Lattice properties of partial orders for complex matrices via orthogonal projectors

ABSTRACT

This paper deals with left star, star, and core partial orders for complex matrices. For each partial order, we present an order-isomorphism between the down-set of a fixed matrix B and a certain set (depending on the partial order) of orthogonal projectors whose matrix sizes can be considerably smaller than that of the matrix B. We study the lattice structure and we give properties of the down-sets. We prove that the down-set of B ordered by the core partial order and by the star partial order are sublattices of the down-set ordered by the left star partial order. We analize the existence of supremum and infimum of two given matrices and we give characterizations of these operations (whenever they exist). Some of the results given in the paper are already known in the literature but we present a different proof based on the previously established order-isomorphism.

KEYWORDS:

• Hartwig–Spindelböck factorization
• left star partial order
• star partial order
• core partial order
• lattice structure

AMS SUBJECT CLASSIFICATIONS:

• 15A09 (primary)
• 06A06 (secondary)

1. Introduction and preliminaries

The set of complex $m\times n$ matrices is denoted by ${\mathbb{C}}^{m\times n}$. The conjugate transpose, range, and rank of $A\in {\mathbb{C}}^{m\times n}$ are denoted by $A^{\ast }$, $\mathcal{R}\left(A\right)$, and $\mathrm{rk}\left(A\right)$, respectively. The identity matrix of order $n\times n$ is denoted by $I_n$ and zero matrices are denoted simply by O.

For each $A\in {\mathbb{C}}^{m\times n}$, there exists a unique matrix $X\in {\mathbb{C}}^{n\times m}$ such that AX and XA are Hermitian, AXA = A, and XAX = X, which is called the Moore-Penrose inverse of A and it is denoted by $A^{†}$. We denote by ${\mathbb{C}}_1^n$ the set of all $n\times n$ complex matrices that have index at most 1, that is, $\mathrm{rk}\left(A^2\right)=\mathrm{rk}\left(A\right)$. If $A\in {\mathbb{C}}_1^n$ then there exists a unique matrix $X\in {\mathbb{C}}^{n\times n}$ that satisfies $\mathrm{AX}=A{A}^{†}$ and $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A\right)$, which is called the core inverse of A and it is denoted by $X=A^{\overline{)\mathrm{\#}}}$. For further properties and applications of these inverses we refer the reader to [1–9].

This paper deals with some matrix partial orders. Specifically, with the star and the left star partial orders defined on the set ${\mathbb{C}}^{n\times n}$ of square complex matrices, and with the core partial order defined on the set ${\mathbb{C}}_1^n$. The star partial order was introduced by Drazin in [10] and it has been studied since then by numerous authors. The left star partial order was introduced by Baksalary and Mitra in [11]. Finally, the core partial order was introduced more recently by Baksalary and Trenkler in [1]. For any $A,B\in {\mathbb{C}}^{n\times n}$, let us recall that (see, for example, [12–14]):

• the left star partial order is defined by: $A\stackrel{{}_{l\ast }}{\le }B$ if and only if $A^{\ast }A=A^{\ast }B\mathrm{and}\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(B\right)$ (or equivalently, $A^{\ast }A=A^{\ast }B$ and $A=B{B}^{†}A$);

• the star partial order is defined by: $A\stackrel{{}_{\ast }}{\le }B$ if and only if $A^{\ast }A=A^{\ast }B$ and $A{A}^{\ast }=B{A}^{\ast }$ (or equivalently, $A^{†}A=A^{†}B$ and $A{A}^{†}=B{A}^{†}$);

and, for any $A,B\in {\mathbb{C}}_1^n$:

• the core partial order is defined by: $A\stackrel{{}_{\overline{)\mathrm{\#}}}}{\le }B$ if and only if $A^{\overline{)\mathrm{\#}}}A=A^{\overline{)\mathrm{\#}}}B$ and $A{A}^{\overline{)\mathrm{\#}}}=B{A}^{\overline{)\mathrm{\#}}}$ (or equivalently, $A^{\ast }A=A^{\ast }B$ and $\mathrm{BA}=A^2$).

For the sake of completeness we recall some basic definitions of structures defined over a partially ordered set that are used throughout the article. Recall that a partially ordered set (poset) $(Q,\le )$ is a lattice if for every $x,y\in Q$ both the least upper bound (or supremum) $x\vee y$ and the greatest lower bound (or infimum) $x\wedge y$ of $\{x,y\}$ exist. A lattice is said to be bounded if it has a first element 0 and a greatest element 1. Two elements a, b of a bounded lattice are complementary if $a\vee b=1$ and $a\wedge b=0$. A complemented lattice is a bounded lattice in which every element has a complement. An orthogonal lattice Q is a bounded lattice with a unary operation ${}^{\prime }$ that satisfies that $x\wedge x^{\prime }=0$, $x\vee x^{\prime }=1$, $(x\vee y{)}^{\prime }=x^{\prime }\wedge y^{\prime }$, $(x\wedge y{)}^{\prime }=x^{\prime }\vee y^{\prime }$, $x^{″}=x$, for all $x,y\in Q$. An orthomodular lattice is an orthogonal lattice that satisfies the law ‘if $x\le y$, then $y=x\vee (y\wedge x^{\prime })$’. A distributive lattice is a lattice which satisfies either (and hence, as it is easy to see, both) of the distributive laws $x\wedge (y\vee z)=(x\wedge y)\vee (x\wedge z)$ and $x\vee (y\wedge z)=(x\vee y)\wedge (x\vee z)$. Finally, a Boolean algebra is a complemented distributive lattice. Every Boolean algebra is an orthogonal lattice but, in general, the converse is not true. We refer the reader to [15] for more information about the different structures defined above.

Let Q and R be two posets. It is said that a map $\varphi :Q\to R$ is order-preserving if $\varphi \left(x\right)\le \varphi \left(y\right)$ holds in R whenever $x\le y$ holds in Q. We say that Q and R are (order-)isomorphic if there exists a bijection ϕ from Q to R such that both ϕ and ${\varphi }^{-1}$ are order-preserving. In that case, ϕ is called an order-isomorphism.

The aim of this paper is to study the down-sets $[O,B{]}^x=\{A|O\stackrel{{}_x}{\le }A\stackrel{{}_x}{\le }B\}$ for each $x\in \{l\ast ,\ast ,\overline{)\mathrm{\#}}\}$ and a fixed matrix B. If $x=\overline{)\mathrm{\#}}$ then it is required that B and all the matrices in $[O,B{]}^{\overline{)\mathrm{\#}}}$ have index at most 1 accordingly. The structure and properties of these down-sets were studied by other authors for rectangular matrices and for the wider case of bounded linear Hilbert space operators. For the case of the left star partial order, $[O,B{]}^{l\ast }$ was studied by Cırulis in [16] where it was proved that $[O,B{]}^{l\ast }$ is a complete orthomodular lattice. Antezana et al. studied in [17] the star partial order on bounded operators on a Hilbert space. In particular, from their results, it can be deduced that $[O,B{]}^{\ast }$ is a lattice. Finally, in [18], Djikić proved that $[O,B{]}^{\overline{)\mathrm{\#}}}$ is also a lattice.

Our approach to the study of $[O,B{]}^x$ is different from the authors abovementioned. In this paper, we prove that $[O,B{]}^x$ is order-isomorphic to a certain ordered set (depending on the partial order we are dealing with) of orthogonal projectors. Our starting point is the characterization given in [19] of matrices which are below a given matrix B by using a Hartwig-Spindelböck decomposition of B. More precisally, given $B\in {\mathbb{C}}^{n\times n}$ (or in ${\mathbb{C}}_1^n$ for $x=\overline{)\mathrm{\#}}$), where $0<r=\mathrm{rk}\left(B\right)$ and the r positive singular values ${\sigma }_1,\dots ,{\sigma }_r$ of B are ordered in decreasing order, we consider a Hartwig-Spindelböck decomposition of B (see [20]) given by (1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) where $U\in {\mathbb{C}}^{n\times n}$ is unitary, $\mathrm{\Sigma }=\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_r)\in {\mathbb{C}}^{r\times r}$, and $K\in {\mathbb{C}}^{r\times r}$ and $L\in {\mathbb{C}}^{r\times (n-r)}$ satisfy $K{K}^{\ast }+L{L}^{\ast }=I_r$ (note that L is absent when $r=n$). It is worth mentioning that this decomposition always exists but it is not necessarily unique, and that $B\in {\mathbb{C}}_1^n$ if and only if K is nonsingular. The predecessors of B are characterized as follows.

Theorem 1.1

[19, Theorems 4, 8, 16]

Let $B\in {\mathbb{C}}^{n\times n}$ (or $B\in {\mathbb{C}}_1^n$ for $x=\overline{)\mathrm{\#}}$) be a nonzero matrix written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) ). The following conditions are equivalent.

(1)

There exists a matrix $A\in {\mathbb{C}}^{n\times n}$ (where $A\in {\mathbb{C}}_1^n$ for $x=\overline{)\mathrm{\#}}$) such that $A\stackrel{{}_x}{\le }B$.

(2)

There exists a unique matrix $T\in {\mathbb{C}}^{r\times r}$ such that (2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) where $T^2=T=T^{\ast }$ and the following conditions hold depending on the partial order: no extra condition for the left star partial order, $T\mathrm{\Sigma }=\mathrm{\Sigma T}$ for the star partial order, and $T\mathrm{\Sigma KT}=\mathrm{\Sigma KT}$ for the core partial order.

According to Theorem 1.1, we define the following posets that play a crucial role in this paper.

Definition 1.2

Let $B\in {\mathbb{C}}^{n\times n}$ (or $B\in {\mathbb{C}}_1^n$ for $x=\overline{)\mathrm{\#}}$) be a nonzero matrix written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) ), let

1. ${\tau }_{\mathrm{\Sigma },K}^{l\ast }=\{T\in {\mathbb{C}}^{r\times r}|T^2=T=T^{\ast }\}$,

2. ${\tau }_{\mathrm{\Sigma },K}^{\ast }=\{T\in {\mathbb{C}}^{r\times r}|T^2=T=T^{\ast }\mathrm{and}T\mathrm{\Sigma }=\mathrm{\Sigma }T\}$, and

3. ${\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}=\{T\in {\mathbb{C}}^{r\times r}|T^2=T=T^{\ast }\mathrm{and}T\mathrm{\Sigma }KT=\mathrm{\Sigma }KT\}$,

endowed each one with the natural partial order given by $T_1\le T_2\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}{T}_1=T_1{T}_2.$

This last relation will be used indistinctly over any of the aformentioned sets.

