A note on star partial order preservers on the set of all variance-covariance matrices

Proposition 1

Let L₁=(y, X₁β, σ²I) and L₂=(y, X₂β, σ²I) be any two (Gauss-Markov) linear models. Then X₁* ⩽ X₂ if and only if the following statements hold.

1. The linear models L₁ and L=(y, (X₂−X₁)β, σ ²I) have no common estimable linear function of β;

2. X₁β is estimable under the model L₂;

3. The BLUE of X₁β under the model L₁ is also its BLUE under L₂, and the variance-covariance matrix of the BLUE of X₁β under the model L₁ is the same as under the model L₂.

Let X ∈ M_{n, p}(ℝ) and let sβ be an estimable linear parametric function (here s is a 1 × p real vector). A linear function of the response, ly (here l is a 1 × n real vector) is called a linear zero function (LZF) if E (ly)=0 for all possible values of β. Here E (ly) denotes the mathematical expectation of the random variable ly. Due to the additivity property of the mathematical expectation, one can easily see that by adding LZFs to LUE, we get other LUEs of sβ. Next we present another application in the theory of comparison of linear models when the model matrices are related via the left-star partial order (see [21: Corollary 15.3.8]).

Proposition 2

Let L₁=(y, X₁β, σ²I) and L₂=(y, X₂β, σ²I) be any two linear models such that X₁* ⩽ X₂. Then the BLUE of every estimable linear function of the parameters under L₁ is a linear zero function under L=(y, (X₂−X₁)β, σ²I) and vice versa.

Let A be some subset of Mn(F) and denote by ⩽ a partial order on A . We say that a map Φ:A→A is a bi-preserver of the order ⩽ (or that it preserves the order ⩽ in both directions) when

for every A,B∈A . Instead of the linear model M=(y, Xβ, σ ²I) one might rather work with the transformed model Mˆ=(y,Xˆβ,σ2I) , where the new matrix Xˆ∈Mn(R) is chosen so that its properties are more attractive than X ∈ M_n(ℝ) (e.g., elements of X that are very close to zero are transformed to zero), and thus it is natural (see [6]) to demand that the transformed model still retains most of the properties of the original model (e.g., has similar relations to other transformed models). Thus, in view of Propositions 1 and 2, it is interesting to know what transformations on M_n(ℝ) or perhaps on some subset of M_n(ℝ) (like Hn+(R)) preserve the left-star partial order. Linear bijective maps on Mn(F) that preserve either the star, or the left-star, or the right-star partial order in one direction (i.e., A ⩽ B implies Φ(A) ⩽ Φ(B), where ⩽ is the appropriate order) were investigated in [12] (see also [11]). In [6], the forms of surjective bi-preservers of the left-star partial order and the right-star partial order on Mn(F) , n=3, were described. Surjective bi-preservers of the star partial order on Mn(F) , n=3, were characterized in [18]. Since surjective bi-preservers of the star, the left-star, and the right-star partial orders on the full matrix algebra Mn(F) , n=3, have already been described (see also [13]) and because the set of all positive semidefinite real matrices has an important role in statistics, we focus our attention on the star partial order bi-preservers on Hn+(R) . In [10] (see also [9]), the following result was proved.

Theorem 3

Let n=3 be an integer. Then Φ:Hn+(R)→Hn+(R) is a surjective, additive bi-preserver of the star partial order if and only if there exist an orthogonal matrix R ∈ M_n(ℝ) and λ>0 such that

for every A∈Hn+(R) .

In this paper, we study bi-preservers of the star partial order on Hn+(R) that may be nonadditive. In Section 2, we give some preliminary results. The main result which describes the form of all surjective bi-preservers of the star partial order on Hn+(R) , n=3, is stated and proved in Section 3.

Proposition 4

Let A,B∈Mn(F) and α∈F with α≠0. Let U,V∈Mn(F) be unitary matrices (i.e., orthogonal when F=R) . The following statements are then equivalent.

