SCHUR IDEALS AND HOMOMORPHISMS OF THE SEMIDEFINITE CONE

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1 SCHUR IDEALS AND HOMOMORPHISMS OF THE SEMIDEFINITE CONE BABHRU JOSHI AND M. SEETHARAMA GOWDA Abstract. We consider the semidefinite cone K n consisting of all n n real symmetric positive semidefinite matrices. A set I in K n is said to be a Schur ideal if it is closed under addition, multiplication by nonnegative scalars, and Schur multiplication by any element of K n. A Schur homomorphism of K n is a mapping of K n to itself that preserves addition, (nonnegative) scalar multiplication and Schur products. This paper is concerned with Schur ideals and homomorphisms of K n. We show that in the topology induced by the trace inner product, Schur ideals in K n need not be closed, all finitely generated Schur ideals are closed, and in K 2, a Schur ideal is closed if and only if it is a principal ideal. We also characterize Schur homomorphisms of K n and, in particular, show that any Schur automorphism of K n is of the form Φ(X) = P XP T for some permutation matrix P. Key words. Semidefinite cone, Schur ideals and homomorphisms AMS subject classifications. 5A86, 5B48, 5B57. Introduction. Consider the Euclidean space R n whose elements are regarded as column vectors. Let S n denote the set of all real n n symmetric matrices. It is known that S n is a finite dimensional real Hilbert space under the trace inner product defined by X, Y = trace(xy ) = i,j x ij y ij, where X = x ij and Y = y ij. The space S n also carries the Schur product (sometimes called Hadamard product) defined by X Y := x ij y ij. With respect to the usual addition, (real) scalar multiplication, and the Schur product, S n becomes a commutative algebra. In this algebra, a nonempty set is an ideal if it is closed under addition, scalar multiplication, and (Schur) multiplication by any element of S n. Also, a (Schur) homomorphism of S n is a linear transformation on S n that preserves Schur products. We note that every ideal in this algebra, being a finite December 0, 203, revised September 29, 204 Department of Computational and Applied Mathematics, Rice University, Houston, TX 77005, USA; babhru.joshi@rice.edu Department of Mathematics and Statistics, University of Maryland Baltimore County, Baltimore, Maryland 2250, USA; gowda@umbc.edu

2 2 B. Joshi and M.S. Gowda dimensional linear subspace of S n, is topologically closed (in the topology induced by the trace inner product) and finitely generated. Also, it is easy to describe ideals and homomorphisms of this algebra, see Sections 3 and 6. The aim of this paper is to study Schur ideals and homomorphisms of the semidefinite cone. A matrix A S n is said to be positive semidefinite if x T Ax 0 for all x R n. The set of all positive semidefinite matrices in S n is the semidefinite cone and is denoted by K n. It is known that K n is a closed convex cone in S n ; this means that K n is topologically closed and X, Y K n, 0 λ R X + Y K n, λx K n. A famous theorem of Schur (see 5, Theorem 7.5.3) says that X, Y K n X Y K n. So K n is closed under addition, multiplication by nonnegative scalars, and the Schur product. Thus, following 2, we say that K n is a semi-algebra (under these operations). We call nonempty subsets of K n that share these properties sub-semi-algebras of K n, and note the following examples: Example. Let DN N n denote the set of all n n doubly nonnegative matrices these are matrices in S n that are both nonnegative and positive semidefinite. This set is a sub-semi-algebra of K n. Example 2. Let CP n denote the set of all completely positive matrices in S n. These are matrices of the form N uu T, where u R n + (the nonnegative orthant of R n ) and N is a natural number. Alternatively, these are matrices of the form BB T, where B is a nonnegative (rectangular) matrix. Since uu T vv T = (u v) (u v) T, where u v is the componentwise product of two vectors u and v, it follows that the Schur product of two completely positive matrices is once again a completely positive matrix. Thus, CP n is a sub-semi-algebra of K n. These two cones, along with K n, have become very important in optimization, particularly in conic linear programming (where one minimizes a linear function subject to linear constraints over a cone); see and 4. Thus, the problem of finding/characterizing such sub-semi-algebras of K n and studying their properties becomes interesting and useful. Motivated by this, we consider special sub-semi-algebras