It is easy to see that $T_1{T}_2=T_1$ implies $T_2{T}_1=T_1$ for any $T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^x$. Note that the set ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$ is the set of all orthogonal projectors in ${\mathbb{C}}^{r\times r}$. It is well-known that if $T_1$ and $T_2$ are orthogonal projectors in ${\mathbb{C}}^{r\times r}$ and we consider the partial order ≤ defined above then $({\tau }_{\mathrm{\Sigma },K}^{l\ast },\le )$ is an orthomodular lattice (see [21, Propositions 1, 2] and [16]) where, for any $T,T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }$, we have that $\begin{array}{rl}{T}_1\vee T_2& =(T_1+T_2)(T_1+T_2{)}^{†}=(T_1+T_2{)}^{†}(T_1+T_2),\\ T_1\wedge T_2& =2T_1(T_1+T_2{)}^{†}{T}_2=2T_2(T_1+T_2{)}^{†}{T}_1,\end{array}$and the complement of T is $T^{\prime }=I_r-T.$By Theorem 1.1, for any $x\in \{l\ast ,\ast ,\overline{)\mathrm{\#}}\}$, we clearly have a bijection $\varphi :[O,B{]}^x\to {\tau }_{\mathrm{\Sigma },K}^x$defined by $\varphi \left(A\right)=T,$ for every $A\in [O,B{]}^x$ and T given as in Theorem 1.1. Furthermore, we prove in Section 2 that ϕ is an order-isomorphism. Taking advantage of this order-isomorphism, we study the ordered structure of $[O,B{]}^x$ by means of the poset ${\tau }_{\mathrm{\Sigma },K}^x$. Matrices $T\in {\tau }_{\mathrm{\Sigma },K}^x$ are orthogonal projectors and, in addition, it can be proved that the Moore-Penrose inverse $T^{†}$ and the core inverse $T^{\overline{)\mathrm{\#}}}$ of T are both equal to T. Moreover, all of them belong to ${\mathbb{C}}^{r\times r}$ (instead of ${\mathbb{C}}^{n\times n}$), with $0<r\le n,$ where r can be considerably smaller than n. So, working with the matrices $T\in {\tau }_{\mathrm{\Sigma },K}^x$ is easier than using the matrices A and this fact brings significant advantages.

In Section 3 we investigate the lattice properties of $[O,B{]}^x$. One of our main goals is to show that there exists a relation between $[O,B{]}^{\ast }$, $[O,B{]}^{l\ast }$, and $[O,B{]}^{\overline{)\mathrm{\#}}}$. More precisally, based on the order-isomorphism proved in Section 2, we show that $[O,B{]}^x$, for each x, is a lattice; and that $[O,B{]}^{\ast }$ and $[O,B{]}^{\overline{)\mathrm{\#}}}$ are sublattices of $[O,B{]}^{l\ast }$. In addition, we find properties of $[O,B{]}^x$, for each partial order. We show that $[O,B{]}^{l\ast }$ and $[O,B{]}^{\ast }$ are orthomodular lattices whose subchains (that is, a subset for which every pair of elements are comparable) are all finite. We give a necessary and sufficient condition for $[O,B{]}^{l\ast }$ to be distributive. We show that if $[O,B{]}^{\ast }$ is distributive then it is a Boolean algebra. Eagambaram et al. showed in [22] that $[O,B{]}^{\ast }$ is a finite lattice if and only if all the positive singular values of B are pairwise distinct. We improve this result by showing that, in that case, not only $[O,B{]}^{\ast }$ is a finite lattice but also a Boolean algebra. Additionally, we derive its cardinality. For the left star and the star partial orders, we prove that if $A_1\stackrel{x}{\le }{A}_2\stackrel{x}{\le }B$ then $[A_1,A_2{]}^x$ and $[O,A_2-A_1{]}^x$ are order-isomorphic. Assuming that $A_2-A_1\stackrel{\overline{)\mathrm{\#}}}{\le }B$, an analogous result is obtained for the core partial order.

As a last application of the order-isomorphism ϕ, we study the supremum and the infimum of two given matrices in ${\mathbb{C}}^{n\times n}$ (or in ${\mathbb{C}}_1^n$ for the core partial order). Xu et al. proved in [23] that there exists the star supremum of $A_1$ and $A_2$ if and only if $A_1$ and $A_2$ have a common upper bound. Moreover, an explicit representation of the supremum was established (whenever it exists). In [24], Hartwig gave necessary and sufficient conditions for the existence of the star supremum in rings with involution and found an expression for that supremum. Later, Djikić gave in [25] a simple necessary and sufficient condition for the existence of the star supremum for two operators on a Hilbert space. Recently, Djikić proved in [18] a similar result to that by Xu et al., for the core partial order by giving necessary and sufficient conditions for the existence of the core supremum in a Hilbert space. In Section 4, we use the order-isomorphism ϕ to present a different proof from those given by Hartwig, Xu et al. and Djikić. Our proof is also valid for the left star partial order. In addition, we compute the supremum (whenever it exists) by means of the same expression for the three orders. Finally, we analyse the infimum of two arbitrary matrices. Hartwig and Drazin proved in [21] that the set of matrices endowed with the star partial order is a lower semilattice, i.e. for every pair of matrices $A_1$ and $A_2$, there exists $A_1\wedge A_2$. The set of matrices that have index at most 1 endowed with the core partial order is also a lower semilattice (see [18]). In Section 4, we compute the infimum of two matrices that have a common upper bound by means of the same expression for the three orders. We would like to highlight that the expressions for the infimum and supremum of two matrices in $[O,B{]}^{l\ast }$ that we provide are different from those given in [16].

If two matrices B (written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) )) and C do not have a common upper bound, we find an expression of the type (2(2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) ) for the infimum and the conditions that the associated orthogonal projectors must satisfy.

2. Isomorphic representation of down-sets

From now on, x will refer to any of the three partial orders we are dealing with, that is, $x\in \{l\ast ,\ast ,\overline{)\mathrm{\#}}\}$. In the case that $x=\overline{)\mathrm{\#}}$, without mentioning it explicitly, we will regard the matrices to be in ${\mathbb{C}}_1^n$.

In this section we state the order-isomorphism between $[O,B{]}^x$ and ${\tau }_{\mathrm{\Sigma },K}^x$. In order to do that, for a fixed a Hartwig-Spindelböck decomposition of B, we consider the posets ${\tau }_{\mathrm{\Sigma },K}^x$ and the bijection $\varphi :[O,B{]}^x\to {\tau }_{\mathrm{\Sigma },K}^x$defined by $\varphi \left(A\right)=T$ given in Section 1. Note that if $T\in {\tau }_{\mathrm{\Sigma },K}^x$ then $T\in {\mathbb{C}}_1^r$; O is the least element and $I_r$ is the greatest element of ${\tau }_{\mathrm{\Sigma },K}^x$. More precisely, we should denote ϕ by ${\varphi }_{\mathrm{\Sigma },K}$ because this map depends on matrices Σ and K of the decomposition used to factorize the matrix B. However, to simplify the notation, from now on, we simply denote it by ϕ.

Theorem 2.1

The posets $[O,B{]}^x$ and ${\tau }_{\mathrm{\Sigma },K}^x$ are order-isomorphic. Moreover, the function rank is preserved under the order-isomorphism ϕ.

Proof.

Let us first prove that ϕ is order-preserving. For that, let $A_1\stackrel{{}_x}{\le }{A}_2\in [O,B{]}^x$ both written as in (2(2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) ), $T_1=\varphi \left(A_1\right)$ and $T_2=\varphi \left(A_2\right)$. From $A_1^{\ast }{A}_1=A_1^{\ast }{A}_2$ and taking into account that $T_1^2=T_1=T_1^{\ast }$, we have that $\left[\begin{array}{cc}{K}^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}& K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}\\ L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}& L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}\end{array}\right]=\left[\begin{array}{cc}{K}^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K}& K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L}\\ L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K}& L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L}\end{array}\right].$Hence we obtain the following system (3) $\begin{array}{rl}& K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K},\end{array}$(3) (4) $\begin{array}{rl}& K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L},\end{array}$(4) (5) $\begin{array}{rl}& L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K},\end{array}$(5) (6) $\begin{array}{rl}& L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L}.\end{array}$(6) Post-multiplicating (3(3) $\begin{array}{rl}& K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K},\end{array}$(3) ) and (4(4) $\begin{array}{rl}& K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L},\end{array}$(4) ) by $K^{\ast }$ and $L^{\ast }$, respectively, and then adding both equations we obtain $K^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma }=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma },$ since $K{K}^{\ast }+L{L}^{\ast }=I_r$. Consequently, (7) $\begin{array}{r}{K}^{\ast }\mathrm{\Sigma }{T}_1=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2.\end{array}$(7) Similarly, from (5(5) $\begin{array}{rl}& L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma K}=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma K},\end{array}$(5) ) and (6(6) $\begin{array}{rl}& L^{\ast }\mathrm{\Sigma }{T}_1\mathrm{\Sigma L}=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2\mathrm{\Sigma L}.\end{array}$(6) ) we obtain (8) $\begin{array}{r}{L}^{\ast }\mathrm{\Sigma }{T}_1=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2.\end{array}$(8) Pre-multiplying (7(7) $\begin{array}{r}{K}^{\ast }\mathrm{\Sigma }{T}_1=K^{\ast }\mathrm{\Sigma }{T}_1{T}_2.\end{array}$(7) ) and (8(8) $\begin{array}{r}{L}^{\ast }\mathrm{\Sigma }{T}_1=L^{\ast }\mathrm{\Sigma }{T}_1{T}_2.\end{array}$(8) ) by K and L respectively, and then adding we have $T_1=T_1{T}_2$ and this means that $T_1\le T_2$.

Let us suppose now that $T_1\le T_2$ with $T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^x$. By $T_1=T_1{T}_2$ and $T_1^2=T_1=T_1^{\ast }$, it is straightforward to see that $A_1^{\ast }{A}_1=A_1^{\ast }{A}_2$. To prove that $A_1\stackrel{{}_x}{\le }{A}_2$, we consider each partial order separately.

