Preservers of the left-star and right-star partial orders

Journal Pre-proof Preservers of the left-star and right-star partial orders

Gregor Dolinar, Sait Halicioglu, Abdullah Harmanci, Bojan Kuzma, Janko Marovt, Burcu Ungor

PII:

S0024-3795(19)30472-0

DOI:

https://doi.org/10.1016/j.laa.2019.11.003

Reference:

LAA 15169

To appear in:

Linear Algebra and its Applications

Received date:

6 June 2019

Accepted date:

3 November 2019

Please cite this article as: G. Dolinar et al., Preservers of the left-star and right-star partial orders, Linear Algebra Appl. (2020), doi: https://doi.org/10.1016/j.laa.2019.11.003.

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PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS GREGOR DOLINAR, SAIT HALICIOGLU, ABDULLAH HARMANCI, BOJAN KUZMA, JANKO MAROVT, AND BURCU UNGOR Abstract. Let Mn (F) be the set of all n × n matrices over the field F, where F = R or F = C. We characterize maps on Mn (F) which are surjective and preserve either the left-star or the right-star order in both directions, but are not necessarily additive.

1. Introduction Let Mm,n (F) be the set of all m × n matrices over F where F denotes the field of complex or real numbers, and let Im A denote the image of A ∈ Mm,n (F). When m = n, we will shortly write Mn (F) for Mn,n (F). Drazin defined in [3] a partial order, named the star partial order, in the following way: A≤B

(1)

∗

if

A∗ A = A∗ B and AA∗ = BA∗ ,

A, B ∈ Mn (F).

Here A∗ stands for the conjugate transpose of A. Let A† denote the Moore-Penrose inverse of A ∈ Mn (F). Drazin noted in [3] that A≤B ∗

if and only if

A† A = A† B and AA† = BA† ,

A, B ∈ Mn (F).

Baksalary and Mitra introduced in [1] notions of order which are related to the star partial order. Definition 1. The left-star partial order on Mm,n (F) is a relation defined by A ∗≤ B

if and only if

A∗ A = A∗ B and Im A ⊆ Im B,

A, B ∈ Mm,n (F).

Definition 2. The right-star partial order on Mm,n (F) is a relation defined by A ≤∗ B

if and only if

AA∗ = BA∗ and Im A∗ ⊆ Im B ∗ ,

A, B ∈ Mm,n (F).

It was shown in [1, Section 2] that both relations ∗≤ and ≤∗ are partial orders and that A ∗≤ B and A ≤∗ B if and only if A ≤ B, ∗

A, B ∈ Mn (F).

2010 Mathematics Subject Classification. 47B49, 15A86, 15A04, 15A09, 06A06. Key words and phrases. Left-star order; right-star order; general preserver; matrix algebra; Moore-Penrose inverse. The authors acknowledge the financial support from the Slovenian Research Agency, ARRS (research core funding No. P1-0222 and No. P1-0288). The authors also acknowledge the project (Rings, Preservers, and Applications, BI-TR/17-19-004) was financially supported by the Slovenian Research Agency, and by the Scientific and Technological Research Council of Turkey, TUBITAK (Grant TUBITAK-116F435). The authors wish to thank ARRS and TUBITAK for financial support. 1

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G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

Also, it is easy to see that for A, B ∈ Mm,n (F), (2)

A ∗≤ B

if and only if

A∗ ≤∗ B ∗ .

Note that the left-star order has important implications in statistics, for example in the theory of linear models. Let us present an example of such an application (see [9, Page 414]). Let y = Xβ +  be the matrix form of a Gauss-Markov linear model. Here y is an n × 1 random vector of observed quantities, X ∈ Mn,p (R) is a matrix which determines the linear model, β is a vector of parameters of the model, and  ∈ Rn is a random vector of errors. It is assumed that all the components of an error vector have the zero mean and the same finite variance σ 2 and that they are uncorrelated. The nonnegative parameter σ 2 and the vector of parameters β ∈ Rp are unspecified. We denote the Gauss-Markov model with the triplet (y, Xβ, σ 2 I) where I denotes the identity matrix. Statistical analysis often focuses on answering questions about certain linear functions of the form Cβ for a specified real matrix C with p columns. We try to estimate Cβ by a linear function Ay of the response y (here A is a real matrix with n columns). We say that Ay is a linear unbiased estimator (LUE) of Cβ if the mathematical expectation of Ay equals Cβ for all possible values of β ∈ Rp . Note that Ay = AXβ + A, so its mean E equals E(Ay) = E(AXβ) + E(A) = E(AXβ) + 0 = AXβ, hence Ay is LUE for Cβ if and only if AX = C. The function Cβ is said to be estimable if it has LUE. The best linear unbiased estimator (BLUE) of an estimable Cβ is defined as LUE having the smallest variance-covariance matrix (”smallest” in terms of the usual (i.e. L¨owner) partial order on the set of positive semidefinite (symmetric) matrices, see, e.g. [9]). Let us consider two models M1 = (y, X1 β, σ 2 I) and M2 = (y, X2 β, σ 2 I) where X1 , X2 ∈ Mn,p (R). Then X1 ∗≤ X2 if and only if (i) the linear models M1 and M = (y, (X2 − X1 )β, σ 2 I) have no common estimable linear function of β; (ii) X1 β is estimable under both models M1 and M2 ; and (iii) BLUE of X1 β under the model M1 is also its BLUE under M2 and the variance-covariance matrix of BLUE of X1 β under the model M1 is the same as under the model M2 , see [9, Theorem 15.3.7.]. Given the model M = (y, Xβ, σ 2 I) one might rather work with the transformed model ˆ = (y, Xβ, ˆ σ 2 I) because the matrix X ˆ ∈ Mn (R) has more attractive properties than M X ∈ Mn (R) (e.g. elements of X that are very close to zero are transformed to zero). It is natural to demand that the transformed model still retains most of properties of the original model (e.g. has similar relations to other transformed models) so in view of the above application of the left-star partial order it would be interesting to know what transformations Φ on Mn (R) preserve the order ∗≤ in both directions. We say that a map Φ : Mn (F) → Mn (F) preserves an order ≤ in one direction when for every A, B ∈ Mn (F), A≤B

implies

Φ(A) ≤ Φ(B).