3 Schur ideals and homomorphisms 3 of K n called Schur ideals. These are nonempty subsets of K n that are closed under addition, multiplication by nonnegative scalars, and Schur multiplication by any element of K n. A nonempty set I K n is a Schur ideal in K n if X, Y I, Z K n, 0 λ R X + Y, λx, X Z I. We also consider related Schur homomorphisms of K n, which are mappings from K n to K n that preserve addition, (nonnegative) scalar multiplication, and Schur products. Now, motivated by the problem of describing Schur ideals and homomorphisms of the semidefinite cone, we raise the following questions (using the term closed to mean topologically closed ): (a) Is every Schur ideal in K n closed? How about finitely generated ones? (b) Is every closed Schur ideal of K n finitely generated? (c) What are the Schur homomorphisms/automorphisms of K n? We show that: () Schur ideals in K n need not be closed, (2) Finitely generated Schur ideals are always closed, (3) In K 2, a Schur ideal is closed if and only if it is a principal ideal, and (4) Any Schur homomorphism of K n is, up to permutation, conjugation by a 0/-matrix with each row containing at most one followed by pinching. In particular, a Schur automorphism of K n (this is a bijection of K n that preserves the three operations on K n ) is of the form where P is a permutation matrix. Φ(X) = P XP T, Section 2 covers some background material. Section 3 deals with two elementary results characterizing Schur ideals in S n and K n. In Section 4, we give an example of a non-closed Schur ideal in any K n, n 2. In Section 5, we show that all finitely generated Schur ideals in K n are closed. Finally, in Section 6, we describe Schur homomorphisms/automorphisms on K n. 2. Preliminaries. In the Euclidean space R n, we let e, e 2,..., e n denote the standard unit vectors (so that e i has one in the i th slot and zeros elsewhere). We let e = e + e e n and E := ee T. We also let E ij be a symmetric matrix with ones in the (i, j) and (j, i) slots and zeros elsewhere; in particular, E ii has a in the (i, i) slot and zeros elsewhere. A matrix is a 0/-matrix if each entry is either 0 or. We let S n denote the space of all n n real symmetric matrices. S n becomes a Hilbert space under the trace product X, Y = trace(xy ) = i,j x ijy ij, where

4 4 B. Joshi and M.S. Gowda X = x ij and Y = y ij. The corresponding norm (called the Frobenius norm) is given by X = ij x2 ij. Thus, convergence in this space is entrywise. From now on, we say that a set in S n is closed if it is topologically closed under the Frobenius norm. We note the following property of the inner product: for all X, Y and Z in S n. We let X Y, Z = X, Y Z K n := {A S n : x T Ax 0 x R n }. It is known that K n is a closed convex cone in S n. In particular, as convergence in S n is entrywise, every bounded sequence in K n has a subsequence that converges to an element of K n. In K n, we have the Löwner ordering: It is known (see 5, Corollary 7.5.4) that Y X or X Y when Y X K n. (2.) U, V 0 U, V 0. Thus, (2.2) 0 X Y X, X X, Y Y, Y X Y. It is known that a real symmetric matrix is positive semidefinite if and only if all its principal minors are nonnegative. In particular, for X = x ij K n, (2.3) x ii 0 and x ii x jj x 2 ij, for all i and j. Thus, if a diagonal entry of X (belonging to K n ) is zero, then all elements in the row/column containing this entry are zero. Also, if some off-diagonal entry is nonzero, then the corresponding diagonal entry is nonzero. Let diag(k n ) denote the set of all diagonal matrices in K n. The interior of K n is denoted by K n. Note that this interior consists of all positive definite matrices of S n these are matrices A in S n satisfying x T Ax > 0 for all 0 x R n. The pinching operation/transformation on S n is defined as follows: By choosing a partition of := {, 2,..., n}, we write any matrix X S n in the block form X = X αβ. Then the pinching operation/transformation (corresponding to this partition)