Consider first the star partial order. Then, $A_2{A}_1^{\ast }=U\left[\begin{array}{cc}{T}_2\mathrm{\Sigma K}& T_2\mathrm{\Sigma L}\\ O& O\end{array}\right]\left[\begin{array}{cc}{K}^{\ast }\mathrm{\Sigma }{T}_1& O\\ L^{\ast }\mathrm{\Sigma }{T}_1& O\end{array}\right]U^{\ast }=U\left[\begin{array}{cc}{T}_2{\mathrm{\Sigma }}^2{T}_1& O\\ O& O\end{array}\right]U^{\ast }.$Since both $T_1$ and $T_2$ commute with Σ, we have that $T_2{\mathrm{\Sigma }}^2{T}_1={\mathrm{\Sigma }}^2{T}_2{T}_1={\mathrm{\Sigma }}^2{T}_1={\mathrm{\Sigma }}^2{T}_1{T}_1=T_1{\mathrm{\Sigma }}^2{T}_1$. Thus, $A_2{A}_1^{\ast }=A_1{A}_1^{\ast }$. Hence, $A_1\stackrel{\ast }{\le }{A}_2$.

Consider now $x=l\ast$. From [19, Lemma 14], we know that $T_2=T_2\mathrm{\Sigma }(T_2\mathrm{\Sigma }{)}^{†}$ since Σ is nonsingular. Then, $T_1=T_2{T}_1=T_2\mathrm{\Sigma }(T_2\mathrm{\Sigma }{)}^{†}{T}_1$. From [19, Lemma 3], we also know that $A_2^{†}=U\left[\begin{array}{cc}{K}^{\ast }(T_2\mathrm{\Sigma }{)}^{†}& O\\ L^{\ast }(T_2\mathrm{\Sigma }{)}^{†}& O\end{array}\right]U^{\ast }$. Now, taking into account this fact, it is easy to see that $A_2{A}_2^{†}{A}_1=A_1$. So, $A_1\stackrel{{}_l\ast }{\le }{A}_2$.

Finally, we consider the core partial order. From $\mathrm{\Sigma K}{T}_1=T_1\mathrm{\Sigma K}{T}_1$ and $T_2{T}_1=T_1$ we have $T_1\mathrm{\Sigma K}{T}_1\mathrm{\Sigma K}=T_2{T}_1\mathrm{\Sigma K}{T}_1\mathrm{\Sigma K}=T_2\mathrm{\Sigma K}{T}_1\mathrm{\Sigma K}$and $T_1\mathrm{\Sigma K}{T}_1\mathrm{\Sigma L}=T_2{T}_1\mathrm{\Sigma K}{T}_1\mathrm{\Sigma L}=T_2\mathrm{\Sigma K}{T}_1\mathrm{\Sigma L}.$Then $A_2{A}_1=A_1^2$ follows. So, $A_1\stackrel{\overline{)\mathrm{\#}}}{\le }{A}_2$.

We have proved that $[O,B{]}^x$ is order-isomorphic to ${\tau }_{\mathrm{\Sigma },K}^x$, for every x.

In order to see that ϕ preserves the rank function, we observe that if $A\in [O,B{]}^x$ and $T=\varphi \left(A\right)$, then $\begin{array}{rl}\mathrm{rk}\left(A\right)& =\mathrm{rk}\left(A{A}^{\ast }\right)=\mathrm{rk}\left(\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]\left[\begin{array}{cc}{K}^{\ast }\mathrm{\Sigma }{T}^{\ast }& O\\ L^{\ast }\mathrm{\Sigma }{T}^{\ast }& O\end{array}\right]\right)=\mathrm{rk}\left(\left[\begin{array}{cc}T{\mathrm{\Sigma }}^2{T}^{\ast }& O\\ O& O\end{array}\right]\right)\\ & =\mathrm{rk}\left(\left[\begin{array}{cc}\left(T\mathrm{\Sigma }\right)(T\mathrm{\Sigma }{)}^{\ast }& O\\ O& O\end{array}\right]\right)=\mathrm{rk}\left(T\mathrm{\Sigma }\right)=\mathrm{rk}\left(T\right)=\mathrm{rk}\left(\varphi \right(A\left)\right).\text{■}\end{array}$

Remark 2.1

By using a Schur's factorization of the matrix $\mathrm{\Sigma K}$, we have that there exists a unitary matrix V and an upper triangular matrix S such that $\mathrm{\Sigma K}=\mathrm{VS}{V}^{\ast }$. It can be proved that the sets ${\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}$ and ${\rho }_B^{\overline{)\mathrm{\#}}}:=\{T\in {\mathbb{C}}^{r\times r}|T=T^2=T^{\ast }\mathrm{and}\mathrm{TST}=\mathrm{ST}\}$, ordered by ≤, are order-isomorphic by using the map $\phi :{\rho }_B^{\overline{)\mathrm{\#}}}\to {\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}$ defined by $\phi \left(T\right)=\mathrm{VT}{V}^{\ast }$. In practice, examples can be constructed more easily with a such matrix S instead of using $\mathrm{\Sigma K}$.

Remark 2.2

Assume $A\stackrel{x}{\le }B$ and $A\ne B$. It is easy to see that $\mathrm{rk}\left(A\right)<\mathrm{rk}\left(B\right)$ and so the maximum length of any subchain in $[O,B{]}^x$ is $\mathrm{rk}\left(B\right)+1$. Moreover, if $\mathrm{rk}\left(B\right)=r$ and we consider the projectors $T_s=\left[t_{\mathrm{ij}}\right]\in {\mathbb{C}}^{r\times r}$ where $t_{\mathrm{ij}}=\{\begin{array}{ll}1& \mathrm{if}i=j\mathrm{and}i\le s\\ 0& \mathrm{otherwise}\end{array},$for each $s\in \{1,\dots ,r\}$, it is straightforward to see that $T_s\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }$, $T_s\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$, and $T_s\in {\rho }_B^{\overline{)\mathrm{\#}}}$. Then, we obtain a chain $O<T_1<\cdots <T_r=I_r.$with r + 1 elements of maximum length.

Lemma 2.2

Let $B_1,B_2\in {\mathbb{C}}^{n\times n}$. If $[O,B_1{]}^x$ is order-isomorphic to $[O,B_2{]}^x$ then $\mathrm{rk}\left(B_1\right)=\mathrm{rk}\left(B_2\right)$. Moreover, if $\mathrm{rk}\left(B_1\right)=\mathrm{rk}\left(B_2\right)$ then $[O,B_1{]}^{l\ast }$ is order-isomorphic to $[O,B_2{]}^{l\ast }$.

Proof.

Let us suppose that $\mathrm{rk}\left(B_1\right)<\mathrm{rk}\left(B_2\right)$. Then, by using Remark 2.2, we can construct a chain in $[O,B_1{]}^x$ of length $\mathrm{rk}\left(B_2\right)+1$ and this contradicts the maximum length of a chain in $[O,B_1{]}^x$. The second statement is immediate from Theorem 2.1.

Remark 2.3

For each $x\in \{l\ast ,\ast ,\overline{)\mathrm{\#}}\}$:

1. if $\mathrm{rk}\left(B\right)=1$, then ${\tau }_{\mathrm{\Sigma },K}^x=\{O,I_1\}$. Hence, $[O,B{]}^x$ is a chain with two elements.

2. if $\mathrm{rk}\left(B\right)=2$, then $\mathrm{rk}\left(T\right)=1$ for each $T\in {\tau }_{\mathrm{\Sigma },K}^x\setminus \{O,I_2\}$; so every distinct $T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^x\setminus \{O,I_2\}$ are incomparable and thus $[O,B{]}^x$ has the aspect presented in Figure 1.

Figure 1. The lattice $[O,B{]}^x$ with $\mathrm{rk}\left(B\right)=2$.

3. Lattice structure of $[O,B{]}^x$

In this section we investigate the lattice structure of $[O,B{]}^x$ for each x by using the order-isomorphism ϕ. We prove that $[O,B{]}^{\ast }$ and $[O,B{]}^{\overline{)\mathrm{\#}}}$ are sublattices of $[O,B{]}^{l\ast }$. For each x, we analyse the structure of $[O,B{]}^x$. We show that $[O,B{]}^{l\ast }$ and $[O,B{]}^{\ast }$ are orthomodular lattices whose subchains are all finite. In addition, we give a necessary and sufficient condition for $[O,B{]}^{l\ast }$ to be distributive. We also state that if $[O,B{]}^{\ast }$ is distributive then it is a Boolean algebra. Finally, we give necessary and sufficient conditions for $[O,B{]}^{\ast }$ to be a finite Boolean algebra.

For the left star and the star partial order, we prove that if $A_1\stackrel{x}{\le }{A}_2\stackrel{x}{\le }B$ then $[A_1,A_2{]}^x$ and $[O,A_2-A_1{]}^x$ are order-isomorphic. An analogous result is obtained for the core partial order, provided that $A_2-A_1\stackrel{\overline{)\mathrm{\#}}}{\le }B$ holds.

We start giving the infimum and the supremum of two matrices in the segment $[O,B{]}^x$ for the case in which their associated orthogonal projectors commute.

Proposition 3.1

Let $T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^x$ such that $T_1{T}_2=T_2{T}_1$. Then $T_1\wedge T_2=T_1{T}_2$ and $T_1\vee T_2=T_1+T_2-T_1{T}_2$.

Proof.

It is clear that $(T_1{T}_2{)}^2=T_1{T}_2=(T_1{T}_2{)}^{\ast }$ and it is well-known that $T_1{T}_2$ is the infimum of $T_1$ and $T_2$ in ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$ (see [16]). In addition, if $x=\ast$, then $\left(T_1{T}_2\right)\mathrm{\Sigma }=T_1\mathrm{\Sigma }{T}_2=\mathrm{\Sigma }\left(T_1{T}_2\right)$; and if $x=\overline{)\mathrm{\#}}$, then $\left(T_1{T}_2\right)\mathrm{\Sigma K}\left(T_1{T}_2\right)=T_1\left(T_2\mathrm{\Sigma K}{T}_2\right)T_1=T_1\left(\mathrm{\Sigma K}{T}_2\right)T_1=\left(T_1\mathrm{\Sigma K}{T}_1\right)T_2=\mathrm{\Sigma K}\left(T_1{T}_2\right)$. So, $T_1\wedge T_2=T_1{T}_2$ in ${\tau }_{\mathrm{\Sigma },K}^x$ for all x.

