ON POSITIVE SEMIDEFINITE PRESERVING STEIN TRANSFORMATION

Abstract

In the setting of semidefinite linear complementarity problems on $S^n$, we focus on the Stein Transformation $S_A(X):=X-AXA^T$ for $A{\in}R^{n{\times}n}$ that is positive semidefinite preserving (i.e., $S_A(S^n_+){\subseteq}S^n_+$) and show that such transformation is strictly monotone if and only if it is nondegenerate. We also show that a positive semidefinite preserving $S_A$ has the Ultra-GUS property if and only if $1{\not\in}{\sigma}(A){\sigma}(A)$.

1. Introduction

In this paper, we focus on the so-called semidefinite linear complementarity problem (SDLCP) introduced by Gowda and Song [4]: Let Sn denote the space of all real symmetric n × n matrices, and be the set of symmetric positive semidefinite matrices in Sn. With the inner product defined by ⟨Z, W⟩ := tr(Z W), ∀Z, W ∈ Sn, the space Sn becomes a Hilbert space. Clearly, is a closed convex cone in Sn. Given a linear transformation L : Sn → Sn and a matrix Q ∈ Sn, the semidefinite linear complementarity problem, denoted by SDLCP(L, Q), is the problem of finding a matrix X ∈ Sn such that

Specializing L to the Stein transformation SA (X) := X −AXAT, various authors tried to characterize GUS-property in terms of the matrix A ∈ Rn×n. The most recent result is by Balaji [1] when A is a 2 × 2 matrix. When we translate the statements of Theorem 6 of [1] to SA : S2 → S2, then SA is GUS if and only if I ± A is positive semidefinite. However, Tao [14] showed that this is not true in general (see Example 4.1 of [14]). The results of this paper states that when SA is positive semidefinite preserving, then SA is Ultra-GUS if and only if I ± A is positive definite.

We list out needed definitions below.

Next, we list out some well known matrix theoretic properties that are needed in the paper [10].

Finally, we state the known results (interpreting for the case of L = SA) that are necessary for the paper. In the following and throughout the paper, σ(A) denotes the spectrum of A, the set of all eigenvalues of an n × n matrix A; and ρ(A) denotes the spectral radius of A, the maximum distance from the origin to an eigenvalue of A in the complex plane.

 

2. Characterization of Ultra-GUS property of a positive semidefinite preserving SA

We start with a Lemma.

Lemma 2.1. For A ∈ Rn×n, suppose SA is nondegenerate and copositive on . Then SA is Cone-Gus.

Proof. Let X be a solution to SDLCP(SA, −Q) where Q ≼ 0. It suffices to show X = 0 to prove SA is Cone-Gus. Since

X(SA(X) − Q) = 0, we have XSA (X) = XQ, and SA (X)X = QX. Since SA is copositive on , ⟨X, SA (X)⟩ ≥ 0, but ⟨X, Q⟩ ≤ 0, and hence ⟨X, Q⟩ = 0 = ⟨X, −Q⟩. Since both X, −Q ≽ 0, XQ = 0 = QX. Then X = 0 follows from the nondegeneracy of SA. □

Note that if SA is positive semidefinite preserving, then SA is Lyapunovlike. (This is because SA ∈ Z and ⟨X, SA (X)⟩ ≥ 0 for all X ≽ 0.) Then by Theorem 3.5 [13], SA is Cone-Gus if and only if SA is GUS. Since every positive semidefinite preserving SA is copositive on , we get the following

Theorem 2.2. If 1 ∉ σ(A)σ(A) and then SA is GUS.

We now show that if SA is nondegenerate and positive semidefinite preserving, then SA is not only GUS, but also Ultra-GUS.

Lemma 2.3. For A ∈ Rn×n, suppose SA is nondegenerate and positive semidefinite preserving. Then SA is Ultra-GUS.