Similarly, we say that a map Φ : Mn (F) → Mn (F) preserves an order ≤ in both directions when A ≤ B if and only if Φ(A) ≤ Φ(B). Let us recall some known results about the maps on Mn (F) that preserve the mentioned orders. In [5], Guterman characterized bijective, linear maps Φ : Mn (F) → Mn (F) that preserve the star partial order in one direction. The same author studied later in [6] additive maps on Mn (C) that preserve either the star partial order or the left-star partial order, or

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

3

the right-star partial order in one direction (see also [2]). Legiˇsa described in [8] the form of all surjective (possibly nonlinear) maps on Mn (F), n ≥ 3, that preserve the star order in both directions. The goal of this paper is to characterize surjective (possibly nonadditive) maps on Mn (F), n ≥ 3, that preserve either the left or the right-star partial order in both directions. 2. Preliminaries and statement of the main result Let T ∈ Mn (F) be an invertible matrix. We will denote by T −∗ the matrix (T −1 )∗ . Let A be the matrix obtained from A ∈ Mn (F) by entrywise conjugation. We will call maps † A → A, A → A† , and A → A , A ∈ Mn (F), the entrywise conjugation, the Moore-Penrose inverse, and the entrywise-conjugated Moore-Penrose inverse, respectively. Let σ : F → F be a function. For an m × n matrix A = (aij ) we define Aσ = (σ(aij )), i.e. we apply σ entrywise. 2.1. Statement of the main results. Theorem 3. Let n ≥ 3 be an integer. Then a surjection Φ : Mn (F) → Mn (F) preserves the left-star order in both directions if and only if there exist invertible T, W ∈ Mn (F) such that Φ has the following form:  †   • • • • • • • • • • • † † −1 −∗ † † −1 −∗ † Φ(X) = T X X + (I − X X ) · T T · X X · X X · T T · X X XW, •

where the map X → X denotes either identity or entrywise conjugation or Moore-Penrose inverse or entrywise-conjugated Moore-Penrose inverse on Mn (F). Corollary 4. Let n ≥ 3 be an integer. Then a surjection Φ : Mn (F) → Mn (F) preserves the right-star order in both directions if and only if there exist invertible T, W ∈ Mn (F) such that Φ has the following form:   †  • • • • • • • • • • • † † −∗ −1 † † −∗ −1 † Φ(X) = T X X X + X X · W W · X X · X X · W W · (I − X X) W, •

where the map X → X is as in Theorem 3. Proof. Apply Theorem 3 on the bijective map Ψ(X) = (Φ(X ∗ ))∗ , X ∈ Mn (F), which by (2) preserves the left-star order in both directions.  Remark 5. Our results easily extend to classify converters from ∗≤ to ≤∗ that is surjective maps Ψ : Mn (F) → Mn (F) with the property A ∗≤ B if and only if Ψ(A) ≤∗ Ψ(B). Namely, given any such Ψ the map Φ(X) = Ψ(X)∗ preserves ∗≤ order. Let U ∈ Mn (F) be a unitary matrix and let S ∈ Mn (F) be an invertible matrix. Observe that by Definition 1 (or see, [2, Lemma 7]), A ∗≤ B

if and only if U AS ∗≤ U BS,

A, B ∈ Mn (F).

Suppose Φ : Mn (F) → Mn (F) preserves the left-star partial order in both directions. In the sequel we will take into account that the map Ψ : Mn (F) → Mn (F) defined by Ψ(A) = U Φ(A)S,

A ∈ Mn (F),

also preserves the left star order in both direction.

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G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

3. Proofs It is almost trivial to see that Φ from Theorem 3 or Corollary 4 is injective and hence an automorphism. Indeed, if Φ(A) = Φ(B), then Φ(A) ∗≤ Φ(B) ∗≤ Φ(A) and therefore A ∗≤ B ∗≤ A, thus A = B. Likewise for the right-star partial order. We proceed with auxiliary results which hold for every integer n ≥ 2. Some of them (Lemma 10 and Corollary 12) were already proven in a more general setting of the algebra of bounded linear operators acting on a complex Hilbert space, see [2]. Nonetheless, we will present below their shorter proofs by taking the advantage of finite-dimensionality. A vector in Fn whose Euclidean norm equals 1 will be called a unimodular vector. Lemma 6. Let rank B ≥ 2. Then, rank-one R ∗≤ B if and only if R = xx∗ B for some unimodular vector x ∈ Im B. Proof. Write R = xf ∗ ; by transferring the appropriate scalar to the second factor we can assume that x = 1. Then R ∗≤ B implies x ∈ Im R ⊆ Im B and f f ∗ = (xf ∗ )∗ (xf ∗ ) = (xf ∗ )∗ B = f x∗ B so that f ∗ = x∗ B. Conversely, one easily deduces that xx∗ B ∗≤ B provided that the unimodular vector x belongs to the image of B.  Lemma 7. Let B, C ∈ Mn (F). The following statements are equivalent. (i) B ∗≤ C; (ii) R ∗≤ B implies R ∗≤ C for every rank-one matrix R ∈ Mn (F). Proof. (i)=⇒(ii): Suppose first B ∗≤ C. So, B ∗ B = B ∗ C and Im B ⊆ Im C. Take any rankone R ∗≤ B. By Lemma 6 we can write xx∗ B ∗≤ B for some unimodular x = By ∈ Im B. Then xx∗ B = Byy ∗ B ∗ B = Byy ∗ B ∗ C = xx∗ C and as x ∈ Im B ⊆ Im C we see that R = xx∗ B = xx∗ C ∗≤ C. (ii)=⇒(i): Let {x1 , x2 , . . . , xk } be an orthonormal basis for Im B. Then Ri = xi x∗i B ∗≤ B so by the assumption also Ri = xi x∗i B ∗≤ C. By the definition of the left-star order it follows that xi ∈ Im xi x∗i B ⊆ Im C for i = 1, 2, . . . , k and hence (3)

Im B ⊆ Im C.

Also,

B ∗ xi x∗i B = (xi x∗i B)∗ xi x∗i B = (xi x∗i B)∗ C = B ∗ xi x∗i C. Summing up with i = 1, 2, . . . , k we obtain B∗P B = B∗P C

 where P = k1 xi x∗i is a projection onto Im B. Since then P B = B and thus also B ∗ = B ∗ P we obtain B ∗ B = B ∗ C. Together with (3) this implies B ∗≤ C.  Lemma 8. The entrywise conjugation map X → X is bijective and preserves the relation ∗≤ in both directions. Proof. The entrywise conjugation map is involutive and hence bijective. To see it preservers ∗≤ we choose a matrix B and let R ∗≤ B be an arbitrary rank-one matrix. Then, R = xx∗ B for some unimodular x = By ∈ Im B. Clearly then, R = xx∗ B and since also x = x = 1 and x = By ∈ Im B, we see that R ∗≤ B. Being involutive, this implies that R ∗≤ B for some  rank-one R if and only if R ∗≤ B. The result then follows from Lemma 7.