5 Schur ideals and homomorphisms 5 takes X to Y = Y αβ, where Y αβ = X αβ when α = β and 0 otherwise. (This means that Y is obtained from X by setting all the off-diagonal blocks to zero.). We write Y = X pinch. For any finite set {C, C 2,..., C N } in K n, we let { } I {C,C 2,...,C N } := Ci X i : X i K n, i =, 2,..., N denote the Schur ideal generated by {C, C 2,..., C N }. We say that a Schur ideal of K n is a principal ideal if it is generated by a single matrix in K n. 3. Two elementary results on Schur ideals. We start with a description of Schur ideals in S n. Let := {, 2, 3..., n}. We say that a subset I of is symmetric if (i, j) I (j, i) I. Proposition 3.. Given any symmetric subset I of, the set I = {X = x ij S n : x kl = 0 for all (k, l) I} is a Schur ideal of S n. Conversely, every Schur ideal of S n arises this way. Proof. The first part of the proposition is easy to verify. We prove the converse part. Let I be any Schur ideal of S n. If I = {0}, we take I =. If I = S n, we take I =. Now assume that {0} I S n. Define I := {(k, l) : x kl = 0 for all X = x ij I} and let J be the complement of I in. Let Ĩ = {X = x ij S n : x kl = 0 (k, l) I}. Then, I Ĩ. Now for any (k, l) J, there is a Y I with y kl 0. For this (k, l), E kl = y kl Y E kl I. Therefore, for any X Ĩ, X = x ij = i j n x ije ij = k l n,(k,l) J x kle kl J R E ij I. It follows that I = Ĩ. The following proposition gives a necessary and sufficient condition for a set to be a Schur ideal in K n. Proposition 3.2. Let I be a set in K n containing zero. Then I is a Schur ideal in K n if and only if (i) I is a convex cone, and (ii) DXD I for every diagonal matrix D S n and X I.

6 6 B. Joshi and M.S. Gowda Proof. Suppose I is a Schur ideal in K n. Then for any λ 0 and X I, we have λx I. From this, it follows that I is a convex cone. Now take any diagonal matrix D = diag(r, r 2,..., r n ) in S n. Then, for r = r, r 2,..., r n T, rr T K n and hence DXD = X rr T I. This proves (ii). The converse is proved by using the fact that any matrix in K n is a finite sum of matrices of the form rr T. 4. A non-closed Schur ideal of K n. In this section, we show that in any K n, n 2, there is a Schur ideal that is not closed. First, we construct such an ideal in the cone K 2 of all 2 2 real symmetric positive semidefinite matrices. Example 3. The set E = diag(k 2 ) K 2 is a non-closed ideal in K 2. To see this, let λ 0, A diag(k 2 ), B K2 and X K 2. Then, A + A diag(k 2 ), K 2 + B K2, λ(a + B) E, and A X diag(k 2 ). If X is diagonal, then B X diag(k 2 ). Suppose X is not diagonal so that x y X =, y z with x > 0, z > 0 and xz y 2 0. Now write a b B = b c, where a > 0, c > 0 and ac b 2 > 0; we see that ax > 0, cz > 0 and (ax)(cz) > (b 2 )(y 2 ). This means that B X K2. Hence in all cases, (A + B) X E. Thus, E is an ideal of K 2. To see that E is not closed, we produce a sequence in E whose limit is not in E: k k. Now, in any K n, n 3, we define the set of block matrices { } X 0 F = : X E. This is a Schur ideal in K n that is not closed. 5. Finitely generated ideals in K n. Example 3 raises the question of which Schur ideals are closed in K n. The following result partially answers this question.