1. A ⩽ ^* B.

2. α A ⩽ ^* α B.

3. UAV ⩽ ^*UBV.

Proposition 5

Let A,B∈Mn(F) . If A ⩽ ^*B and A≠B, then rank(A)<rank(B) .

The next lemma states that any Hermitian matrix A is idempotent if and only if the identity matrix I is its upper bound with respect to the star partial order (see [15: Theorem 3.4]).

Lemma 6

For A∈Hn(F) , A ⩽ ^*I if and only if A²=A.

Proof. Suppose A²=A, A∈Hn(F) . Then A^*A=A²=A=AI=A^*I and similarly AA^*=IA^*. So, A ⩽ ^*I. Conversely, suppose A ⩽ ^*I for A∈Hn(F) . By, e.g., [20: Theorem 8], A²=A. □

Let x,y∈Fn be nonzero. We denote by xy^* a rank-one matrix in Mn(F) . Note that it represents a rank-one linear operator on Fn defined with (xy^*)z=〈z, y〉 x for every z∈Fn (here 〈z, y〉=y^*z is an inner product in Fn) . It is known that every rank-one matrix in Mn(F) may be written in this form and that P∈Pn(F) is of rank-one if and only if P=x x^* for some x∈Fn with ∣ x∣=1. The following result was proved in [18] (see also [10]).

Proposition 7

Let A∈Mn(F) and let x,y∈Fn be nonzero. Then

Corollary 8

Let A∈Hn(F) and let x∈Fn be nonzero. Then

Denote by Ei,j∈Mn(F) the matrix with all entries equal to zero except the (i, j)-entry which is equal to one, and by e1,e2,…,en∈Fn the standard basis vectors.

Lemma 9

Let xx^*, yy∗∈Hn+(F) be rank-one matrices with different spectra. Then x x^* and y y^* have a common upper bound with respect to the star partial order if and only if vectors x and y are orthogonal.

Proof By using unitary similarity, we may by Proposition 4 assume that x=λ e₁ for some nonzero λ∈R and y=α e₁+β e₂, α,β∈F , where 0<α2+β2 . Note that λ ²≠α ²+β ² since x x^* and y y^* have different spectra.

Suppose first that x and y are orthogonal. Then α=0 and hence A=xx∗+yy∗=λ2e1e1∗+β2e2e2∗ is a common upper bound of x x^* and y y^*.

Suppose now x and y are not orthogonal, and let us assume that there exists A=(ai,j)∈Hn+(F) which is a common upper bound of x x^* and y y^*. By Corollary 8, we have A(λ e₁)=λ ³e₁ and thus Ae₁=λ ²e₁. Also, A(α e₁+β e₂)=(α ²+β ²)(α e₁+β e₂). It follows that

and therefore,

If α=0, then y=β e₂ and thus x and y are orthogonal, a contradiction. If β=0, then we get a contradiction by (2) since α≠ 0 and α ²+β ²−λ ²≠ 0. By Ae₁=λ ²e₁, we obtain a_2,1=0, and since β≠ 0, we get (see (2)) that a_1,2≠ 0. So, A∉Hn(F) , a contradiction. □

Let A∈Mn(F) be nonzero. As a direct corollary of the singular value decomposition there is a unique decomposition

where t1>t2>…>tk>0 and V₁, V₂, ⋅, V_k are pairwise orthogonal nonzero partial isometries. This decomposition is known as the Penrose decomposition. Here rank A=∑j=1k rank (V_j) . In [14: page 371] (see also [18: Theorem 3.3]), the following result was proved.

Proposition 10

Let A, B∈Mn(F) have the Penrose decompositions

Then A ⩽ ^*B if and only if there exists a function φ:{1, 2, ⋅, k} → {1, 2, ⋅, l} such that t_j=u_φ(j) and V_j ⩽ ^*W_φ(j) for all j ∈ {1, 2, ⋅, k}.