It is also known that if $T_1$ and $T_2$ commute then $T_1\vee T_2=T_1+T_2-T_1{T}_2\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }$ (see [16]). In addition, if $x=\ast$, then $(T_1+T_2-T_1{T}_2)\mathrm{\Sigma }=\mathrm{\Sigma }(T_1+T_2-T_1{T}_2)$; and if $x=\overline{)\mathrm{\#}}$, then $(T_1+T_2-T_1{T}_2)\mathrm{\Sigma K}(T_1+T_2-T_1{T}_2)=T_1\mathrm{\Sigma K}{T}_1+T_1\mathrm{\Sigma K}{T}_2-T_1\mathrm{\Sigma K}{T}_1{T}_2+T_2\mathrm{\Sigma K}{T}_1+T_2\mathrm{\Sigma K}{T}_2-T_2\mathrm{\Sigma K}{T}_1{T}_2-T_1{T}_2\mathrm{\Sigma K}{T}_1-T_1{T}_2\mathrm{\Sigma K}{T}_2+T_1{T}_2\mathrm{\Sigma K}{T}_1{T}_2=\mathrm{\Sigma K}{T}_1+T_1\mathrm{\Sigma K}{T}_2-\mathrm{\Sigma K}{T}_1{T}_2+T_2\mathrm{\Sigma K}{T}_1+\mathrm{\Sigma K}{T}_2-\mathrm{\Sigma K}{T}_2{T}_1-T_2\mathrm{\Sigma K}{T}_1-T_1\mathrm{\Sigma K}{T}_2+\mathrm{\Sigma K}{T}_1{T}_2=\mathrm{\Sigma K}{T}_1+\mathrm{\Sigma K}{T}_2-\mathrm{\Sigma K}{T}_1{T}_2=\mathrm{\Sigma K}(T_1+T_2-T_1{T}_2)$. Therefore, $T_1+T_2-T_1{T}_2\in {\tau }_{\mathrm{\Sigma },K}^x$ for every x.

As an immediate consequence of the above result and the fact that ϕ is an order-isomorphism we have the following result.

Corollary 3.2

Let $A_1,A_2\in [O,B{]}^x$ be written as in (2(2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) ) such that $T_1{T}_2=T_2{T}_1$, where $T_i=\varphi \left(A_i\right)$ for every $i\in \{1,2\}$. Then:

(a)	$A_1\wedge A_2=U\left[\begin{array}{cc}{T}_1{T}_2\mathrm{\Sigma K}& T_1{T}_2\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$ and
(b)	$A_1\vee A_2=A_1+A_2-A_1\wedge A_2$.

We now investigate $[O,B{]}^x$ separately for each order.

3.1. Left star partial order

In this section we show that $[O,B{]}^{l\ast }$ is an orthomodular lattice of finite height and nondistributive provided that $\mathrm{rk}\left(B\right)\ge 2$. It is worth mentioning that the fact that $[O,B{]}^{l\ast }$ is an orthomodular lattice was proved by Cırulis in [16] for the more general case of a bounded operator X over a complex Hilbert space H, by setting an isomorphism between every down-set $[O,X{]}^{l\ast }$ of the set of all bounded linear operators over H and the down-set $[O,P_X{]}^{l\ast }$ of projectors where $P_X$ is the projector onto the closure of the range $\mathcal{R}\left(X\right)$ ($P_X=X{X}^{†}$ for $X,P_X\in {\mathbb{C}}^{n\times n}$). Our proof is based on the order-isomorphism ϕ and the advantage of this technique is that allows us to work with orthogonal projectors whose sizes can be considerably smaller than those of matrix B itself.

Our first objective is to show that $[O,B{]}^{l\ast }$ is a nondistributive lattice if $\mathrm{rk}\left(B\right)\ge 2$. In order to do that, we have to observe that ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$ is exactly the set of all $r\times r$ orthogonal projectors. So, we only need to find an example where the distributive property does not hold and this example will serve in general.

Example 3.3

Let B be any matrix in ${\mathbb{C}}^{n\times n}$ such that $\mathrm{rk}\left(B\right)\ge 2$ and $A_1,A_2,A_3\in [O,B{]}^{l\ast }$ such that $T_i=\varphi \left(A_i\right)=\left[\begin{array}{cc}{X}_i& O\\ O& O\end{array}\right]\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }$ for every $i\in \{1,2,3\}$, where $X_1=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, $X_2=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$, and $X_3=\left[\begin{array}{cc}1/2& 1/2\\ 1/2& 1/2\end{array}\right]$. Let us see that $A_3\wedge (A_1\vee A_2)\ne (A_3\wedge A_1)\vee (A_3\wedge A_2)$. Indeed, by Proposition 3.1, $T_3\wedge (T_1\vee T_2)=T_3(T_1+T_2-T_1{T}_2)=T_3$. On the other hand, by Remark 2.3, $(T_3\wedge T_1)\vee (T_3\wedge T_2)=O\vee O=O$, since $\mathrm{rk}\left(T_i\right)=1$, for each $i\in \{1,2,3\}$.

Since ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$ is an orthomodular lattice (see [16]), by Theorem 2.1, $[O,B{]}^{l\ast }$ is an orthomodular lattice too and, by considering a rank argument, it is clear that all its subchains are finite. In this case, it is said that the lattice has finite height. We summarize these reasonings in the following theorem.

Theorem 3.4

If $B\in {\mathbb{C}}^{n\times n}$ then $[O,B{]}^{l\ast }$ is an orthomodular lattice of finite height. In addition, if $\mathrm{rk}\left(B\right)\ge 2$ then $[O,B{]}^{l\ast }$ is nondistributive.

Remark 3.1

Let $A_1,A_2\in [O,B{]}^{l\ast }$ be written as in (2(2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) ), where $T_i=\varphi \left(A_i\right)$ for each $i\in \{1,2\}$. Then:

1. $A_1\wedge A_2=U\left[\begin{array}{cc}(T_1\wedge T_2)\mathrm{\Sigma K}& (T_1\wedge T_2)\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$ and

2. $A_1\vee A_2=U\left[\begin{array}{cc}(T_1\vee T_2)\mathrm{\Sigma K}& (T_1\vee T_2)\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$.

Remark 3.2

If $\mathrm{rk}\left(B\right)\ge 2$ then $[O,B{]}^{l\ast }$ is an infinite lattice. For example, if $X=\left[\begin{array}{cc}a& b\\ \overline{b}& 1-a\end{array}\right]$, with a in the real interval $[0,1]$, $b\in \mathbb{C}$, and $|b{|}^2=a-a^2$, then $T=\left[\begin{array}{cc}X& O\\ O& O\end{array}\right]\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }$.

Note also that if A is any matrix such that $A\stackrel{l\ast }{\le }B$ and $\mathrm{rk}\left(A\right)=2$ then, in general, $[O,A{]}^{l\ast }$ is an infinite lattice order-isomorphic to the one given in Figure 1.

Remark 3.3

Let $P,T,Q\in {\mathbb{C}}^{r\times r}$ be orthogonal projectors such that $P\le T$ and TQ = O. It is easy to see that PQ = O.

Lemma 3.5

Let $A,A_1,A_2,B\in {\mathbb{C}}^{n\times n}$. If $A_1\stackrel{l\ast }{\le }{A}_2\stackrel{l\ast }{\le }B$ then $[O,A_2-A_1{]}^{l\ast }$ and $[A_1,A_2{]}^{l\ast }$ are order-isomorphic. In particular, if $A\stackrel{l\ast }{\le }B$ then $[O,B-A{]}^{l\ast }$ and $[A,B{]}^{l\ast }$ are order-isomorphic.

Proof.

Assume that $A_1\stackrel{l\ast }{\le }{A}_2\stackrel{l\ast }{\le }B$ and set $T_1,T_2$ such that $\varphi \left(A_i\right)=T_i$, for each $i\in \{1,2\}$. If P satisfies $P^2=P^{\ast }=P\le T_2-T_1$, by $(T_2-T_1)T_1=O$ and Remark 3.3, we have that $P{T}_1=O$. Moreover, $T_{1}P=(P^{\ast }{T}_1^{\ast }{)}^{\ast }=(P{T}_1{)}^{\ast }=O$. It is easy to see that $P+T_1$ is idempotent, Hermitian, and $T_1\le P+T_1$. Now, again from $P\le T_2-T_1$, we have that $P=P(T_2-T_1)=P{T}_2$ and then $(P+T_1)T_2=P{T}_2+T_1{T}_2=P+T_1$; i.e. $P+T_1\le T_2$. Thus, the map $\phi :[O,T_2-T_1]\to [T_1,T_2]$ given by $\phi \left(P\right)=P+T_1$ is well-defined. Let us prove that φ is an order-isomorphism. Indeed, let $Q\in [T_1,T_2]$ and $P=Q-T_1$. Since $P(T_2-T_1)=(Q-T_1)(T_2-T_1)=Q{T}_2-T_1{T}_2-Q{T}_1+T_1^2=Q-T_1-T_1+T_1=Q-T_1=P$, we get $P\in [O,T_2-T_1]$ and $\phi \left(P\right)=Q$. Thus, φ is surjective. Let $P_1,P_2\in [O,T_2-T_1]$. Since $T_1{P}_2=O$ and $P_1{T}_1=O$, we have $\phi \left(P_1\right)\le \phi \left(P_2\right)$ if and only if $(P_1+T_1)(P_2+T_1)=P_1+T_1$, that is equivalent to $P_1{P}_2+T_1{P}_2+P_1{T}_1+T_1^2=P_1+T_1$, which simplifies to $P_1{P}_2=P_1$, that is, $P_1\le P_2$. Then, φ is an order-isomorphism.

The second statement follows by setting $A_1=A$ and $A_2=B$.

Lemma 3.5 allows us to realize the complexity of the down-set $[O,B{]}^{l\ast }$ when $\mathrm{rk}\left(B\right)\ge 2$. For instance, if we choose a matrix A such that $\mathrm{rk}(B-A)=2$ then the Figure 1 will appear repeated at the top (down-set $[A,B{]}^{l\ast }$) and at the bottom (down-set $[O,B-A{]}^{l\ast }$) of the Hasse diagram of the whole down-set $[O,B{]}^{l\ast }$.

3.2. Star partial order

We now need the following technical result.

Lemma 3.6

[3, Theorem 1.4.2]

Let $A\in {\mathbb{C}}^{m\times n}$ and $B\in {\mathbb{C}}^{n\times p}$. Then $(\mathrm{AB}{)}^{†}=B^{†}{A}^{†}$ if and only if $A^{†}\mathrm{AB}{B}^{\ast }{A}^{\ast }=B{B}^{\ast }{A}^{\ast }$ and $B{B}^{†}{A}^{\ast }\mathrm{AB}=A^{\ast }\mathrm{AB}$.

Theorem 3.7

If $B\in {\mathbb{C}}^{n\times n}$ then $[O,B{]}^{\ast }$ is a sublattice of $[O,B{]}^{l\ast }$.

Proof.

It is immediate that ${\tau }_{\mathrm{\Sigma },K}^{\ast }\subseteq {\tau }_{\mathrm{\Sigma },K}^{l\ast }$. Then $[O,B{]}^{\ast }\subseteq [O,B{]}^{l\ast }$.