7 Schur ideals and homomorphisms 7 Theorem 5.. Every finitely generated Schur ideal in K n is closed. The proof is based on the following two lemmas. Lemma 5.2. Let A K n with diag(a) > 0. If X k is a sequence in K n with A X k convergent, then X k is a bounded sequence. Proof. Let X k =, A = a ij and A X k Y. Then, by componentwise x (k) ij convergence, for all i, i n, a ii x (k) ii positive semidefinite and so by (2.3), y ii and so x (k) ii x (k) ii x(k) jj (x (k) ij )2 yii a ii. Now, each X k is for all i, j =, 2, 3,..., n. As x (k) ii and x (k) jj are bounded sequences, it follows that x (k) ij is also bounded for all i, j. Hence X k is a bounded sequence. Lemma 5.3. Let X k be a sequence in K n and A K n with A X k Z K n. Then there is an X K n such that A X = Z. Proof. If A = 0, we can take X = 0. So, assume A 0 and without loss of A 0 generality write A =, where A is positive semidefinite and diag(a ) > 0. If we partition each X k and Z conformally to A, we get A 0 X (k) X (k) 2 (X (k) 2 )T X (k) 3 Z Z 2 Z2 T Z 3. Thus, A X (k) Z, Z 2 = 0, and Z 3 = 0. Now, from Lemma 5.2, X (k) is a bounded sequence; we may assume that it converges to X. Then Hence Since X 0 A X (k) 0 A X 0 X 0 A K n, the result follows. = Z. = Z 0 Proof of Theorem 5.. Let E be an ideal in K n generated by a finite number of matrices. For simplicity, we assume that E is generated by two, say, A and B, so that any element of E is of the form A X + B Y for some X, Y K n..

8 8 B. Joshi and M.S. Gowda To show that E is closed, let X k and Y k be sequence of matrices in K n such that A X k + B Y k Z. We show that Z = A X + B Y for some X, Y K n. Now, with respect to the Löwner ordering, 0 A X k A X k + B Y k. From (2.2), we get A X k A X k + B Y k. Thus, the sequence A X k is bounded. We may assume (by going through a subsequence, if necessary) that A X k U. By Lemma 5.3, we may write U = A X for some X K n. Similarly, we may assume that B Y k V, where V = B Y for some Y K n. It follows that Z = A X + B Y with X, Y K n. Thus, E is closed. Remarks. By modifying the above proof, one can show the following: If I and J are two closed ideals in K n, then so is I + J. In particular, if I is a closed ideal in K n and I := {Y K n : Y, X = 0 X I}, then I + I is a closed ideal in K n. We note, however, that I may be just {0}, as in the case of the ideal consisting of all diagonal matrices in K n. Remarks. Every finitely generated ideal in K n is closed, but we do not know if the converse holds. In the following theorem, we show that the converse does hold for K 2. Theorem 5.4. Every closed ideal in K 2 is a principal ideal. Conversely, every principal ideal in K 2 is closed. We begin with a lemma. Lemma 5.5. Let E be a closed ideal in K 2 that contains a matrix with an offdiagonal entry nonzero. Let (5.) Then, u u { β α β u := sup : αγ β γ E. Moreover, a b E b c } E and α, β, γ > 0. a b u c b u K 2. α β Proof. Let E with β 0; we may, if necessary, consider the Schur β γ product of this matrix with the positive semidefinite matrix and assume that β > 0. Then α, β, and γ are positive. Since αγ β 2, it follows that u in (5.)

9 Schur ideals and homomorphisms 9 αk β is well defined. Now consider a sequence k E such that α k, β k, γ k > 0 β k γ k and β k αk γ k u as k. Since α k αk γ k αk γ k γ k is positive semidefinite (as it is symmetric and all its principal minors are nonnegative) and E is an ideal, it follows that β αk β k α k αk γ k αk k γ = k E. β k γ k αk γ k γ k Since E is closed, letting k, we have u u E. β αk k γ k Now for the second part of the lemma. The implication is obvious if b = 0. If a b b 0, then a, c > 0 and E and so (by definition of u), u b b c ac. This a b shows that the symmetric matrix u b belongs to K 2 as its principal minors are u c nonnegative. Proof of Theorem 5.4. If E is a principal ideal, then it is closed by Theorem 5.. For the converse, let E be a closed ideal in K 2. Suppose first that E contains a matrix with a nonzero off-diagonal entry. Then from the preceding lemma there exists a u nonzero number u with U := E. Now, the principal ideal generated by U u is contained in E. The identity a b u a b = u b c u c shows that every element of E is in this principal ideal. Hence, E is a principal ideal generated by U. Now suppose that every off-diagonal entry of every matrix in E is zero. Then E consists of diagonal matrices. In this case, E is generated by one of the following: 0 0,,,. 0 0 Thus, we have shown that E is a principal ideal. b u 6. Schur homomorphisms of K n. In this section, we characterize Schur homomorphisms of K n. A map f : K n K n is a Schur homomorphism if for all X, Y K n