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

5

The lemma below may be proved by composing the previous lemma with Theorem 2.5 in [1]. For the sake of completeness we will present an alternative proof. †

Lemma 9. The map X → X is bijective and preserves the relation ∗≤ in both directions. †

†

†

††

Proof. For bijectivity note that X → (X ) = X = X i.e., the map is involutive. To see it preservers ∗≤ we choose a matrix B and let R ∗≤ B be arbitrary rank-one. Then, R = xx∗ B for some unimodular x = By ∈ Im B. Define the unimodular u=

B t By Btx = . B t x B t x

†

†

Since B(B ) is a (self-adjoint) projection onto Im B  x, we have x∗ B(B ) = x∗ . Combined ∗

†

∗

∗

(B x)x with (B t )∗ = B and the fact that (xy ∗ )† = xyx = 2 ·y2 , so R = x2 ·B ∗ x2 deduce † † B t xx∗ (B t )∗ B B t xx∗ BB B t xx∗ † † = = =R . uu∗ B = B t x 2 B t x 2 B t x 2 †

†

B t xx∗ x2 ·B t x2

we

†

If we show that u ∈ Im B , then this already implies that R ∗≤ B . To this end, let B = U DV ∗ be a singular value decomposition. Then B = U DV t so B t B = V DU t U DV t

and

†

∗

B = (V t )∗ D† U = V D† U t . †

Thus, u ∈ Im B t B ⊆ Im V D = Im V D† = Im B where we used the fact that for a diagonal † † matrix D one has Im D = Im D† . Indeed, R ∗≤ B for rank-one R implies R ∗≤ B . The † implication is actually an equivalence because X → X is an involution. The rest of the claim now follows from Lemma 7.  Let us recall that a projection is a matrix P ∈ Mn (F) such that P = P 2 = P ∗ . Lemma 10. The following statements are equivalent for a matrix A ∈ Mn (F). (i) A is a projection. (ii) A ∗≤ I. Proof. (ii) =⇒ (i). By the definition of the left-star order we have A∗ A = A∗ I = A∗ . Applying adjoint on the both sides we get A∗ A = (A∗ )∗ = A. So A∗ = A which implies A2 = A∗ A = A. (i) =⇒ (ii). Trivial.  Recall that a nonempty subset S of a poset is a chain if every two elements in S are comparable. Lemma 11. In a poset (Mn (F), ∗≤ ) every chain has at most n+1 elements and every matrix is a part of a chain of maximal length. Moreover, if X0 ∗≤ X1 ∗≤ . . . ∗≤ Xn is a chain of maximal length, then rank Xi = i for i = 0, 1, . . . , n. Proof. First we will show that every matrix A ∈ Mn (F) is an element of a chain with n+1 elements. Let A = U diag (d1 , . . . , dr , 0, . . . , 0)V , d1 ≥ d2 ≥ . . . ≥ dr > 0 be the singular value decomposition of A. Define X0 = 0 and let Xi = diag (d1 , . . . , di , 0, . . . , 0)V, i = 1, 2, . . . n where dr+1 = . . . = dn = 1. Then it is easy to see that A = Xr and that X0 ∗≤ X1 ∗≤ . . . ∗≤ Xn .

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G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

Now we will prove that A ∗≤ B and Im A = Im B implies A = B. Indeed, from A ∗≤ B we have A∗ A = A∗ B and hence A∗ (A − B) = 0. So Im(A − B) ⊆ Ker A∗ = (Im A)⊥ . Since Im A = Im B, we have Im(A − B) ⊆ Im A and therefore Im(A − B) ⊆ Im A ∩ (Im A)⊥ = 0. It follows that if A = B and A ∗≤ B, then rank A < rank B. The rest is easy to see.  Corollary 12. The map Φ preserves rank and hence also invertibility. Let S = Φ(I)−1 . It is easy to see that X → XS is an automorphism with respect to the left star order in both directions on Mn (F). So in the sequel we may and we will replace the map Φ with a map X → Φ(X)Φ(I)−1 . This new map which we will still denote by Φ is a unital automorphism with respect to the left-star order. Note that by Lemma 10 this redefined map Φ maps the set of all projections bijectively onto itself. 3.1. Action of Φ on rank-one projection matrices. Note first that for projections P and Q the following two identities hold P ∗≤ Q ⇐⇒ P Q = QP = P ⇐⇒ Im P ⊆ Im Q.

(4)

Lemma 13. Let P = xx∗ , Q = yy ∗ ∈ Mn (F) be distinct rank-one projections. There exists a unique rank-two projection Z ∈ Mn (F) which is a common upper bound of P and Q with respect to the left-star order. In addition, if a rank-one projection R ∗≤ Z, then there exist α, β ∈ F such that R = (αx + βy)(αx + βy)∗ . Proof. Assume a rank-two projection Z majorizes P and Q. By the definition of the left star partial order x ∈ Im P ⊆ Im Z and similarly, y ∈ Im Z. Since x, y are linearly independent we see that Z must be a projection onto span {x, y} so Z, if exist, must be unique. One easily verifies that this Z actually does majorizes P and Q (with respect to ∗≤ order), whence its existence. For the last part if R ∗≤ Z where Z is a rank-two projection onto span {x, y} then clearly Im R ⊆ Im Z = span {x, y}.  Let Pn (F) = {[x] = Fx : x ∈ Fn \ {0}} denote the n − 1 dimensional projective space over the field F. Recall that a map χ : Pn (F) → Pn (F) is a projective morphism if for every [x] , [y] , [z] ∈ Pn (F) where [x] ∈ [y] + [z], we have χ([x]) ∈ χ([y]) + χ([z]). Lemma 14. There exist an invertible S ∈ Mn (F) and a field automorphism σ : F → F such that (Sxσ )(Sxσ )∗ Φ(xx∗ ) = ; x = 1. Sxσ 2 Proof. By Lemma 10 and Corollary 12, Φ maps rank-one projections bijectively onto itself. Let us define the map φ on a projective space Pn (F) = {[x] = Fx : x ∈ Fn \ {0}} in the following way: yy ∗ xx∗ φ([x]) = [y] if Φ( x 2 ) = y2 . Let us show that φ is a morphism of the projective space. To this end, let [w] ⊆ [x] + [y] for some [x], [y], [w] ∈ Pn (F). Without loss of generality [x] = [y]. By Lemma 13 there exists a xx∗ yy ∗ ∗ unique rank-two projection Z with x 2 , y2 ≤ Z. Since w = αx + βy for suitable scalars ww∗ ∗ xx∗ α, β then also w2 ≤ Z (see (4)). Since Φ is injective, the rank-one projections Φ( x 2 ) and ∗

∗

yy ww Φ( y 2 ) are different, and Φ(Z) is their common upper bound as well as for Φ( w2 ). By

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS ∗

∗

7 ∗

yy ww xx Corollary 12 rank Φ(Z) = rank Z = 2 so Im Φ( w 2 ) ⊆ Im Φ(Z) = Im Φ( x2 ) + Im Φ( y2 ), as claimed. We now use the nonsurjective version of Fundamental Theorem of projective geometry (see [4]) to deduce that φ([x]) = [Sxσ ] for some invertible matrix S and some field automorphism σ : F → F. 