Let $A_1,A_2\in [O,B{]}^{\ast }$. We know that $A_1\vee A_2$ and $A_1\wedge A_2$ exist in $[O,B{]}^{l\ast }$. Now we prove that $A_1\vee A_2,A_1\wedge A_2\in [O,B{]}^{\ast }$. By Theorem 3.4, $\varphi \left(A_1\right)\vee \varphi \left(A_2\right)$ and $\varphi \left(A_1\right)\wedge \varphi \left(A_2\right)$ exist in ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$. So, we only need to see that $\left(\varphi \right(A_1)\vee \varphi (A_2\left)\right)\mathrm{\Sigma }=\mathrm{\Sigma }\left(\varphi \right(A_1)\vee \varphi (A_2\left)\right)$ and $\left(\varphi \right(A_1)\wedge \varphi (A_2\left)\right)\mathrm{\Sigma }=\mathrm{\Sigma }\left(\varphi \right(A_1)\wedge \varphi (A_2\left)\right)$.

Let $T_1=\varphi \left(A_1\right)$ and $T_2=\varphi \left(A_2\right)$. Taking into account that ${\mathrm{\Sigma }}^{†}={\mathrm{\Sigma }}^{-1}$ and ${\mathrm{\Sigma }}^{\ast }=\mathrm{\Sigma }$, the equalities ${\mathrm{\Sigma }}^{†}\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{\ast }{\mathrm{\Sigma }}^{\ast }=(T_1+T_2\left)\right(T_1+T_2{)}^{\ast }{\mathrm{\Sigma }}^{\ast }$, and $(T_1+T_2)(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{\ast }\mathrm{\Sigma }(T_1+T_2)=(T_1+T_2\left)\right(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma \Sigma }=(T_1+T_2)\mathrm{\Sigma \Sigma }={\mathrm{\Sigma }}^{\ast }\mathrm{\Sigma }(T_1+T_2)$ imply, by Lemma 3.6, that (9) $\begin{array}{r}\left(\mathrm{\Sigma }\right(T_1+T_2){)}^{†}=(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}.\end{array}$(9) Now, the equalities $\begin{array}{rl}& \left(\mathrm{\Sigma }\right(T_1+T_2\left){)}^{†}\mathrm{\Sigma }\right(T_1+T_2\left){\mathrm{\Sigma }}^{-1}\right({\mathrm{\Sigma }}^{-1}{)}^{\ast }\left(\mathrm{\Sigma }\right(T_1+T_2){)}^{\ast }\\ & =(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}\mathrm{\Sigma }(T_1+T_2\left){\mathrm{\Sigma }}^{-1}\right(T_1+T_2{)}^{\ast }\\ & =\left(\right(T_1+T_2{)}^{†}(T_1+T_2){)}^{\ast }(T_1+T_2{)}^{\ast }{\mathrm{\Sigma }}^{-1}\\ & =\left(\right(T_1+T_2\left)\right(T_1+T_2{)}^{†}(T_1+T_2){)}^{\ast }{\mathrm{\Sigma }}^{-1}=(T_1+T_2){\mathrm{\Sigma }}^{-1}\\ & ={\mathrm{\Sigma }}^{-1}\left({\mathrm{\Sigma }}^{-1}{)}^{\ast }\right(\mathrm{\Sigma }(T_1+T_2){)}^{\ast }\end{array}$and $\begin{array}{rl}& {\mathrm{\Sigma }}^{-1}\left({\mathrm{\Sigma }}^{-1}{)}^{†}\right(\mathrm{\Sigma }(T_1+T_2){)}^{\ast }\mathrm{\Sigma }(T_1+T_2){\mathrm{\Sigma }}^{-1}={\mathrm{\Sigma }}^{-1}\mathrm{\Sigma }\left(\mathrm{\Sigma }\right(T_1+T_2\left){)}^{\ast }\mathrm{\Sigma }\right(T_1+T_2){\mathrm{\Sigma }}^{-1}\\ & =\left(\mathrm{\Sigma }\right(T_1+T_2\left){)}^{\ast }\mathrm{\Sigma }\right(T_1+T_2){\mathrm{\Sigma }}^{-1},\end{array}$again by Lemma 3.6 and (9(9) $\begin{array}{r}\left(\mathrm{\Sigma }\right(T_1+T_2){)}^{†}=(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}.\end{array}$(9) ), imply that $\left(\right(\mathrm{\Sigma }(T_1+T_2)){\mathrm{\Sigma }}^{-1}{)}^{†}=\mathrm{\Sigma }(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}.$Finally, from $(T_1+T_2{)}^{†}=(T_1\mathrm{\Sigma }{\mathrm{\Sigma }}^{-1}+T_2\mathrm{\Sigma }{\mathrm{\Sigma }}^{-1}{)}^{†}=(\mathrm{\Sigma }{T}_1{\mathrm{\Sigma }}^{-1}+\mathrm{\Sigma }{T}_2{\mathrm{\Sigma }}^{-1}{)}^{†}=(\mathrm{\Sigma }(T_1+T_2){\mathrm{\Sigma }}^{-1}{)}^{†}=\mathrm{\Sigma }(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}$, we get that $\begin{array}{rl}& (T_1\wedge T_2)\mathrm{\Sigma }=2T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma }=2T_1\mathrm{\Sigma }(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}{T}_2\mathrm{\Sigma }\\ & =2\mathrm{\Sigma }{T}_1(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}\mathrm{\Sigma }{T}_2=\mathrm{\Sigma }2T_1(T_1+T_2{)}^{†}{T}_2=\mathrm{\Sigma }(T_1\wedge T_2)\end{array}$and $\begin{array}{rl}& (T_1\vee T_2)\mathrm{\Sigma }=(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma }=(T_1+T_2\left)\mathrm{\Sigma }\right(T_1+T_2{)}^{†}{\mathrm{\Sigma }}^{-1}\mathrm{\Sigma }\\ & =\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{†}=\mathrm{\Sigma }(T_1\vee T_2).\end{array}$Hence, $[O,B{]}^{\ast }$ is a sublattice of $[O,B{]}^{l\ast }$.

Proposition 3.8

The lattice $[O,B{]}^{\ast }$ is an orthomodular lattice of finite height. Moreover, if $[O,B{]}^{\ast }$ is distributive then $[O,B{]}^{\ast }$ is a Boolean algebra.

Proof.

Let $T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$. Let us see that $I_r-T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$. Indeed, it is clear that $(I_r-T{)}^2=I_r-T=(I_r-T{)}^{\ast }$. Since $T\mathrm{\Sigma }=\mathrm{\Sigma T}$, then $(I_r-T)\mathrm{\Sigma }=\mathrm{\Sigma }(I_r-T)$. So $I_r-T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$. Thus, ${\tau }_{\mathrm{\Sigma },K}^{\ast }$ is closed under the unary operation of complementation of ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$. Taking into account Theorems 3.4 and 3.7, we have that ${\tau }_{\mathrm{\Sigma },K}^{\ast }$ is an orthomodular lattice.

If ${\tau }_{\mathrm{\Sigma },K}^{\ast }$ is a distributive lattice then ${\tau }_{\mathrm{\Sigma },K}^{\ast }$ is a Boolean algebra. So, $[O,B{]}^{\ast }$ is a Boolean algebra.

The next example illustrates the existence of matrices B such that $[O,B{]}^{\ast }$ are distributive lattices.

Example 3.9

Let us consider the matrix $B=\left[\begin{array}{ccc}2& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$. Some computations give ${\tau }_{\mathrm{\Sigma },K}^{\ast }=\{O,\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right],I_2\}$. The Hasse diagram associated to $[O,B{]}^{\ast }$ is given in Figure 2.

Figure 2. The Boolean algebra $[O,B{]}^{\ast }$.

Eagambaram et al. showed in [22] that $[O,B{]}^{\ast }$ is a finite lattice if and only if all the positive singular values of B are pairwise distinct. The next theorem improves this result by showing that, in that case, $[O,B{]}^{\ast }$ is not only a finite lattice but also a Boolean algebra. Additionally, we find its cardinality.

Theorem 3.10

Let $B\in {\mathbb{C}}^{n\times n}\mathrm{\setminus }\left\{O\right\}$. The lattice $[O,B{]}^{\ast }$ is a Boolean algebra if and only if all the positive singular values of B are pairwise distinct.

Proof.

Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_r\in {\mathbb{R}}^{+}$ be pairwise distinct and $\mathrm{\Sigma }=\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_r)$. If $T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$ then $T=T^2=T^{\ast }$ and $T\mathrm{\Sigma }=\mathrm{\Sigma T}$. Thus, $T=\mathrm{diag}(a_1,\dots ,a_r)$, where $a_j\in \{0,1\}$. Let $A_1,A_2,A_3\in [O,B{]}^{\ast }$ and $T_i=\varphi \left(A_i\right)$, for each $i\in \{1,2,3\}$. Then $T_i=\mathrm{diag}(a_{i1},\dots ,a_{\mathrm{ir}})$, where $a_{\mathrm{ij}}\in \{0,1\}$ for all i. Note that $T_i{T}_j=T_j{T}_i$, for any i, j, and all the supremum and infimum obtained from these projectors also commute with $T_i$, for all i. Then, by Proposition 3.1, we have that $T_1\wedge (T_2\vee T_3)=T_1(T_2+T_3-T_2{T}_3)=T_1{T}_2+T_1{T}_3-T_1{T}_2{T}_3=(T_1\wedge T_2)\vee (T_1\wedge T_3)$. Thus, $[O,B{]}^{\ast }$ is a distributive lattice and, by Proposition 3.8, it is Boolean algebra.

Conversely, suppose that $\mathrm{\Sigma }=\mathrm{diag}({\sigma }_1{I}_{r_1},\dots ,{\sigma }_t{I}_{r_t})$ with $r_i>1$ for some $i\in \{1,\dots ,t\}$. Let us consider the matrices $X_1,X_2,X_3$ constructed in Example 3.3 and $Y_j=\left[\begin{array}{cc}{X}_j& O\\ O& O\end{array}\right]\in {\mathbb{C}}^{r_i\times r_i}$, for every $j\in \{1,2,3\}$. Now, take $T_j\in {\mathbb{C}}^{r\times r}$ partitioned in blocks like the matrix Σ where the block $(i,i)$ is the matrix $Y_j$ and the rest is completed with null matrices of the corresponding order. Now, we can choose $A_1,A_2,A_3\in [O,B{]}^{\ast }$ such that $T_j=\varphi \left(A_j\right)$. Then $A_3\wedge (A_1\vee A_2)\ne (A_3\wedge A_1)\vee (A_3\wedge A_2)$. Hence, $[O,B{]}^{\ast }$ is nondistributive.