Remark 11

Note that the function φ from Proposition 10 is injective. Indeed, if φ(i)=φ(j), then t_i=u_φ(i)=u_φ(j)=t_j which implies by the definition of the Penrose decomposition that i=j.

Let now A∈Hn+(F) be nonzero. By the spectral theorem there exists a unique (eigenvalue) decomposition

where λ1>λ2>⋯>λk>0 are (nonzero) eigenvalues of A and P₁, P₂, …, P_k are pairwise orthogonal nonzero projectors. Observe that this is in fact also the Penrose decomposition of A. We thus have the following direct corollary of Proposition 10.

Let ℝ⁺=[0, ∞) denote the set of all nonnegative real numbers. For A∈Hn+(F) , let Sp(A) denote the spectrum of A, i.e., the set of all eigenvalues of A, and let #Sp(A) denote the number of (pairwise distinct) eigenvalues of A.

Definition 1

Let f:ℝ⁺ → ℝ⁺ be a function. For A∈Hn+(F) , let r=#Sp(A) and let p_A be the polynomial of the degree r − 1 such that the restriction of p_A to Sp(A) equals the restriction of f to Sp(A) . For this A, we define

Remark 13

Note that for every fixed A∈Hn+(F) the polynomial p_A, introduced in Definition 1, exists and is unique. Moreover, we could take any polynomial q_A such that its restriction to Sp(A) equals the restriction of f to Sp(A) . Namely, for any λ∈Sp(A) , we have (q_A−p_A)(λ)=f(λ)−f(λ)=0 and thus (q_A−p_A)(A)=0.

Remark 14

Note that when f is analytic or even measurable, f(A) as defined in Definition 1 matches f(A) obtained via the functional calculus.

Remark 15

Let f:ℝ⁺ → ℝ⁺ be a function with f(0)=0, and let Φ:Hn+(F)→Hn+(F) be defined as Φ(A)=f(A). Let A=∑j=1kλjPj be the eigenvalue decomposition of A∈Hn+(F) . Observe, by the orthogonality of projectors involved, that then Φ(A)=∑j=1kf(λj)Pj and thus Φ(Hn+(F))⊆Hn+(F) .

Lemma 16

Let f:ℝ⁺ → ℝ⁺ be a bijective function with f (0)=0 and let Φ:Hn+(F)→Hn+(F) be defined as

Then Φ is a bijective star partial order bi-preserver.

Proof. Let B∈Hn+(F) be nonzero and let B=∑i=1lμiQi be the eigenvalue decomposition of B. Let A=∑i=1lf−1(μi)Qi . This is up to a permutation of indices the eigenvalue decomposition of A. Thus

Since f (0)=0, we have Φ (0)=0 and thus Φ is surjective. Let now Φ(A)=Φ(B)≠0 and let ∑j=1kλjPj be the eigenvalue decomposition of Φ(A). Then A=∑j=1kf−1(λj)Pj=B . If Φ(A)=Φ(B)=0, then A=B=0 since f (0)=0 and f is injective. We may conclude that Φ is injective and thus bijective.

Suppose now A ⩽ ^*B for some nonzero A, B∈Hn+(F) . Let A=∑j=1kλjPj , B=∑i=1lμiQi be the eigenvalue decompositions of A and B, respectively. By Corollary 12, there exists a function φ:{1, 2, …, k} → {1, 2, …, l} such that λ_j=μ_φ(j) and P_j ⩽ ^*Q_φ(j) for all j ∈ {1, 2, …, k} . Also,

and therefore, since λ_j=μ_φ(j) implies f(λ_j)=f(μ_φ(j)), we may again by Corollary 12 conclude that Φ(A) ⩽ ^*Φ(B). If A=0, then Φ(A)=f (0)=0 and so (see (\refdef_star)) A ⩽ ^*B and Φ(A) ⩽ ^*Φ(B) for every B∈Hn+(F) . We proved that A ⩽ ^*B implies Φ(A) ⩽ ^*Φ(B) for every A,B∈Hn+(F) . Since f is bijective, we may similarly show that Φ(A) ⩽ ^*Φ(B) implies A ⩽ ^*B. □

Theorem 17

Let n=3 be an integer. Then Φ:Hn+(R)→Hn+(R) is a surjective bi-preserver of the star partial order if and only if there exist an orthogonal matrix Q ∈ M_n(ℝ), a positive λ∈R , and a bijective function f:ℝ⁺↦ℝ⁺ with f (0)=0 such that

for every A∈Hn+(R) .

