cal H}_{0}$ with trivial nullspace and dense range, and $P_{\Gamma}$, $Q_{\Gamma}$ the above projections in ${\cal H}_{0}\times{\cal H}_{0}$.

If $\Gamma$ is invertible, then

[TABLE] 2. 2.

If $\Gamma$ is non invertible, then

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