Corollary 3.11

Let $B\in {\mathbb{C}}^{n\times n}$ be a nonzero matrix of rank r. The following conditions are equivalent.

(a)	$[O,B{]}^{\ast }$ is a finite lattice.
(b)	All positive singular values of B are pairwise distinct.
(c)	$[O,B{]}^{\ast }$ is a Boolean algebra with $2^r$ elements.

Corollary 3.12

If $A,B\in {\mathbb{C}}^{n\times n}$ are nonzero matrices such that all positive singular values of B are pairwise distinct and $A\in [O,B{]}^{\ast }\mathrm{\setminus }\{O\}$, then all the positive singular values of A are pairwise distinct as well.

Proof.

It follows from Theorem 3.10, because every down-set of a Boolean algebra is a Boolean algebra too.

Remark 3.4

1. If $\mathrm{\Sigma }=\sigma I_r$ for some $\sigma \in {\mathbb{R}}^{+}$ then ${\tau }_{\mathrm{\Sigma },K}^{\ast }={\tau }_{\mathrm{\Sigma },K}^{l\ast }$. If, in addition, $r\ge 2$, then $[O,B{]}^{\ast }$ is an infinite nondistributive lattice by Theorem 3.4 and Remark 3.2.

2. If $\mathrm{\Sigma }=\mathrm{diag}({\sigma }_1{I}_{r_1},\dots ,{\sigma }_t{I}_{r_t})$, for some ${\sigma }_1,\dots ,{\sigma }_t\in {\mathbb{R}}^{+}$, then the condition $T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }$ is equivalent to $T=\mathrm{diag}(X_1,\dots ,X_t)$ where $X_i\in {\mathbb{C}}^{r_i\times r_i}$ and $X_i^2=X_i=X_i^{\ast }$ for every $i\in \{1,\dots ,t\}$.

If $A\stackrel{\ast }{\le }B$, $A_1\stackrel{\ast }{\le }{A}_2\stackrel{\ast }{\le }B$, and we consider the map φ defined in the proof of Lemma 3.5, then we have the following result since $\phi \left(P\right)$ commutes with Σ.

Lemma 3.13

Let $A,A_1,A_2,B\in {\mathbb{C}}^{n\times n}$. If $A_1\stackrel{\ast }{\le }{A}_2\stackrel{\ast }{\le }B$ then $[O,A_2-A_1{]}^{\ast }$ and $[A_1,A_2{]}^{\ast }$ are order-isomorphic. In particular, if $A\stackrel{\ast }{\le }B$ then $[O,B-A{]}^{\ast }$ and $[A,B{]}^{\ast }$ are order-isomorphic.

3.3. Core partial order

We now investigate the lattice structure of $[O,B{]}^{\overline{)\mathrm{\#}}}$ for any $B\in {\mathbb{C}}_1^n$. Once again, we take advantadge of the order-isomorphism ϕ to prove that $[O,B{]}^{\overline{)\mathrm{\#}}}$ is a sublattice of $[O,B{]}^{l\ast }$. Inspired by some examples, we highlight that the behaviour of the core partial order is rather different from the others. For instance, $[O,B{]}^{\overline{)\mathrm{\#}}}$ is not necessarily an orthogonal lattice (see Example (c)). Moreover, under the natural assumptions $A_1\stackrel{\overline{)\mathrm{\#}}}{\le }{A}_2\stackrel{\overline{)\mathrm{\#}}}{\le }B$ and $A_2-A_1\stackrel{\overline{)\mathrm{\#}}}{\le }B$, we demonstrate that $[A_1,A_2{]}^{\overline{)\mathrm{\#}}}$ and $[O,A_2-A_1{]}^{\overline{)\mathrm{\#}}}$ are order-isomorphic.

Theorem 3.14

If $B\in {\mathbb{C}}_1^n\mathrm{\setminus }\left\{O\right\}$ then $[O,B]{}^{\overline{)\mathrm{\#}}}$ is a sublattice of $[O,B{]}^{l\ast }$.

Proof.

It is immediate that ${\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}\subseteq {\tau }_{\mathrm{\Sigma },K}^{l\ast }$. Then $[O,B{]}^{\overline{)\mathrm{\#}}}\subseteq [O,B{]}^{l\ast }$.

Let $A_1,A_2\in [O,B{]}^{\overline{)\mathrm{\#}}}$. We know that $A_1\vee A_2$ and $A_1\wedge A_2$ exist in $[O,B{]}^{l\ast }$. Now we prove that $A_1\vee A_2,A_1\wedge A_2\in [O,B{]}^{\overline{)\mathrm{\#}}}$. By Theorem 3.4, $\varphi \left(A_1\right)\vee \varphi \left(A_2\right)$ and $\varphi \left(A_1\right)\wedge \varphi \left(A_2\right)$ exist in ${\tau }_{\mathrm{\Sigma },K}^{l\ast }$. So, it remains to prove:

1. $\left(\varphi \right(A_1)\vee \varphi (A_2\left)\right)\mathrm{\Sigma K}\left(\varphi \right(A_1)\vee \varphi (A_2\left)\right)=\mathrm{\Sigma K}\left(\varphi \right(A_1)\vee \varphi (A_2\left)\right)$ and

2. $\left(\varphi \right(A_1)\wedge \varphi (A_2\left)\right)\mathrm{\Sigma K}\left(\varphi \right(A_1)\wedge \varphi (A_2\left)\right)=\mathrm{\Sigma K}\left(\varphi \right(A_1)\wedge \varphi (A_2\left)\right)$.

Indeed, let $T_1=\varphi \left(A_1\right)$ and $T_2=\varphi \left(A_2\right)$.

Replacing the supremum expressions in (a), we have $\begin{array}{rl}& (T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}(T_1+T_2\left)\right(T_1+T_2{)}^{†}\\ & =\left(\right(T_1+T_2\left)\right(T_1+T_2{)}^{†}{T}_1\mathrm{\Sigma K}{T}_1+(T_1+T_2)(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_2)(T_1+T_2{)}^{†}\\ & =(T_1\mathrm{\Sigma K}{T}_1+T_2\mathrm{\Sigma K}{T}_2)(T_1+T_2{)}^{†}=\mathrm{\Sigma K}(T_1+T_2\left)\right(T_1+T_2{)}^{†}.\end{array}$So, $(T_1\vee T_2)\mathrm{\Sigma K}(T_1\vee T_2)=\mathrm{\Sigma K}(T_1\vee T_2)$. Therefore, (a) is proved.

To show (b), notice first that: (10) $\begin{array}{rl}& T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=T_1(T_1+T_2{)}^{†}(T_2\mathrm{\Sigma K}{T}_2\left)\right(T_1+T_2{)}^{†}{T}_1\\ & =T_1(T_1+T_2{)}^{†}\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2\end{array}$(10) and (11) $\begin{array}{rl}& T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=T_2(T_1+T_2{)}^{†}(T_1\mathrm{\Sigma K}{T}_1\left)\right(T_1+T_2{)}^{†}{T}_2\\ & =T_2(T_1+T_2{)}^{†}\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2.\end{array}$(11) By adding (10(10) $\begin{array}{rl}& T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=T_1(T_1+T_2{)}^{†}(T_2\mathrm{\Sigma K}{T}_2\left)\right(T_1+T_2{)}^{†}{T}_1\\ & =T_1(T_1+T_2{)}^{†}\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2\end{array}$(10) ) and (11(11) $\begin{array}{rl}& T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=T_2(T_1+T_2{)}^{†}(T_1\mathrm{\Sigma K}{T}_1\left)\right(T_1+T_2{)}^{†}{T}_2\\ & =T_2(T_1+T_2{)}^{†}\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2.\end{array}$(11) ), $\begin{array}{rl}& 2T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2\\ & =(T_1+T_2)(T_1+T_2{)}^{†}{T}_1\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2=T_1\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2\\ & =\mathrm{\Sigma K}{T}_1(T_1+T_2{)}^{†}{T}_2.\end{array}$Then, $(T_1\wedge T_2)\mathrm{\Sigma K}(T_1\wedge T_2)=\mathrm{\Sigma K}(T_1\wedge T_2)$. Thus, (b) is proved.

Remark 3.5

1. $[O,B]{}^{\overline{)\mathrm{\#}}}$ may be a nondistributive lattice. For example, if $\mathrm{\Sigma K}=\sigma I_r$, for some $\sigma \in \mathbb{C}$, then ${\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}={\tau }_{\mathrm{\Sigma },K}^{l\ast }$.

2. The next example is constructed by using the set ${\rho }_B^{\overline{)\mathrm{\#}}}$ defined in Remark 2.1 and it shows that $[O,B]{}^{\overline{)\mathrm{\#}}}$ may be a Boolean algebra. Indeed, consider the matrix $B=\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]=\left[\begin{array}{cccc}3/2& -1/2& 0& \sqrt{6}/2\\ -1/2& 3/2& 0& \sqrt{6}/2\\ 0& 0& -1& 0\\ 0& 0& 0& 0\end{array}\right]$where $\begin{array}{rl}\mathrm{\Sigma K}& =\left[\begin{array}{ccc}3/2& -1/2& 0\\ -1/2& 3/2& 0\\ 0& 0& -1\end{array}\right]=\mathrm{VS}{V}^{\ast },V=\left[\begin{array}{ccc}\sqrt{2}/2& \sqrt{2}/2& 0\\ -\sqrt{2}/2& \sqrt{2}/2& 0\\ 0& 0& 1\end{array}\right]\mathrm{and}\\ S& =\left[\begin{array}{ccc}2& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right].\end{array}$Some computations lead to ${\rho }_B^{\overline{)\mathrm{\#}}}=\left\{\mathrm{diag}\right(a_1,a_2,a_3)|a_i\in \{0,1\left\}\right\}$ and $\begin{array}{rl}& {\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}=\{\mathrm{VT}{V}^{\ast }|T\in {\rho }_B^{\overline{)\mathrm{\#}}}\}=\{O,\underset{\varphi \left(A_1\right)}{\underbrace{V\mathrm{diag}(1,0,0)V^{\ast }}},\underset{\varphi \left(A_2\right)}{\underbrace{V\mathrm{diag}(0,1,0)V^{\ast }}},\\ & \underset{\varphi \left(A_3\right)}{\underbrace{V\mathrm{diag}(0,0,1)V^{\ast }}},\underset{\varphi \left(A_4\right)}{\underbrace{I_3-\varphi \left(A_3\right)}},\underset{\varphi \left(A_5\right)}{\underbrace{I_3-\varphi \left(A_2\right)}},\underset{\varphi \left(A_6\right)}{\underbrace{I_3-\varphi \left(A_1\right)}},I_3\}.\end{array}$The associated Hasse diagram of $[O,B]{}^{\overline{)\mathrm{\#}}}$ is given in Figure 3.