Proof. Let Φ:Hn+(R)→Hn+(R) be of the form Φ(A)=λ Qf (A)Q^t, A∈Hn+(R) , where Q, λ and f are as above. From Proposition 4 and Lemma 16, it follows that Φ is a bijective bi-preserver of the star partial order.

Conversely, let Φ:Hn+(R)→Hn+(R) be a surjective bi-preserver of the star partial order. We will split the proof into several steps. The first four steps are standard and we include them for the sake of completeness.

Step 1. Φ is bijective. Let Φ(A)=Φ(B) for A,B∈Hn+(R) . Thus Φ(A) ⩽ ^*Φ(B) and Φ(B) ⩽ ^*Φ(A) and since Φ is a bi-preserver of the star partial order, we have A ⩽ ^*B and B ⩽ ^*A. So A=B, i.e., Φ is injective and hence bijective.

Step 2. Φ (0)=0. By (\refdef_star), 0 ⩽ ^*A for every A∈Hn+(R) . So on the one hand 0 ⩽ ^*Φ (0) and on the other hand, since Φ ⁻¹ has the same properties as Φ, 0 ⩽ ^*Φ ⁻¹(0). Thus Φ (0) ⩽ ^*0 and therefore Φ (0)=0.

Step 3. Φ preserves rank.Let A∈Hn+(R) with rank(A)=k . There exists an orthogonal matrix U ∈ M_n(ℝ) such that

where D=diag(di) with d_i⩾0 for every i=1, 2, ⋅, n . Without loss of generality we may assume that d1≥d2≥…≥dk>0=dk+1=dk+2=…=dn and write D=d₁E_1,1+d₂E_2,2+⋅+d_kE_{k, k}. Clearly,

By Proposition 4 and since congruence preserves rank, we have

Since Φ (0)=0 and since Φ preserves the order ⩽ ^* and is injective, we obtain

By Proposition 5, every successor in the above chain of matrices is of rank strictly greater than its predecessor. Since rank of any matrix in Hn+(R) can not be greater than n, it follows that rankΦ(A)=k .

Step 4. We may without loss of generality assume that Φ(I)=I. By the previous step, Φ(I) is of rank n. It follows that there exist an orthogonal matrix U ∈ M_n(ℝ) and λi>0 , i=1, 2, ⋅, n, with Φ(I)=Udiag(λi)Ut . We may thus without loss of generality assume that Φ(I)=diag(λi) . Suppose there are λ_i, λ_j with λ_i≠λ_j, i, j ∈ {1, 2, ⋅, n} . By Corollary 12, there are exactly two rank-one matrices that are below λ_iE_{i, i}+λ_jE_{j, j} with respect to the star partial order, namely λiEi,i,λjEj,j∈Hn+(R) . Since Φ preserves the order in both directions, we have Φ ⁻¹(λ_iE_{i, i}+λ_jE_{j, j}) ⩽ ^*I. By Lemma 6 and the previous step, Φ ⁻¹(λ_iE_{i, i}+λ_jE_{j, j}) is a rank-two self-adjoint idempotent matrix. This is a contradiction, since on the one hand Φ ⁻¹(λ_iE_{i, i}) and Φ ⁻¹(λ_jE_{j, j}) are the only two rank-one matrices that are below Φ ⁻¹(λ_iE_{i, i}+λ_jE_{j, j}) but on the other hand, since Φ ⁻¹(λ_iE_{i, i}+λ_jE_{j, j}) is a self-adjoint idempotent, for every rank-one self-adjoint idempotent P ∈ M_n(ℝ) with ImP⊆ImΦ−1(λiEi,i+λjEj,j) , we have P ⩽ ^*Φ ⁻¹(λ_iE_{i, i}+λ_jE_{j, j}), and there are infinitely many such matrices P (see (3)). It follows that Φ(I)=λ I for some λ>0 . Let now Ψ:Hn+(R)→Hn+(R) be the map defined with