10 0 B. Joshi and M.S. Gowda and λ 0 in R, we have f(x + Y ) = f(x) + f(y ), f(λ X) = λf(x), and f(x Y ) = f(x) f(y ). These are special cases of Schur multiplicative maps (which preserve the Schur products) studied in 3. To motivate our results, we consider an n n 0/-matrix Q with each row containing at most one. Then, the mapping X QXQ T is a Schur homomorphism of K n. The process of pinching (corresponding to a partition of ) followed by a permutation continues to retain the Schur homomorphism property, that is, the mapping X P (QXQ T ) pinch P T is a Schur homomorphism of K n, where P denotes a permutation matrix. We now show that every Schur homomorphism arises this way. Let f be a Schur homomorphism on K n. Then f can be extended to S n in the following way. As S n = K n K n, any element X of S n can be written as X = A B with A, B K n. Then we define (6.) F (X) := f(a) f(b). Note that F (X) is well defined: If X = A B = C D with A, B, C, D K n, then A + D = B + C and so f(a) + f(d) = f(b) + f(c) yielding F (X) = f(a) f(b) = f(c) f(d). Then, F : S n S n is a Schur homomorphism of S n, that is, F is linear and for all X, Y S n, F (X Y ) = F (X) F (Y ). We now characterize Schur homomorphisms of S n, such as F. First, a definition. We say that a mapping φ : is symmetric if φ(i, j) = φ(j, i) for all (i, j); for notational convenience, we sometimes write φ(ij) in place of φ(i, j). Corresponding to such a mapping, we define Φ : S n S n by (6.2) Φ(X) = Y where y ij = x φ(ij). Theorem 6.. Given any symmetric mapping φ on and a symmetric 0/-matrix Z, the mapping F : S n S n defined by F (X) = Z Φ(X) (X S n ) is a Schur homomorphism on S n. arises in this way. Conversely, every Schur homomorphism of S n Proof. The first part in the theorem is easily verified. We prove the converse. Consider a Schur homomorphism F on S n. In order to simplify the proof and make it more transparent, we think of matrices X and Y as vectors x and y in R d, where d = n2 +n 2. Then F (X) can be regarded as a linear transformation f(x) = Ax on R d

11 Schur ideals and homomorphisms (for some square matrix A) satisfying A(x y) = A(x) A(y). This equation splits into d independent equations of the form a T (x y) = (a T x)(a T y), where a is any row vector of A. From this we see that a is a 0/-vector with at most one. Thus, A is a 0/-matrix with at most one in each row. Reverting back to F, we get the stated result. Remarks. The preceding theorem is analogous to Theorem. in 3, proved for Schur multiplicative maps on R m n under a nonsingularity assumption, but without any linearity assumptions. Before we present our general result on the Schur homomorphisms of K n, we introduce some notation and prove a lemma. For m n, let α := {, 2,..., m}. Corresponding to a symmetric mapping ψ : α α, we define Ψ : S n S m by (6.3) Ψ(X) = Y, where y ij = x ψ(ij). Lemma 6.2. Consider ψ and Ψ defined in (6.3). Then Ψ(K n ) K m if and only if there exists an m n 0/-matrix Q such that each row of Q contains exactly one and Ψ(X) = QXQ T for all X K n. Proof. Since the if part is obvious, we prove the only if part. Suppose Ψ(K n ) K m. We prove the following statements: (a) If ψ(i, i) = (k, l), then k = l. (b) If i j, ψ(i, i) = (k, k) and ψ(j, j) = (l, l), then ψ(i, j) {(k, l), (l, k)}. (This implies that y ij = x ψ(ij) = x kl.) (a) If possible, let ψ(i, i) = (k, l) with k l. Then there exists a matrix X K n such that x kl =. But then, for Y = Ψ(X), y ii = x ψ(ii) = x kl =. This cannot happen as Y, being positive semidefinite, has all nonnegative diagonal entries. Hence k = l. (b) Let i j, ψ(i, i) = (k, k) and ψ(j, j) = (l, l). We consider two cases. Case : k = l. Without loss of generality, let k = l =. Suppose ψ(i, j) (, ). Consider u =, 2, 3,..., n T and X = uu T K n. Then every entry in X other than the (, ) is bigger than one. Moreover, y ii = y jj = x = and y ij = x ψ(ij) >. As Y = Ψ(X) K m, these values of Y violate (2.3). Hence, ψ(i, j) = (, ) = (k, l). Case 2: k l. Without loss of generality (using (a)), let ψ(i, i) = (, ) and ψ(j, j) = (2, 2). We need to show that ψ(i, j) {(, 2), (2, )}. Assuming this does not hold, we consider the following 2 2 principal submatrix of Y : yii y ij xψ(ii) x = ψ(ij) x x (6.4) = ψ(ij). y ij y jj x ψ(ij) x 22 x ψ(ij) x ψ(jj)