3.2. Action of Φ on general rank-one matrices. Lemma 15. Let x, y ∈ Fn be different nonzero vectors and let z ∈ Fn be nonzero. Then the set {zx∗ , zy ∗ } does not have an upper bound with respect to the left star order. Proof. Assume otherwise, i.e. zx∗ ∗≤ B and zy ∗ ∗≤ B for some B ∈ Mn . By definition (zx∗ )∗ (zx∗ ) = (zx∗ )∗ B and (zy ∗ )∗ (zy ∗ ) = (zy ∗ )∗ B. This implies z 2 xx∗ = xz ∗ B and z 2 yy ∗ = yz ∗ B. In particular, if y = αx then we get |α|2 xz ∗ B = z 2 |α|2 xx∗ = z 2 yy ∗ = yz ∗ B = αxz ∗ B which due to α ∈ / {0, 1} gives z ∗ B = 0 a contradiction. Similarly, if x, y are linearly indepen∗ dent, then z B ∈ (F \ {0})x∗ ∩ (F \ {0})y ∗ = ∅ a contradiction.  Lemma 16. Let x, y, z, w ∈ Fn be such that rank (zx∗ + wy ∗ ) = 2. Then the set {zx∗ , wy ∗ } has an upper bound with respect to the left star order. Moreover, there exists a unique ranktwo operator which is its upper bound. Proof. It follows from rank (zx∗ + wy ∗ ) = 2 that z, w are linearly independent vectors. Let B ∈ Mn be such a matrix that z ∗ B = z 2 x∗ and w∗ B = w 2 y ∗ . Since x and y are also linearly independent we can extend B ∗ to an invertible linear operator. Then, a condition on image from the definition of the left star partial order, i.e., Im zx∗ , Im wy ∗ ⊆ Im B is satisfied. By the definition of B also the conditions (5)

(zx∗ )∗ (zx∗ ) = (zx∗ )∗ B

and

(wy ∗ )∗ (wy ∗ ) = (wy ∗ )∗ B

are satisfied, hence B is an upper bound. We next prove the uniqueness of the rank-two upper bound. Suppose B is a rank-two matrix which is an upper bound of the set {zx∗ , wy ∗ }. By transferring an appropriate scalar to the other factor of the product we may assume that z = w = 1. Using unitary similarity we may further assume that z = e1 and w = ce1 + se2 where c, s ∈ F, |c|2 + |s|2 = 1 and s = 0. By (5) we have (6)

B ∗ e1 = x

and

B ∗ (ce1 + se2 ) = y.

It follows that sB ∗ e2 = y − cB ∗ e1 = y − cx and therefore 1 c (7) B ∗ e2 = y − x. s s Since rank B ∗ = 2 and x, y are linearly independent vectors in Im B ∗ , y c  − x (β1 e1 + β2 e2 + · · · + βn en )∗ B ∗ = x(α1 e1 + α2 e2 + · · · + αn en )∗ + s s for appropriate scalars α1 , α2 , . . . , αn and β1 , β2 , . . . , βn . By (6) and (7) we easily deduce that α1 = β2 = 1 and α2 = β1 = 0. If follows that  y c ∗ B = (e1 + α3 e3 + · · · + αn en )x∗ + (e2 + β3 e3 + · · · + βn en ) − x . s s

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G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

Clearly, e1 + α3 e3 + · · · + αn en ∈ Im(B) = Lin{z, w} = Lin{e1 , e2 }. Thus, α3 = . . . = αn = 0. Similarly, β3 = . . . = βn = 0 and hence  y c ∗ (8) B = e1 x ∗ + e2 − x . s s So there is at most one rank-two upper bound of the set {zx∗ , wy ∗ }. One can easily see that the matrix B from equation (8) is of rank-two and an upper bound for the set {zx∗ , wy ∗ }.  Lemma 17. Suppose w, z ∈ Fn are linearly independent and x ∈ Fn is a nonzero vector. Then the set {wx∗ , zx∗ } does not have an upper bound with respect to the left star order. Proof. By transferring the appropriate scalars from w to x and from z to x we may write ∗ and z the two rank-one operators from the formulation of the lemma as w(αx) ˆ ˆ(βx)∗ where w ˆ = ˆ z = 1. We can find a unitary matrix U ∈ Mn such that U w ˆ = e1 and U zˆ = ce1 + se2 where c, s ∈ F satisfy |c|2 + |s|2 = 1 and s = 0. Let B be an upper bound for the set {wx∗ , zx∗ }. Then C = U BU ∗ is an upper bound for {e1 f ∗ , (ce1 + se2 )f ∗ }, where f = U x. In particular f f ∗ = (e1 f ∗ )∗ (e1 f ∗ ) = (e1 f ∗ )∗ C and f f ∗ = ((ce1 + se2 )f ∗ )∗ ((ce1 + se2 )f ∗ ) = ((ce1 + se2 )f ∗ )∗ C. In particular, e∗1 C = λf ∗ and (ce1 + se2 )∗ C = μf ∗ for some scalars λ, μ ∈ F. It follows that the first two rows of C are linearly dependent and so e1 and ce1 + se2 can not both be in the image of C. This contradicts the fact that e1 ∈ Im(e1 f ∗ ) ⊆ Im C and (ce1 + se2 ) ∈ Im((ce1 + se2 )f ∗ ) ⊆ Im C.  Lemma 18. Let rank R = 1 and let λ ∈ F be nonzero. Then Φ(λR) and Φ(R) are linearly dependent rank one operators. Proof. By Corollary 12, rank Φ(R) = 1. With no loss of generality assume that λ = 1 and write R = xf ∗ . By Lemma 15 the set {R, λR} has no common upper bound and the same holds for their Φ-images. Hence, by Lemma 16 either Φ(R), Φ(λR) ∈ Ly = {yg ∗ : g ∈ Fn \{0}} for some nonzero vector y or Φ(R), Φ(λR) ∈ Rg = {yg ∗ : y ∈ Fn \ {0}} for some nonzero vector g. We consider only the first case; the other is similar. Suppose, therefore, that Φ(R) = yg ∗ and Φ(λR) = yh∗ for some linearly independent g, h. Choose any w ∈ F, linearly independent from f . Then, the sets {xf ∗ , xw∗ } as well as {λxf ∗ , xw∗ } have no common upper bound. We see that Φ(xw∗ ) ∈ Ly ∪ Rg and Φ(xw∗ ) ∈ Ly ∪ Rh which is possible only if Φ(xw∗ ) ∈ Ly . It follows that Φ(Lx ) ⊆ Ly . Similarly we deduce that Φ(Rf ) ⊆ Ly . Consider now vectors u, v ∈ Fn with u, x are linearly independent and the same for f, v. Then rank (xv ∗ + uf ∗ ) = 2, so by Lemma 16 the set {xv ∗ , uf ∗ } has an upper bound with respect to the left-star partial order. On the one hand this implies that the same holds for {Φ(xv ∗ ), Φ(uf ∗ )}, while on the other hand Φ(xv ∗ ) ∈ Φ(Lx ) ⊆ Ly and Φ(uf ∗ ) ∈ Φ(Rf ) ⊆ Ly , which by Lemma 15 implies that the set {Φ(xv ∗ ), Φ(uf ∗ )} has no upper bound unless Φ(xv ∗ ) = Φ(uf ∗ ), a contradiction to injectivity.  Recall that Φ preserves rank-one matrices, see Corollary 12. Let us define the map on Pn (F) × Pn (F) with ([x], [f ]) → ([y], [g]) if Φ(xf ∗ ) = yg ∗ . Notice that by Lemma 18 this is