Then Ψ is a surjective map that by Proposition 4 preserves the order ⩽ ^* in both directions. Moreover, Ψ(I)=I.

We will thus from now on assume that

Step 5. There exists a bijective functionf:ℝ⁺ → ℝ⁺ such that f (0)=0 and Φ(λ x x^*)=f(λ)Φ(x x^*) for every x ∈ ℝⁿ with ∣ x∣=1 and every λ>0 .

To see this, let x ∈ ℝⁿ with ∣ x∣=1 and let λ>0 . By unitary similarity we may without loss of generality assume that x=e₁, i.e., x x^*=E_1,1. The matrices λ E_1,1, (λ +2) E_2,2, (λ +3) E_3,3, …, (λ +n) E_{n, n} are pairwise orthogonal and have pairwise different spectra. Similarly, the matrices E_1,1, (λ +2) E_2,2, (λ +3) E_3,3, …, (λ +n) E_{n, n} are pairwise orthogonal and have pairwise different spectra. By Lemma 9 and since Φ preserves the star partial order, it follows that matrices

are pairwise orthogonal and similarly

are also pairwise orthogonal. Since Φ(λ E_1,1) and Φ(E_1,1) are pairwise orthogonal to the same set of n − 1 pairwise orthogonal rank-one matrices, they must be linearly dependent, i.e., Φ(λE1,1)=λe1Φ(E1,1) , λe1>0 . So, Φ(λ x x^*)=λ_xΦ(x x^*), λx>0 , for every x ∈ ℝⁿ with ∣ x∣=1 and λ>0 .

Suppose now x, y ∈ ℝⁿ with ∣ x∣=∣ y∣=1 are linearly independent and let λ>0 . We have Φ(λ x x^*)=λ_xΦ(x x^*) and Φ(λ y y^*)=λ_yΦ(y y^*). To show that λ_x=λ_y, let P ∈ P_n(ℝ) be of rank-two with x,y∈ImP . By Corollary 12, λ x x^*, λ y y^* ⩽ ^*λ P. Note that Φ(λ P) is of rank-two and suppose Φ(λP)=μ1z1z1∗+μ2z2z2∗ is the eigenvalue decomposition of Φ(λ P) with μ₁≠μ₂. By Corollary 12, μ1z1z1∗ , μ2z2z2∗ are the only two rank-one matrices below Φ(λ P) with respect to the star partial order. This is a contradiction, since Φ preserves the order in both directions and there are infinitely many rank-one matrices below λ P. So, Φ(λ P)=μ Q, where μ>0 and Q ∈ P_n(ℝ) is of rank-two. It follows that Φ(λ x x^*), Φ(λ y y^*) ⩽ ^*μ Q and thus by Corollary 12, λ_x=μ=λ_y.

Recall that Φ (0)=0. We may conclude that there exists a function \f:ℝ⁺ → ℝ⁺ such that f (0)=0 and Φ(λ x x^*)=f(λ)Φ(x x^*) for every x ∈ ℝⁿ with ∣ x∣=1 and every λ>0 . Function f is bijective since Φ is bijective.

Step 6. Letx x^*, y y^* ∈ P_n(ℝ) be pairwise orthogonal. ThenΦ(x x^*), Φ(y y^*) are pairwise orthogonal.