12 2 B. Joshi and M.S. Gowda Case 2.: Suppose ψ(i, j) = (, ). Then, taking u = 2,, 0,..., 0 T, X = uu T, we see that the principal submatrix (6.4) of Y violates (2.3). Hence this case cannot arise. Case 2.2: Suppose ψ(i, j) = (2, 2). Then, taking u =, 2, 0,..., 0, X = uu T, we see that the principal submatrix (6.4) of Y violates (2.3). Hence this case cannot arise. Case 2.3: Suppose ψ(i, j) {(, ), (2, 2), (, 2), (2, )}. Then, taking X = uu T, where u =,, 2, 2,..., 2 T, we see that even in this case, the principal submatrix (6.4) of Y violates (2.3). We conclude that ψ(i, j) {(, 2), (2, )}. Thus, we have (b). Having proved Items (a) and (b), we now define the m n 0/-matrix Q = q ij as follows: For any pair (i, k) with i {, 2,..., m} and k {, 2,..., n}, q ik = when ψ(i, i) = (k, k) and q ik = 0 otherwise. Then each row of Q contains exactly one and Ψ(X) = QXQ T for all X S n. We now come to the description of Schur homomorphisms of K n. Theorem 6.3. Let Q be an n n 0/ matrix with each row containing at most one and P be an n n permutation matrix. Given a (specific) pinching operation, f(x) := P (QXQ T ) pinch P T (X S n ) defines a Schur homomorphism of K n. Conversely, every Schur homomorphism of K n arises this way. Proof. We have shown that f(x) := P (QXQ T ) pinch P T (X S n ) defines a Schur homomorphism. We prove the converse. Suppose f is a nonzero Schur homomorphism of K n. We extend f to F on S n via (6.). By Theorem 6., there exists a nonzero 0/ symmetric matrix Z and a symmetric mapping φ such that F (X) = Z Φ(X) X S n, where Φ is defined by (6.2). Since Φ(E) = E (for any mapping φ), we see that Z = Z E = Z Φ(E) = F (E) = f(e) K n. Thus, the 0/ symmetric matrix Z is also positive semidefinite. Now by Proposition 2 in 6, we see that up to a permutation, Z is of the form Z 0 0 Z Z = (6.5) 0,. 0 Z k 0 0 where k n and each Z i is a matrix of ones. We assume without loss of generality that Z is of the form (6.5) (this induces our permutation P ) and F (X) = Z Φ(X) for