Figure 3. The Boolean algebra $[O,B]{}^{\overline{)\mathrm{\#}}}$.

1. $[O,B]{}^{\overline{)\mathrm{\#}}}$ may be a non-Boolean distributive lattice as the following example shows. Consider $B=\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]$ where $\mathrm{\Sigma }=2I_3$, $K=\left[\begin{array}{ccc}1/2& -1/2& 0\\ 0& 1/2& 0\\ 0& 0& 1\end{array}\right],\mathrm{and}L=\left[\begin{array}{ccc}1/2& 1/2& 0\\ 0& 1/2& \sqrt{2}/2\\ 0& 0& 0\end{array}\right].$Some computations lead to ${\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}=\{O,\underset{\varphi \left(A_1\right)}{\underbrace{\mathrm{diag}(1,0,0)}},\underset{\varphi \left(A_2\right)}{\underbrace{\mathrm{diag}(1,1,0)}},\underset{\varphi \left(A_3\right)}{\underbrace{\mathrm{diag}(0,0,1)}},\underset{\varphi \left(A_4\right)}{\underbrace{\mathrm{diag}(1,0,1)}},I_3\},$and the associated Hasse diagram of $[O,B]{}^{\overline{)\mathrm{\#}}}$ is given in Figure 4.

As we can observe in the last example, not always $B-A\stackrel{\overline{)\mathrm{\#}}}{\le }B$ holds whenever $A\stackrel{\overline{)\mathrm{\#}}}{\le }B$. When $B-A\stackrel{\overline{)\mathrm{\#}}}{\le }B$, the following result is valid.

Figure 4. The lattice $[O,B]{}^{\overline{)\mathrm{\#}}}$.

Lemma 3.15

Let $A,A_1,A_2,B\in {\mathbb{C}}_1^n$. If $A_1\stackrel{\overline{)\mathrm{\#}}}{\le }{A}_2\stackrel{\overline{)\mathrm{\#}}}{\le }B$ and $A_2-A_1\stackrel{\overline{)\mathrm{\#}}}{\le }B$ then $[A_1,A_2{]}^{\overline{)\mathrm{\#}}}$ and $[O,A_2-A_1{]}^{\overline{)\mathrm{\#}}}$ are order-isomorphic. In particular, if $A,B-A\stackrel{\overline{)\mathrm{\#}}}{\le }B$ then $[O,B-A{]}^{\overline{)\mathrm{\#}}}$ and $[A,B{]}^{\overline{)\mathrm{\#}}}$ are order-isomorphic.

Proof.

Take φ as in Lemma 3.5. Let us see that φ is surjective. Since $T_1\le T_2$ where $T_1,T_2\in {\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}$, from $(T_2-T_1)\mathrm{\Sigma K}(T_2-T_1)=\mathrm{\Sigma K}(T_2-T_1)$, we obtain that $\mathrm{\Sigma K}{T}_1=T_1\mathrm{\Sigma K}{T}_2$. Now, consider $T_1\le Q\le T_2$. Then $(Q-T_1)\mathrm{\Sigma K}(Q-T_1)=Q\mathrm{\Sigma KQ}-Q\mathrm{\Sigma K}{T}_1-T_1\mathrm{\Sigma KQ}+T_1\mathrm{\Sigma K}{T}_1$. But $Q\mathrm{\Sigma K}{T}_1=Q{T}_1\mathrm{\Sigma K}{T}_1=T_1\mathrm{\Sigma K}{T}_1=\mathrm{\Sigma K}{T}_1$ and $T_1\mathrm{\Sigma KQ}=T_1\mathrm{\Sigma K}{T}_{2}Q=\mathrm{\Sigma K}{T}_{1}Q=\mathrm{\Sigma K}{T}_1$. So, $Q\mathrm{\Sigma KQ}-Q\mathrm{\Sigma K}{T}_1-T_1\mathrm{\Sigma KQ}+T_1\mathrm{\Sigma K}{T}_1=\mathrm{\Sigma KQ}-\mathrm{\Sigma K}{T}_1-\mathrm{\Sigma K}{T}_1+\mathrm{\Sigma K}{T}_1=\mathrm{\Sigma KQ}-\mathrm{\Sigma K}{T}_1=\mathrm{\Sigma K}(Q-T_1)$. The rest of conditions for φ to be an order-isomorphism can be proved as in Lemma 3.5.

4. Supremum and infimum of two arbitrary matrices

In this section we first demonstrate that there exists the supremum (for all three partial orders) of two given matrices $A_1$ and $A_2$ if and only if $A_1$ and $A_2$ have a common upper bound. Our main tools are Theorems 3.4, 3.7, and 3.14. In addition, we find an expression for this supremum. Secondly, we analyse the infimum of two given matrices. In the case where the matrices have a common upper bound, we obtain an expression for their infimum. If two matrices B and C do not have a common upper bound, we already know that $B\wedge C$ exists for the three partial orders (see [16, 18, 21]). If B is written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) ) then the infimum can be written as in (2(2) $\begin{array}{r}A=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(2) ) and we find the conditions that the associated orthogonal projector must satisfy.

Theorem 4.1

Let $B\in {\mathbb{C}}^{n\times n}$ be a nonsingular matrix, and $A_1,A_2\in [O,B{]}^x$ such that $S=A_1\vee A_2\in [O,B{]}^x$. If $A_1,A_2\stackrel{x}{\le }\tilde{B}$, for some $\tilde{B}\in {\mathbb{C}}^{n\times n}$, then $S\stackrel{x}{\le }\tilde{B}$.

Proof.

Let $T_1=\varphi \left(A_1\right)$ and $T_2=\varphi \left(A_2\right)$. The fact that B is nonsingular yields that $B=U\mathrm{\Sigma K}{U}^{\ast }$ and $A_i=U{T}_i\mathrm{\Sigma K}{U}^{\ast }$, for every $i\in \{1,2\}$, with L = O and $K{K}^{\ast }=I_n$. By Theorems 3.4, 3.7, or 3.14, depending on the corresponding partial order x, and by Remark 3.1 we know that $S=U(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }$.

Since $A_i\stackrel{x}{\le }\tilde{B}$, we have that $A_i^{\ast }{A}_i=A_i^{\ast }\tilde{B}$. Then $U{K}^{\ast }\mathrm{\Sigma }{T}_i\mathrm{\Sigma K}{U}^{\ast }=U{K}^{\ast }\mathrm{\Sigma }{T}_i{U}^{\ast }\tilde{B}$ and consequently $T_i\mathrm{\Sigma K}{U}^{\ast }=T_i{U}^{\ast }\tilde{B}.$Taking into account this last fact, (12) $\begin{array}{rl}{S}^{\ast }S& =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)U^{\ast }\tilde{B}=S^{\ast }\tilde{B}.\end{array}$(12) Now we need to study each order separately.

• From $A_i\stackrel{{}_{\ast }}{\le }\tilde{B}$ we know that $A_i{A}_i^{\ast }=\tilde{B}{A}_i^{\ast }$. Then $U{T}_i\mathrm{\Sigma \Sigma }{T}_i{U}^{\ast }=\tilde{B}U{K}^{\ast }\mathrm{\Sigma }{T}_i{U}^{\ast }$ and consequently $U{\mathrm{\Sigma }}^2{T}_i=\tilde{B}U{K}^{\ast }\mathrm{\Sigma }{T}_i$. Since $\mathrm{\Sigma T}=T\mathrm{\Sigma }$, for every $T\in {\tau }_{\mathrm{\Sigma }K}^{\ast }$, and $(T_1+T_2)(T_1+T_2{)}^{†}(T_1+T_2)=T_1+T_2$, we have $\begin{array}{rl}S{S}^{\ast }& =U(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma \Sigma }(T_1+T_2\left)\right(T_1+T_2{)}^{†}{U}^{\ast }\\ & =U{\mathrm{\Sigma }}^2(T_1+T_2)(T_1+T_2{)}^{†}{U}^{\ast }=\tilde{B}U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2\left)\right(T_1+T_2{)}^{†}{U}^{\ast }=\tilde{B}{S}^{\ast }\end{array}$and by (12(12) $\begin{array}{rl}{S}^{\ast }S& =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)U^{\ast }\tilde{B}=S^{\ast }\tilde{B}.\end{array}$(12) ), we get $S\stackrel{{}_{\ast }}{\le }\tilde{B}$.

• If $A_i\stackrel{{}_{l\ast }}{\le }\tilde{B}$ then $A_i=\tilde{B}{\tilde{B}}^{†}{A}_i$. So, $U{T}_i\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}{\tilde{B}}^{†}U{T}_i\mathrm{\Sigma K}{U}^{\ast }$. Thus, $U{T}_i=\tilde{B}{\tilde{B}}^{†}U{T}_i$. Then $S=U(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}{\tilde{B}}^{†}U(T_1+T_2\left)\right(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}{\tilde{B}}^{†}S$and from (12(12) $\begin{array}{rl}{S}^{\ast }S& =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)U^{\ast }\tilde{B}=S^{\ast }\tilde{B}.\end{array}$(12) ) we obtain that $S\stackrel{{}_{l\ast }}{\le }\tilde{B}$.

• Finally, if $A_i\stackrel{{}_{\overline{)\mathrm{\#}}}}{\le }\tilde{B}$ then $A_i^2=\tilde{B}{A}_i$. Thus, $U\left(T_i\mathrm{\Sigma K}{T}_i\right)\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}U{T}_i\mathrm{\Sigma K}{U}^{\ast }$ or equivalently $U\mathrm{\Sigma K}{T}_i=\tilde{B}U{T}_i$. Taking into account that $T_1\vee T_2\in {\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}$, we have $\begin{array}{rl}{S}^2& =U\left(\right(T_1+T_2\left)\right(T_1+T_2{)}^{†}\mathrm{\Sigma K}(T_1+T_2)(T_1+T_2{)}^{†})\mathrm{\Sigma K}{U}^{\ast }\\ & =U\mathrm{\Sigma K}(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}U(T_1+T_2\left)\right(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=\tilde{B}S.\end{array}$Therefore, by (12(12) $\begin{array}{rl}{S}^{\ast }S& =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2)(T_1+T_2{)}^{†}\mathrm{\Sigma K}{U}^{\ast }=U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =U{K}^{\ast }\mathrm{\Sigma }(T_1+T_2{)}^{†}(T_1+T_2)U^{\ast }\tilde{B}=S^{\ast }\tilde{B}.\end{array}$(12) ), we have that $S\stackrel{{}_{\overline{)\mathrm{\#}}}}{\le }\tilde{B}$.