By unitary similarity, we may without loss of generality assume that x x^*=E_1,1 and y y^*=E_2,2. Lemma 9 implies Φ(E_1,1) and Φ (2E_2,2) are pairwise orthogonal, and Step 5 yields Φ (2E_2,2)=f (2)Φ(E₂₂). Thus, Φ(E_2,2) is orthogonal to Φ(E_1,1).

Step 7. We may without loss of generality assume thatΦ(P)=P for everyP ∈ P_n(ℝ). Let P ∈ P_n(ℝ) . It follows that P ⩽ ^*I by Lemma 6. Therefore Φ(P) ⩽ ^*Φ(I)=I. Again, by Lemma 6, Φ(P) ∈ P_n(ℝ). Since Φ ⁻¹ has the same properties as Φ, we may conclude that

i.e., Φ preserves the set of all orthogonal projector matrices. Recall that we may identify subspaces of ℝⁿ with elements of P_n(ℝ). Let C(Rn) be the lattice of all subspaces of ℝⁿ. It follows that the map Φ induces a lattice automorphism, i.e., a bijective map τ:C(Rn)→C(Rn) such that

for all M,N∈C(Rn) . It was proved in [19: page 246] (see also [7: pages 820 and 823]) that for n=3 every such a map is induced by an invertible linear operator, i.e., there exists an invertible linear operator T:ℝⁿ → ℝⁿ such that τ(M)=T(M) for every M∈C(Rn) . For the map Φ, we thus have

for every P=PImP∈Pn(R) . Let x x^*, y y^* ∈ P_n(ℝ) be pairwise orthogonal. By Step 6, Φ(x x^*), Φ(y y^*) are pairwise orthogonal and hence by (4),

and therefore 〈T^tTx, y〉=0. This equation holds for every y ∈ ℝⁿ with ∣ y∣=1 and 〈x, y〉=0. Since TtTx,y=xyTtTxx,yy , we may conclude that for any fixed x ∈ ℝⁿ, we have 〈T^tTx, y〉=0 for every y ∈ ℝⁿ with 〈x, y〉=0. So T^tTx is a scalar multiple of x, i.e., T^tT and I are locally linearly dependent. It is known that for linear operators of rank at least 2 local linear dependence implies (global) linear dependence (see, e.g., [23: page 1869]). Note that TtT∈Hn+(R) . Therefore,

for some α>0 . Let now Q=1αT . It follows that QtQ=1αTtT=I . So, Q is a linear isometry and since it is also invertible (and thus surjective), it is also a coisometry (QQ^t=I). For any P ∈ P_n(ℝ), we thus have Φ(P)=PQ(ImP) , where Q is an orthogonal operator,\i.e., it may be represented with an orthogonal matrix Q, where QQ^t=Q^tQ=I. Therefore for every P ∈ P_n(ℝ)

Since clearly QPQ^t ∈ P_n(ℝ), we conclude that

for every P ∈ P_n(ℝ).

Without loss of generality, we assume from now on that

for every P ∈ P_n(ℝ).

Step 8. Conclusion of the proof.Let A∈Hn+(R) . Then we may write

where λ_i⩾0, i=1, 2, ⋅, n, are the eigenvalues of A and {x₁, x₂, ⋅, x_n} is an orthonormal basis of ℝⁿ. Then λixixi∗ ⩽ ^*A for every i=1, 2, ⋅, n and therefore

where f:ℝ⁺ → ℝ⁺ is the function obtained in Step 5. By Corollary 12 and since Φ preserves rank, it follows that

Recall Definition 1 to conclude that Φ(A)=f(A) for every A∈Hn+(R) . Taking into account the assumptions from Steps 4 and 7, we may conclude that if Φ:Hn+(R)→Hn+(R) is a surjective bi-preserver of the star partial order, there exist an orthogonal matrix Q ∈ M_n(ℝ), a positive λ, and a bijective function f:ℝ⁺↦ℝ⁺ with f (0)=0 such that

for every A∈Hn+(R) . □