13 Schur ideals and homomorphisms 3 all X. For each i =, 2,..., k, let α i be a subset of {, 2,..., n} such that Z αiα i = Z i. We let m i := α i (the number of elements in α i ). These sets are disjoint. Now (F (X)) αiα i = Z i (Φ(X)) αiα i = (Φ(X)) αiα i. We know that for each X K n, F (X) = f(x) K n. Thus, for each i =, 2,..., k, (Φ(X)) αiα i K mi. Now we define ψ i : α i α i by ψ i (k, l) := φ(k, l). Then, for the corresponding mapping Ψ i we have Ψ i (K n ) K mi. By Lemma 6.2, for each i =, 2,..., k, there exists an m i n 0/-matrix Q i (with exactly one in each row) such that Ψ i (X) = Q i XQ T i. Now, let Q k+ be the zero matrix and define the n n 0/-matrix Q with blocks Q, Q 2,..., Q k+. A computation shows that QXQ T has diagonal blocks Q i XQ T i, i =, 2,..., k +. By setting the offdiagonal blocks to zero (pinching), We see that F (X) = (QXQ T ) pinch. By taking into consideration the permutation used to rearrange rows and columns of Z, we finally arrive at f(x) = P (QXQ T ) pinch P T. The following result characterizes Schur automorphisms of K n. Corollary 6.4. Let f : K n K n be a Schur homomorphism. Then the following are equivalent: (i) f is one-to-one (or onto) on K n. (ii) There exists a permutation matrix P such that f(x) = P XP T K n. (iii) f is a Schur automorphism of K n. for all X Proof. Let f : K n K n be a Schur homomorphism. Then, Theorem 6. ensures that its extension F to S n can be written in the form F (X) = Z Φ(X) for all X S n. (i) (ii): Suppose that f is one-to-one. Then, F is also one-to-one. (When f is onto, that is, f(k n ) = K n, its extension F satisfies F (K n ) = K n. Since F is linear and its range contains an open set, namely, the interior of K n, F is also onto and hence one-to-one.) We now show that Z = E (the matrix of all ones). Suppose that Z has a zero entry, say z ij = 0. Then, for any X, F (X), E ij = 0. Thus, X, F T (E ij ) = 0 for all X S n. This implies that F T (E ij ) = 0. As F is one-to-one (hence invertible), F T is also one-to-one. Thus, E ij = 0, which is a contradiction. Hence, Z has no zero entries. As Z is a 0/-matrix, we have Z = E. Now, this implies that F (X) = E Φ(X) = Φ(X) so that Φ(= F ) is a one-to-one Schur homomorphism on S n such that Φ(K n ) K n. By Lemma 6.2, there exists a (square) 0/-matrix Q with each row containing exactly one such that F (X) = QXQ T for all X S n. As F is one-to-one, it is necessarily invertible. This implies (for example, by looking at F (X) = I) that Q is invertible. This means that Q cannot have a pair of identical

14 4 B. Joshi and M.S. Gowda rows. Thus, Q has exactly one in each column/row. Hence, Q is a permutation matrix. The implications (ii) (iii) (i) are easy to verify or obvious. Concluding Remarks. Schur homomorphisms of K n are completely characterized, but the problem of describing all closed Schur ideals of K n remains open. In particular, it is not known if every closed Schur ideal of K n is finitely generated. Acknowledgments. We thank the referee and an Advisory Editor for their comments and suggestions which simplified the proofs (particularly in Section 6) and the exposition. REFERENCES M.F. Anjos and J.B. Lasserre (Eds). Handbook on Semidefinite, Conic and Polynomial Optimization. International Series in Operations Research & Management Science, Vol. 66, Springer, New York, F.F. Bonsall. Lectures on some fixed point theorems in Functional Analysis. TIFR Lecture Notes, S. Clark, C.-K. Li, and A. Rastogi. Schur multiplicative maps on matrices. Bulletin of the Australian Mathematical Society, 77 (2008) H. Dong and K. Anstreicher. Separating doubly nonnegative and completely positive matrices. Mathematical Programming, Series A, 37 (3-54) R.A. Horn and C.R. Johnson. Matrix Analysis. Cambridge University Press, Cambridge, A.N. Letchford and M.M. Sorensen. Binary positive semidefinite matrices and associated integer polytopes. Mathematical Programming, Series A, 3 (202)