Proof. First we show that SA ∈ P′2. Assume the contrary and let 0 ≠ X ≽ 0 be such that XSA (X)X ≼ 0. But SA is positive semidefinite preserving, so tr (XSA (X)X) = 0. Let X = UDUT where with D+ ≻ 0 diagonal and U orthogonal. Then

Let Note that M ≽ 0. Then the matrix product Thus,

0 = tr(XSA (X)X) = tr(D+ MD+) = tr(M(D+)2)

with D+ nonsingular, so M = 0, which implies N = 0. Therefore, So D and UTSA (X)U commute with the product 0 where both are in . Hence XSA (X) = 0 = SA (X)X. Then X = 0 by nondegeneracy of SA.

As we noted earlier (right after Lemma 1), SA is Lyapunov-like. So by Theorem 6.1 [14], P'2 = P2, that is, Ultra Cone-Gus = Ultra GUS. This completes the proof. □

Now we characterize Ultra-GUS property of a positive semidefinite preserving SA.

Lemma 2.4. For A ∈ Rn×n, let SA be positive semidefinite preserving. Then the following are equivalent.

Proof. The statement (a) ⇒ (b) is exactly Theorem 3.

Assume (b). Since P2 ⇒ nondegeneracy of SA, we get (a).

Finally, (b) and (c) are equivalent because SA is Lyapunov-like, and so by Theorem 6.1 of [14], SA ∈ P2 if and only if SA is strictly monotone. This completes the proof. □

Remark 2.1. In our previous paper [12], the strict monotonicity of SA was characterized in terms of its real numerical radius (Theorem 2.1 of [12]). Hence if SA is positive semidefinite preserving, then 1 ∉ σ(A)σ(A) if and only if νr (UAUT ◦ UAUT) < 1 for all U orthogonal. We now show that under the assumption of positive semidefinite preservedness, both of these are equivalent to the (easier-to-check) statement, I ± A positive definite.

Theorem 2.5. If SA is positive semidefinite preserving, then the following are all true or all false:

Proof. Assume (a). Note that I ± UTAU = UT (I ± A)U is also positive definite for all orthogonal matrices U, and hence the (k, k)-entry of UTAU ([UTAU]kk) is less than 1 in absolute value. We will show first that SA is strictly copositive on . Suppose there exists 0 ≠ X ≽ 0 with ⟨X, SA (X)⟩ = 0. Let X = UDUT = U (d1E11 + · · · + dnEnn)UT, where di ≥ 0 for all i and dk > 0 for some k. The matrix Eii is a diagonal matrix with all entries being 0 except the unit (i, i)-entry. Then

Since SA is positive semidefinite preserving, so is SUTAU, then

didj ⟨SUTAU (Eii), Ejj ⟩ ≥ 0 for each i and j. In particular,

but the last term is positive because ⟨SUTAU (Ekk), Ekk⟩ = 1 − ([UTAU]kk)2 > 0. Then ⟨X, SA (X)⟩ > 0 which is a contradiction. Hence SA is strictly copositive on . Then by Theorem 3.2 of [14], SA ∈ P'2. Since P'2 = P2 for this SA (see the proof of Theorem 3) and P2 ⇒ P, we get (b).

Since P ⇒ nondegenerate, we have (b) ⇒ (c).

The statement (c) ⇔ (d) is done in Theorem 4.

Finally, assume (d). Then ⟨X, SA (X)⟩ > 0 for all 0 ≠ X ∈ Sn. So, without loss of generality, 0 < ⟨uuT, SA (uuT)⟩ for all 0 ≠ u ∈ Rn with ∥u∥ = 1. Then, ⟨uuT, SA (uuT)⟩ = 1 − (⟨u, Au⟩)2 > 0. So, I ± A is positive definite and the proof is complete. □

Remark 2.2. Theorem 6 offers a way of checking when SA is not positive semidefinite perserving. For example,

satisfies (b), but not (a) of Theorem 6, so SA is not positive semidefinite preserving.

 

3. Conclusion

In an attempt to find a characterization of GUS-property of the Stein transformation, Balaji showed that for SA : S2 → S2, SA is GUS if and only if I ± A is positive semidefinite (Theorem 6 [1]). Nevertheless, this does not generalize to Sn as Tao showed in his Example 4.1 [13]. In this paper, we showed SA : Sn → Sn that is positive semidefinite preserving is Ultra-GUS if and only if I ±A is positive definite. Still much to be done to characterize the GUS-property of a general Stein transformation and that is the author’s future work.