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

9

a well-defined map. It is also bijective and preserves adjacency as defined by Westwick [11]. By introducing the notation [y][g]∗ := Fyg ∗ and using [11] we obtain Φ(xf ∗ ) ∈ ψ([x])χ([f ])∗ for every rank-one matrix xf ∗ ∈ Mn (F), or Φ(xf ∗ ) ∈ ψ([f ])χ([x])∗ for every rank-one matrix xf ∗ ∈ Mn (F), for some maps ψ, χ : Pn (F) → Pn (F). Inserting f = x with x unimodular and comparing with Lemma 14 we obtain in both cases that ψ([x]) = χ([x]) = [Sxσ ], where S and σ are as in Lemma 14, in particular S is invertible. This gives that (9)

Φ(xf ∗ ) = γxf ∗ Sxσ (fσ )∗ S ∗ ,

or (10)

Φ(xf ∗ ) = γxf ∗ Sfσ (xσ )∗ S ∗

where γxf ∗ is a nonzero scalar that depends on R = xf ∗ (for example if R = xx∗ is a rank-one projection, then γR = Sx1σ 2 ). Using polar decomposition for S = U |S|, one may replace Φ by a map X → U ∗ Φ(X)U and assume without loss of generality that S = |S| = S ∗ is positive definite, so also Hermitian. We will thus assume for the next lemma that the matrix S from (9) or (10) is positive definite. Lemma 19. The restriction of the map Φ on rank-one matrices has one of the following four forms (i) Φ(xy ∗ ) = γxy∗ Sxy ∗ S, (ii) Φ(xy ∗ ) = γxy∗ Sxy ∗ S, (iii) Φ(xy ∗ ) = γxy∗ S(xy ∗ )t S, (iv) Φ(xy ∗ ) = γxy∗ S(xy ∗ )∗ S, where x is unimodular, y is nonzero, and γxy∗ ∈ F \ {0}. In particular, in case (i) we have Sxy ∗ S 1 ∗ γxy∗ = Sx 2 and Φ(xy ) = Sx2 . Proof. Assume Φ takes the form (9). Let B be invertible, and let x be a nonzero vector. Then, by Lemma 6, ∗ x ∗ x x x∗ ∗ x (B x ) = x x B ≤ B and applying Φ gives

∗  ∗ Sxσ xσ γx σ(x) ) S ∗≤ Φ(B), (B )σ ( σ(x)

where we shortly denoted γx instead of γxx∗ B . Denote Z := Φ(B) and use Lemma 6 to see that ∗  ∗ Sxσ xσ ) S = uu∗ Z γx σ(x) (B )σ ( σ(x) for some unimodular u ∈ Im Z. Comparing both sides and using that, for rank-one matrices, xf ∗ = yg ∗ if and only if y = λx and g = (1/λ)f for some λ ∈ F, gives after slight rearranging ∗ ·Sxσ   Sxσ xσ and γxσ(x) ) S = δu∗ Z (B ∗ )σ ( σ(x) (11) u = δ Sx σ

10

G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

for some unimodular scalar δ ∈ F (here we use that u = 1). Inserting the left equation into the right one and simplifying gives (S −1 Z ∗ S − ξx (B ∗ )σ )xσ = 0 2

·Sxσ  where the scalar ξx = γ| xσ(x) depends on x. Since σ is surjective, we see that S −1 Z ∗ S and |2 (B ∗ )σ are locally linearly dependent and as (B ∗ )σ is invertible, there exists a nonzero scalar γB such that Φ(B) = Z = γB S −1 ((B ∗ )σ )∗ S.

It follows that ξx = γB for every nonzero x. In particular, recalling the shortcut notation γx instead of γxx∗ B we have (12)

γxx∗ B =

| σ( x ) |2 · ξx | σ( x ) |2 · γB = . Sxσ 2 Sxσ 2

Insert x = (δ, 0, . . . , 0)∗ where |δ| = 1 runs over a complex unit circle. Then x = 1 for every δ so that the numerator constantly equals γB . Clearly, the left-hand side, γxx∗ B = γe1 e∗1 B is also a constant while the denominator on the right equals |σ(δ)|2 · Se1 2 . Thus |σ(δ)| is a constant on unimodular numbers δ and since σ(1) = 1 we see that |σ(δ)| = 1;

|δ| = 1.

Clearly 1+(−1) = 0 while 1+1 = 2. By continuity the expression |1+δ| takes all the values in the line segment [0, 2] as δ varies over the unit circle. By rotating, i.e., simultaneously multiplying 1 and δ with a suitable unimodular number we see from the polar decomposition of a complex number that the sum δ1 + δ2 covers the whole disk of radius 2 centered at 0 as δ1 , δ2 vary over the unit circle. Since the automorphism σ is additive and maps the unit circle into itself we see that it is bounded on the closed unit disk Δ; actually, σ(Δ) ⊆ 2Δ. Note that σ(r) = r for rational numbers, so by multiplicativity, σ(rΔ) ⊆ 2rΔ. Hence, σ is continuous and as it is well-known (see, e.g. [10, page 54]), σ is either the identity or the complex conjugation. It follows that when (9) holds, Φ is of the form (i) or (ii). Similarly we prove that if (10) holds, then Φ takes the form (iii) or (iv). Moreover, with x = 1, (12) implies that γB γxx∗ B = . Sxσ 2 Choose an invertible B such that B ∗ fixes x. Then, xx∗ B = xx∗ is a rank-one projection, ∗S which is by Lemma 14 mapped by Φ into another rank-one projection Φ(xx∗ ) = Sxx . Sx2 Hence, by letting Sp B denote the set of all eigenvalues of the matrix B we obtain γB = 1

if

1 ∈ Sp B.

Since the size of matrices is at least three, it follows that for each two unimodular vectors x, y we can find an invertible matrix B such that 1 is its eigenvalue and Bx = y. It follows that each rank-one R can be written as R = xx∗ B for some unimodular x and some invertible B with an eigenvalue 1. Thus, γB 1 γR = = 2 Sxσ Sxσ 2

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

11

for every rank-one R = xx∗ B. Note that R = xx∗ B implies that x ∈ Im R and thus there exists y ∈ Fn such that R = xy ∗ . In particular, in case (i) where Φ(xy ∗ ) = γxy∗ Sxy ∗ S, we 1  have γxy∗ = Sx 2. Remark 20. Note that if Φ(R) = γR SRS for every rank-one R ∈ Mn (F), then we may compose Φ with X → X to achieve that the new map sends rank-one matrices R = xy ∗ into Φ(R) = Φ(xy ∗ ) = γR Sxy ∗ S = γR Sxy ∗ S. Thus by composing Φ with X → X we can by Lemma 8 reduce item (ii) of Lemma 19 to item (i). Moreover, if Φ(R) = γR SRt S for † every rank-one R, then we may compose Φ with X → X to achieve that the new map sends γ † ∗ † ∗ R rank-one matrices R = xy ∗ into Φ(R ) = Φ( xyx 2 y2 ) = x2 y2 Sxy S. Thus by composing †