Let us observe that if $\tilde{B}\in {\mathbb{C}}^{n\times n}$ (or $\tilde{B}\in {\mathbb{C}}_1^n$ for the core partial order) then there exists a nonsingular matrix B such that $\tilde{B}\stackrel{x}{\le }B$. Indeed:

• If $x=l^{\ast }$ and $\tilde{B}=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },$ then it is enough to consider $B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& I_{n-r}\end{array}\right]U^{\ast }$.

• If $x=\ast$ then consider a singular value decomposition of $\tilde{B}$ given by $\tilde{B}=U\left[\begin{array}{cc}\mathrm{\Sigma }& O\\ O& O\end{array}\right]V^{\ast }$ and we can choose $B=U\left[\begin{array}{cc}\mathrm{\Sigma }& O\\ O& I_{n-r}\end{array}\right]V^{\ast }$.

• If $x=\overline{)\mathrm{\#}}$ and we consider again $\tilde{B}=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$ then, by Baksalary and Trenkler [1, Lemma 3], we can take $B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& I_{n-r}\end{array}\right]U^{\ast }$.

Proposition 4.2

Let $A_1,A_2\in {\mathbb{C}}^{n\times n}$ (or $A_1,A_2\in {\mathbb{C}}_1^n$ for the core partial order). Then, $A_1\vee A_2$ exists if and only if $A_1$ and $A_2$ have a common upper bound. In that case, $A_1\vee A_2=(A_1{A}_1^{†}+A_2{A}_2^{†}{)}^{†}(A_1+A_2)$.

Proof.

The first statement is immediate from Theorem 4.1 taking into account that if $A_1$ and $A_2$ have a common upper bound $\tilde{B}$ then there exists a nonsingular matrix B such that $\tilde{B}\stackrel{x}{\le }B$, for all partial order x. For the second statement, assume that $A_1,A_2$ have a common upper bound and take B a nonsingular matrix such that $A_1,A_2\le B$. Consider a Hartwig-Spindelböck decomposition of B given by $B=U\mathrm{\Sigma K}{U}^{\ast }$, where $U,K\in {\mathbb{C}}^{n\times n}$ are unitary and $\mathrm{\Sigma }=\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)\in {\mathbb{C}}^{n\times n}$. Let $T_1$ and $T_2$ be the orthogonal projectors such that $\varphi \left(A_i\right)=T_i$, for each i, that is $A_i=U{T}_i\mathrm{\Sigma K}{U}^{\ast }$. Then, $\begin{array}{rl}{A}_1\vee A_2& =U(T_1\vee T_2)\mathrm{\Sigma K}{U}^{\ast }=U(T_1+T_2{)}^{†}(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =U(T_1+T_2{)}^{†}{U}^{\ast }U(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }=(U(T_1+T_2)U^{\ast }{)}^{†}U(T_1+T_2)\mathrm{\Sigma K}{U}^{\ast }\\ & =(U{T}_1{U}^{\ast }+U{T}_2{U}^{\ast }{)}^{†}(A_1+A_2).\end{array}$From $K{K}^{\ast }=I_n$, by using the facts that Σ is nonsingular and $T_i\mathrm{\Sigma }(T_i\mathrm{\Sigma }{)}^{†}=T_i$ for each i (see [19, Lemma 14]), we have that $U{T}_i{U}^{\ast }=U{T}_i\mathrm{\Sigma K}{U}^{\ast }U{K}^{\ast }(T_i\mathrm{\Sigma }{)}^{†}{U}^{\ast }=A_i{A}_i^{†}.$Hence, $A_1\vee A_2=(A_1{A}_1^{†}+A_2{A}_2^{†}{)}^{†}(A_1+A_2)$.

Proposition 4.3

Let $A_1,A_2\in {\mathbb{C}}^{n\times n}$ (or $A_1,A_2\in {\mathbb{C}}_1^n$ for the core partial order). If $A_1$ and $A_2$ have a common upper bound then $A_1\wedge A_2=2A_1{A}_1^{†}(A_1{A}_1^{†}+A_2{A}_2^{†}{)}^{†}{A}_2$.

Proof.

Proceeding as in the proof of the Proposition 4.2, consider a nonsingular matrix B and a Hartwig-Spindelböck decomposition $B=U\mathrm{\Sigma K}{U}^{\ast }$ such that $A_i\stackrel{x}{\le }B$ and $T_i$ the orthogonal projectors such that $\varphi \left(A_i\right)=T_i$. Then, $A_i=U{T}_i\mathrm{\Sigma K}{U}^{\ast }$, $A_i{A}_i^{†}=U{T}_i{U}^{\ast }$, and $(A_1{A}_1^{†}+A_2{A}_2^{†}{)}^{†}=U(T_1+T_2{)}^{†}{U}^{\ast }$. By Theorems 3.4, 3.7 or 3.14, depending on the corresponding partial order x, we have that $A_1\wedge A_2=U(T_1\wedge T_2)\mathrm{\Sigma K}{U}^{\ast }=U2T_1(T_1+T_2{)}^{†}{T}_2\mathrm{\Sigma K}{U}^{\ast }=2U{T}_1{U}^{\ast }U(T_1+T_2{)}^{†}{U}^{\ast }U{T}_2\mathrm{\Sigma K}{U}^{\ast }=2A_1{A}_1^{†}(A_1{A}_1^{†}+A_2{A}_2^{†}{)}^{†}{A}_2$.

In general, if B and C do not have a common upper bound, we know that there exists $B\wedge C$ for the three partial orders. If B is written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) ) and we write $C=U\left[\begin{array}{cc}{C}_1& C_2\\ C_3& C_4\end{array}\right]U^{\ast }$, where $C_1\in {\mathbb{C}}^{r\times r}$, then for the infimum $J=B\wedge C$ there exists $T\in {\tau }_{\mathrm{\Sigma },K}^x$ such that $J=U\left[\begin{array}{cc}T\mathrm{\Sigma K}& T\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$. It is straightforward to see that $J^{\ast }J=J^{\ast }C$ if and only if $T\mathrm{\Sigma K}=T{C}_1$ and $T\mathrm{\Sigma L}=T{C}_2$. Moreover, $\mathcal{R}\left(J\right)\subseteq \mathcal{R}\left(C\right)$ if and only if $\mathcal{R}\left(T\right)\subseteq \mathcal{R}\left(\right[C_1{C}_2\left]\right)$. Indeed, $\mathcal{R}\left(J\right)=\mathcal{R}\left(J{J}^{†}\right)=U\mathcal{R}\left(\left[\begin{array}{c}T\mathrm{\Sigma }(T\mathrm{\Sigma }{)}^{†}\\ O\end{array}\right]\right)$. Thus, $\mathcal{R}\left(J\right)\subseteq \mathcal{R}\left(C\right)$ if and only if $\mathcal{R}\left(\left[\begin{array}{c}T\mathrm{\Sigma }(T\mathrm{\Sigma }{)}^{†}\\ O\end{array}\right]\right)\subseteq \mathcal{R}\left(\left[\begin{array}{cc}{C}_1& C_2\\ C_3& C_4\end{array}\right]\right)$, and this is equivalent to $\mathcal{R}\left(T\right)\subseteq \mathcal{R}\left(\right[C_1{C}_2\left]\right)$ because Σ is nonsingular and so $\mathcal{R}\left(T\mathrm{\Sigma }\right(T\mathrm{\Sigma }{)}^{†})=\mathcal{R}(T\mathrm{\Sigma })=\mathcal{R}(T)$.

For the star partial order we have that $J{J}^{\ast }=C{J}^{\ast }$ if and only if $T\mathrm{\Sigma }=(C_1{K}^{\ast }+C_2{L}^{\ast })T$ and $(C_3{K}^{\ast }+C_4{L}^{\ast })T=O$.

Finally, for the core partial order we obtain that $\mathrm{CJ}=J^2$ if and only if $C_{1}T=\mathrm{\Sigma KT}$ and $C_{3}T=O$.

We summarize the last reasoning in the following proposition.

Proposition 4.4

Let $B,C\in {\mathbb{C}}^{n\times n}$ (or $B,C\in {\mathbb{C}}_1^n$ for the core partial order) where B is written as in (1(1) $\begin{array}{r}B=U\left[\begin{array}{cc}\mathrm{\Sigma K}& \mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast },\end{array}$(1) ) and C as above. Then the infimum is given by $B\wedge C=U\left[\begin{array}{cc}{T}_m\mathrm{\Sigma K}& T_m\mathrm{\Sigma L}\\ O& O\end{array}\right]U^{\ast }$, where $T_m$ is the maximum of the following set.

(a)	For the left star partial order, $\{T\in {\tau }_{\mathrm{\Sigma },K}^{l\ast }|T\mathrm{\Sigma K}=T{C}_1,T\mathrm{\Sigma L}=T{C}_2,\mathcal{R}(T)\subseteq \mathcal{R}(\left[C_1{C}_2\right]\left)\right\}.$
(b)	For the star partial order, $\begin{array}{rl}& \{T\in {\tau }_{\mathrm{\Sigma },K}^{\ast }|T\mathrm{\Sigma K}=T{C}_1,T\mathrm{\Sigma L}=T{C}_2,\\ & T\mathrm{\Sigma }=(C_1{K}^{\ast }+C_2{L}^{\ast })T,(C_3{K}^{\ast }+C_4{L}^{\ast })T=O\}.\end{array}$
(c)	For the core partial order, $\{T\in {\tau }_{\mathrm{\Sigma },K}^{\overline{)\mathrm{\#}}}|T\mathrm{\Sigma K}=T{C}_1,T\mathrm{\Sigma L}=T{C}_2,C_{1}T=\mathrm{\Sigma KT},C_{3}T=O\}.$

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No potential conflict of interest was reported by the author(s).

Additional information

Funding

The first, the second, and the fourth author were partially supported by Departamento de Matemática, Universidad Nacional del Sur (UNS), Argentina [project number PGI 24/L108]. The third author was partially supported by FWF Austria [project numbers I 4427 and P 31955]. The fourth author was partially supported by Ministerio de Economía y Competitividad of Spain [grant Red de Excelencia, MTM2017-90682-REDT], by Universidad Nacional de Río Cuarto [grant number PPI 083/2020], and by Universidad Nacional de La Pampa, Facultad de Ingeniería, Argentina [grant number Resol. Nro. 135/19].