Φ with X → X we can by Lemma 9 reduce item (iii) to item (i). Lastly, item (iv) is reduced to item (iii) by composing Φ with the map X → X, and then reduced to (i) using Lemma 9. Let Ω ⊆ Mn (F), let r ≥ 0 be an integer, and suppose there exists the unique matrix Z ∈ Mn (F) of rank r which is the least upper bound for Ω with respect to the ∗≤ partial order. More precisely, A ∗≤ Z for every A ∈ Ω, rank Z = r, and if A ∗≤ W for every A ∈ Ω, then Z = W or rank W = r. Then we denote this fact by writing Z = supr Ω. Note that there are subsets of matrices for which such a matrix Z = supr Ω does not exist, even if no bound on r is imposed, e.g., a subset of invertible matrices with at least two elements. Lemma 21. Let B ∈ Mn (F) and let S ∈ Mn (F) be a positive definite matrix. Then  

Sxx∗ BS −1 † † 2 † † 2 † † BB +(I −BB )S (BB )·[BB S BB ] BS. : x ∈ Im B, x = 1 = S sup Sx 2 rank B ∗ Proof. If B = xy is of rank at ∗most one, then the set on the left-hand side is a singleton  2 x SBS ∗ 2 2 . Now, use B † = yyx 2 x2 and use that, with S self-adjoint, x S x = Sx to get Sx2 that the matrix on the right-hand side equals 

∗ †  −1 xx∗ 2 xx∗ 2 xx∗ xx∗ xx S BS + (I − x2 )S x2 · x2 S x2 x2     2 ∗ x 2 Sxy ∗ S Sx2 xx∗ −1 xx∗ −1 S 2 xx∗ S xx xx∗ =S BS = S + − BS = · 2 2 4 2 2 x x x Sx Sx Sx 2 which clearly equals the supremum of the singleton from the left-hand side. Let now rank B ≥ 2. We first prove that any upper bound Z ∈ Mn (F) of the set on the left-hand side and satisfying rank Z = rank B, if exists, must equal to the matrix on the right-hand side. To this end, Z, being an upper bound, implies that for each unimodular x ∈ Im B we have Sxx∗ BS ∗≤ Z. Sx 2 By Lemma 6 this implies that there exists unimodular u = ux ∈ Im Z such that Sxx∗ BS = uu∗ Z. Sx 2 It follows that, for suitable δ ∈ F, (13)

u=

δ Sx Sx2

and

δu∗ Z = x∗ BS.

12

G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

Since u ∈ Im Z and x ∈ Im B and rank Z = rank B we deduce that (14)

Im Z = S Im B.

Also, the first equality in (13) is equivalent to u∗ = which combining with the second one gives |δ|2 x∗ SZ Sx2

= x∗ BS;

δ x∗ S ∗ Sx2

=

δ x∗ S Sx2

(because S = S ∗ )

x ∈ Im B, x = 1. 2

|δ| By writing x = By and by temporarily denoting αy := Sx 2 we can rewrite the above equation into y ∗ B ∗ (αy SZ − BS) = 0; y ∈ Fn .

It follows that B ∗ SZ and B ∗ BS are locally linearly dependent (when acting from the right) and since rank (B ∗ BS) = rank B ≥ 2 there exists a scalar γ = γB such that B ∗ SZS −1 = γB B ∗ B. Therefore, Im(SZS −1 − γB B) ⊆ Ker B ∗ = (Im B)⊥ . With respect to the orthogonal decomposition Fn = (Im B) ⊕ (Im B)⊥ we thus have a block decomposition     B 1 B2 0 0 −1 . , B= SZS − γB B = 0 0 W1 W2 ˆ r = rank B such that Moreover, since rank Z = rank B there exists matrix C,   an (n − r)-by-r γ I 0 r B ˆ 1 B2 ) and therefore C := ∈ Mn (F) satisfies (W1 W2 ) = C(B ˆ 0n−r C   γ B B 1 γ B B2 −1 SZS = CB = ˆ ˆ 2 . CB1 CB Recall from (14) that Im(SZ) = S 2 Im B = Im(S 2 B). With this in mind, and again using orthogonal decomposition Fn = Im B ⊕ (Im B)⊥ we write   S 1 S2 S2 = . S2∗ S3 Recall that S 2 is invertible and positive-definite. Clearly then, its compression, S1 , to Im B is also positive-definite, hence invertible and thus maps Im B bijectively onto itself. Thus, Im B= ImS1 B and hence there exists T ∈ Mn (F) with B = S1 BT . So, by decomposing T = TT13 TT24 and using the fact that (B1 | B2 ) : Fn → Im B is surjective, we are forced to let Cˆ := γB S2∗ S1−1 = γB (I − P )S 2 P |(Im P ) · ((P S 2 P )|Im P )−1 ;

P := BB †

where B † is a Moore-Penrose inverse (i.e., P is a projection onto Im B) and with this unique choice of Cˆ we have   I 0 (15) Z = γB S −1 S ∗ Sr−1 0 BS = γB S −1 C  BS. 2

1

Using Moore-Penrose pseudoinverse (and noting that BB † is a projection onto Im B) we can rewrite this into   −1 † † 2 † † 2 † † (16) Z = γB S BB + (I − BB )S (BB ) · [BB S BB ] BS.

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

We claim that γB = 1. Namely, given a unimodular x ∈ Im B we have Sxx∗ BS Sx2

= and only if complex number δ, (17)

uu∗ Z

Sxx∗ BS Sx2

13

∗≤ Z if

for some unimodular u ∈ Im Z or equivalently, for a unimodular x∗ BS = δu∗ Z. Sx

Sx = δu, Sx

Insert Z = γB S −1 C  BS and recall that (17) holds for every unimodular x = By ∈ Im B. So after simplification we may rewrite this into B ∗ (B − γB C  B) = 0, which, relative to decomposition Fn = Im B ⊕ (Im B)⊥ , can be written as a 2-by-2 block matrix   ∗ B B B∗B 0 = (1 − γB ) B1∗ B11 B1∗ B22 = (1 − γB )B ∗ B, 2

2

which is possible only if 1 − γB = 0.  ∗ BS It remains to see that this matrix Z is an upper bound for Sxx : x ∈ Im B, x = 1 . 2 Sx First of all, let us show that S Im B ⊆ Im Z. ⊥ To this end, choose any x = x1 ⊕0 ∈ ImB ⊕ (Im B)  choose y with BSy = (S1 x1 ) ⊕ 0 ∈  and S1 S2 S1 x1 x1 ⊥ 2 Im B ⊕ (Im B) . Then, S x = S2∗ S3 ( 0 ) = S2∗ x1 = SZy, as claimed. Since clearly

rank Z = rank B, we actually have S Im B = Im Z. Now, given Sxx∗ BS Sx2

unimodular one easily checks that Lemma 6, Z is indeed an upper bound.

= uu∗ Z for u =

Sx Sx

Sxx∗ BS Sx2

with x ∈ Im B

∈ S Im B = Im Z. So, by 

3.3. Proof of the sufficiency. In this subsection we will prove the nontrivial fact that the maps defined in Theorem 3 actually do preserve the left-star partial order in both directions and are bijective. This will be done in another series of lemmas. For the sole purpose of this subsection, let us explicitly state the four forms of maps defined in Theorem 3. They are  †   † † −1 −∗ † † −1 −∗ † (i) Ψ(X) = T XX + (I − XX ) · T T · XX · XX · T T · XX XW ; †    † † −1 −∗ † † −1 −∗ † (ii) Ψ(X) = T XX + (I − XX ) · T T · XX · XX · T T · XX XW ; †    (iii) Ψ(X † ) = T XX † + (I − XX † ) · T −1 T −∗ · XX † · XX † · T −1 T −∗ · XX † XW ; †    † (iv) Ψ(X ) = T XX † + (I − XX † ) · T −1 T −∗ · XX † · XX † · T −1 T −∗ · XX † XW . Let T = U |T | be its polar decomposition. Clearly, in each of the four items we may replace Ψ by the map X → U ∗ Ψ(X)W −1 |T |−1 , which for brevity we still denote by Ψ. Let us denote S = |T |−1 . Then, T −1 T −∗ = S 2 . The form of the map Ψ from item (i) thus transforms to  †   −1 † † 2 † † 2 † Ψ(X) = S XX + (I − XX )S (XX ) · XX S XX XS.

14

G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

We can now prove that the four maps Ψ defined in Theorem 3 do preserve the ∗≤ order in both directions. In view of Lemmas 8 and 9 (see also Remark 20) it suffices to show this for the map  †   −1 † † 2 † † 2 † XX + (I − XX )S (XX ) · XX S XX XS (18) Ψ(X) = S where S ∈ Mn (F) is a positive definite matrix. Also, note that in this case for every unimodular x ∈ Fn and f ∈ Fn , SRS ; R = xf ∗ . Ψ(R) = Sx 2 We proceed to show that the map Ψ from (18) preserves left-star partial order in both directions, and then that it is surjective. The first part will follow immediately from Lemma 7 once we show the lemma below. Lemma 22. Let R ∈ Mn (F) be of rank-one. Then R ∗≤ B for some B ∈ Mn (F) if and only if Ψ(R) ∗≤ Ψ(B). Proof. Let R ∗≤ B. Without loss of generality we may assume that B = R so B is at least of rank two. There exists a unimodular x ∈ Im B such that R = xx∗ B. Using orthogonal decomposition Fn = Im B ⊕ (Im B)⊥ write   S 1 S2 2 S = S2∗ S3 and note that then by (18) (see also (15)) Ψ(B) = S

−1



 Ir 0 BS. S2∗ S1−1 0

We have Ψ(R)∗ Ψ(R) =

1 1 SB ∗ xx∗ SSxx∗ BS = SB ∗ xx∗ S 2 xx∗ BS. 4 Sx Sx 4

n Note that x∗ S 2 x = Sx 2 and since  x ∈ Im B so, relative to decomposition F = Im B ⊕ I 0 (Im B)⊥ we have x∗ = x∗ S ∗ Sr−1 0 . It follows that 2 1   1 1 Ir 0 ∗ ∗ ∗ ∗ ∗ −1 BS = Ψ(R)∗ Ψ(B). SB xx BS = SB xx SS Ψ(R) Ψ(R) = S2∗ S1−1 0 Sx 2 Sx 2

It remains to show that Im Ψ(R) ⊆ Im Ψ(B). Decompose x = x1 ⊕ 0 ∈ Im B ⊕ (Im B)⊥ . Then      S 1 x1 x1 S 1 S2 2 S x= = . S2∗ S3 0 S2∗ x1 There exists Now use the fact that (B1 | B2 ) : Fn = Im B ⊕ (Im B)⊥ → ImB is surjective.  ⊥ such that Bz = B1 z1 +B2 z2 S1 x1 . It follows that z = z1 ⊕ z2 ∈ Im B ⊕ (Im B) = 0 0    B1 B2 S1 x1 z1 2 x ∈ Im(SΨ(B)) so that Sx ∈ Im Ψ(B). But ) = ( . Therefore S ∗ −1 −1 z2 S2 x1 S2∗ S1 B1 S2∗ S1 B2 Im Ψ(R) = span Sx ⊆ Im Ψ(B) and thus Ψ(R) ∗≤ Ψ(B).

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

15

∗ Conversely, let Ψ(R)  = uu Ψ(B) ∗≤ Ψ(B) where u ∈ Im Ψ(B) is unimodular. Recall I 0 1 Ψ(B) = S −1 S ∗ Sr−1 0 BS and Ψ(R) = Sx 2 SRS. It follows that 2 1      B1 0 B2 Ir B1 B 2 . = Im Su ∈ Im 0 0 S2∗ S1−1 B1 S2∗ S1−1 B2 S2∗ S1−1 0   B1 B2 . Since S is invertible, there exists x = ( xx12 ) such that u = Sx. So S 2 x ∈ Im S ∗ S −1 −1 ∗ 2 1 B 1 S2 S1 B 2 z1 For some z = ( z2 ) we have      B1 z1 + B2 z2 S 1 S2 x1 = S2∗ S3 x2 S2∗ S1−1 (B1 z1 + B2 z2 )

and therefore S1 x1 + S2 x2 = B1 z1 + B2 z2 and S2∗ x1 + S3 x2 = S2∗ S1−1 (S1 x1 + S2 x2 ). So S3 x2 = S2∗ S1−1 S2 x2 . Note that S3 − S2∗ S1−1 S2 is the Schur complement of a positive definite matrix   S 1 S2 2 S = S2∗ S3

and hence it is also positive definite. Since x2 ∈ Ker (S3 − S2∗ S1−1 S2 ), it follows that x2 = 0 and therefore x = x1 ⊕ 0 ∈ Im B. Note that span u = Im Ψ(R) and therefore span x = span (S −1 u) = Im R. Thus Im R ⊆ Im B. It remains to show that R∗ R = R∗ B. Write R = xf ∗ with x = 1. Note that Ψ(R) ∗≤ Ψ(B) implies Ψ(R)∗ Ψ(R) = Ψ(R)∗ Ψ(B). Since S is invertible this easily simplifies to   1 Ir 0 ∗ 2 ∗ R S R=R B. (19) S2∗ S1−1 0 Sx 2 Since x∗ S 2 x = Sx 2 the left side of (19) simplifies to 1 f x∗ S 2 xf ∗ = f f ∗ = f x∗ xf ∗ = R∗ R. Sx 2 Since Im R ⊆ Im B we may, relative to the orthogonal decomposition Fn = Im B ⊕ (Im B)⊥ , write R = (x1 ⊕ 0)f ∗ . It easily follows that the right side of equation (19) equals R∗ B. So equation (19) implies R∗ R = R∗ B. Together with Im R ⊆ Im B this gives R ∗≤ B.  Lemma 23. The maps Ψ defined in Theorem 3 are surjective. Proof. It suffices to show this for the map  †   −1 † † 2 † † 2 † XX + (I − XX )S (XX ) · XX S XX XS Ψ(X) = S where S ∈ Mn (F) is a positive definite matrix. Recall that if Ψ is of this form, then for every ∗S unimodular x ∈ Fn and f ∈ Fn , Ψ(xf ∗ ) = Sxf . Let R ∈ Mn (F) be of rank-one. Then there Sx2 n n exists a unimodular x ∈ F and y ∈ F such that R = xy ∗ . It is an easy calculation that  −1 ∗ S −1  (20) Ψ S Sxy = xy ∗ −1 x2 so Ψ is bijective on rank-one matrices. Consider any B ∈ Mn (F). Then the set of all rank-one matrices which are majorized by B with respect to the left-star order takes the form ΩB := {xx∗ B : x ∈ Im B, x = 1}.

16

G. DOLINAR, S. HALICIOGLU, A. HARMANCI, B. KUZMA, J. MAROVT, AND B. UNGOR

By equation (20) the set

 S −1 xx∗ BS −1

Ω=

S −1 x2

: x ∈ Im B



is mapped bijectively to ΩB . With respect to the decomposition Fn = Im B ⊕ (Im B)⊥ define    Ir 0 B 1 B2 S −1 A = S ˆ∗ ˆ−1 0 0 S2 S1 0 ˆ ˆ  where Sˆi , i = 1, 2, are blocks from S −2 = SSˆ1∗ SSˆ2 . Clearly, rank A = rank B, so by Lemma 21 2

3



Ω = ΩA . Since also Ψ(ΩA ) = Ψ(Ω) = ΩB , we have by Lemma 7 that Ψ(A) = B.

3.4. Proof of Theorem 3. As explained after Corollary 12, Φ(I) is invertible, and the map X → Φ(X)Φ(I)−1 is a unital automorphism of ∗≤ . As explained in text before (9)–(10) we can find a unitary matrix U such that the map X → U Φ(X)Φ(I)−1 U ∗ , which we still denote by Φ, takes one of the forms (i)–(iv) stated in Lemma 19 for some positive definite matrix S. † By composing Φ with the maps X → X or X → X or X → X † we can translate the forms (ii)–(iv) from Lemma 19 to the form (i). So without loss of generality we may assume in the sequel that Φ takes the form (i) on rank-one matrices. By Lemma 21, the matrix B is, among all matrices of rank r = rank B, the supremum of the set {xx∗ B: x ∈ Im B, x = 1}. This  Sxx ∗ : x ∈ Im B, x = 1 . Recall that by Corollary set is mapped by Φ onto the set SxBS 2 12, rank Φ(X) = rank X for every X ∈ Mn (F). Since Φ is an automorphism of the left-star partial order we see that   ∗ BS Φ(B) = sup Sxx : x ∈ Im B, x = 1 . Sx2 r



By Lemma 21 again, Φ(B) =

S −1



BB †

+ (I −

BB † )S 2 (BB † )

·

[BB † S 2 BB † ]†

BS.



Example. Suppose Φ assumes both forms from Theorem 3 and simultaneously from Corollary 4. Say,  †   Φ(X) = T XX † + (I − XX † ) · T −1 T −∗ · XX † · XX † · T −1 T −∗ · XX † XW (21)   †  •

•

•

•

•

•

•

= Z X X † X + X † X · W −∗ W −1 · X † X

•

•

•

•

· X † X · W −∗ W −1 · (I − X † X) N. •

By inserting invertible X and using XX † = I we get that T XW = Z XN for each invertible • X. It follows easily that X = X and that (T, W ) = (λZ, λ1 N ). By inserting X = P = P ∗ = P 2 we obtain, after simplification (P + (I − P )|T −∗ |2 P · (P |T −∗ |2 P )† )P = P (P + (P |W −1 |2 P )† · P |W −1 |2 (I − P )) which further simplifies into (I − P )|T −∗ |2 P · (P |T −∗ |2 P )† = (P |W −1 |2 P )† · P |W −1 |2 (I − P ) It follows that both sides must vanish identically for every self-adjoint projector P , which is possible only if |T −∗ |2 = μI and |W −1 |2 = νI are positive-definite scalar matrices. This is, by polar decomposition, equivalent to the fact that T = μU and W = νV are scalar multiples of unitary matrices.

PRESERVERS OF THE LEFT-STAR AND RIGHT-STAR PARTIAL ORDERS

17

With this in mind, using T −1 T −∗ = μI in (21) it simplifies into •

•

•

(XX † )X = X(X † X) for every X, which is clearly satisfied. Therefore, such Φ takes the form X → ξU XV for some unitary matrices U, V and some scalar ξ. Remark that each bijective Φ which assumes both forms from Theorem 3 and simultaneously from Corollary 4 preserves the star partial order in both directions. The form of all bijective maps on Mn (F), n ≥ 3, that preserve the star partial order in both directions was determined by Legiˇsa [7] and is richer than X → ξU XV . References [1] J. K. Baksalary, S. K. Mitra, Left-star and right-star partial orderings, Linear Algebra Appl. 149 (1991), 73–89. [2] G. Dolinar, A. E. Guterman, J. Marovt, Monotone transformations on B(H) with respect to the left-star and the right-star partial order, Math. Inequal. Appl. 17 (2014), No. 2, 573–589. [3] M. P. Drazin, Natural structures on semigroups with involution, Bull. Amer. Math. Soc. 84 (1978), 139–141. [4] C-A. Faure, An elementary proof of the fundamental theorem of projective geometry, Geom. Dedicata 90 (2002), No. 1, 145–151. [5] A. E. Guterman, Linear preservers for Drazin star partial order, Comm. Algebra 29 (2001), 3905–3917. [6] A. E. Guterman, Monotone additive transformations on matrices, Mat. Zametki 81 (2007), 681–692. [7] P. Legiˇsa, Automorphisms of Mn , partially ordered by rank substractivity ordering, Linear Algebra Appl. 389 (2004), 147–158. [8] P. Legiˇsa, Automorphisms of Mn , partially ordered by the star order, Linear Multilinear Algebra 54 (2006), No. 3, 157–188. [9] S. K. Mitra, P. Bhimasankaram, S. B. Malik, Matrix partial orders, shorted operators and applications, Word Scientific, London, 2010. [10] A. M. Robert, A course in p-adic analysis, Springer-Verlag, New York, 2000. [11] R. Westwick, Transformations on tensor spaces, Pacific J. Math. 23 (1967), No. 3, 613–620. (Gregor Dolinar) University of Ljubljana, Faculty of Electrical Engineering, Trˇ zaˇ ska cesta 25, SI-1000 Ljubljana, Slovenia, and IMFM, Jadranska 19, SI-1000 Ljubljana, Slovenia E-mail address: [email protected] (Sait Halicioglu) Department of Mathematics, Ankara University, 06100, Ankara, Turkey E-mail address: [email protected] (Abdullah Harmanci) Department of Mathematics, Hacettepe University, 06800, Ankara, Turkey E-mail address: [email protected] (Bojan Kuzma) University of Primorska, Glagoljaˇ ska 8, SI-6000 Koper, Slovenia, and IMFM, Jadranska 19, SI-1000 Ljubljana, Slovenia E-mail address: [email protected] (Janko Marovt) University of Maribor, Faculty of Economics and Business, Razlagova 14, SI-2000 Maribor, Slovenia, and IMFM, Jadranska 19, SI-1000 Ljubljana, Slovenia E-mail address: [email protected] (Burcu Ungor) Department of Mathematics, Ankara University, 06100, Ankara, Turkey E-mail address: [email protected]