GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION

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1 J. OPERATOR THEORY 00:0(0000, Copyright by THETA, 0000 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION CHI-KWONG LI and YIU-TUNG POON Communicated by Editor ABSTRACT. For a noisy quantum channel, a quantum error correcting code of dimension k exists if and only if the joint rank-k numerical range associated with the error operators of the channel is non-empty. In this paper, geometric properties of the joint rank k-numerical range are obtained and their implications to quantum computing are discussed. It is shown that for a given k if the dimension of the underlying Hilbert space of the quantum states is sufficiently large, then the joint rank k-numerical range of operators is always star-shaped and contains the convex hull of the rank ˆk-numerical range of the operators for sufficiently large ˆk. In case the operators are infinite dimensional, the joint rank -numerical range of the operators is a convex set closely related to the joint essential numerical ranges of the operators. KEYWORDS: Quantum error correction, joint higher rank numerical range, joint essential numerical range, self-adjoint operator, Hilbert space. MSC (2000: 47A12, 15A60, 15A90, 81P INTRODUCTION In quantum computing, information is stored in quantum bits, abbreviated as qubits. Mathematically, a qubit is represented by a 2 2 rank one Hermitian matrix Q = vv, where v C 2 is a unit vector. A state of N-qubits Q 1,..., Q N is represented by their tensor products in M n with n = 2 N. A quantum channel for states of N-qubits corresponds to a trace preserving completely positive linear map Φ : M n M n. By the structure theory of completely positive linear map [2], there are T 1,..., T r M n with r j=1 T j T j = I n such that (1.1 Φ(X = r T j XTj. j=1 In the context of quantum error correction, T 1,..., T r are known as the error operators.

2 102 CHI-KWONG LI AND YIU-TUNG POON Let V be a k-dimensional subspace of C n and P the orthogonal projection of C n onto V. Then V is a quantum error correcting code for the quantum channel Φ if there exists another trace preserving completely positive linear map Ψ : M n M n such that Ψ Φ(A = A for all A PM n P. By the results in [9], this happens if and only if there are scalars γ ij with 1 i, j r such that PT i T jp = γ ij P. Let P k be the set of rank k orthogonal projections in M n. Define the joint rank k-numerical range of an m-tuple of matrices A = (A 1,..., A m M m n by Λ k (A = {(a 1,..., a m C m : there is P P k such that PA j P = a j P for j = 1,..., m}. Then the quantum channel Φ defined in (1.1 has an error correcting code of k- dimension if and only if Λ k (T 1 T 1, T 1 T 2,..., T r T r =. Evidently, (a 1,..., a m Λ k (A if and only if there exists an n k matrix U such that U U = I k, and U A j U = a j I k for j = 1,..., m. Let x, y C n. Denote by Ax, y the vector ( A 1 x, y,..., A m x, y C m. Then a Λ k (A if and only if there exists an orthonormal set {x 1,..., x k } in C n such that Ax i, x j = δ ij a, where δ ij is the Kronecker delta. When k = 1, Λ 1 (A reduces to the (classical joint numerical range W(A = {(x A 1 x,..., x A m x : x C n, x x = 1} of A, which is quite well studied; see [11] and the references therein. It turns out that even for a single matrix A M n, the study of Λ k (A is highly non-trivial, and the results are useful in quantum computing, say, in constructing binary unitary channels; see [3, 4, 5, 6, 7, 13, 15, 18]. More generally, let B(H be the algebra of bounded linear operators acting on a Hilbert space H, which may be infinite dimensional. One can extend the definition of Λ k (A to A B(H. If H is infinite dimensional, one may allow k = by letting P k be the set of infinite rank orthogonal projections in B(H in the definition; see [14, 16]. There are a number of reasons to consider rank k-numerical range of infinite dimensional operators. First, many quantum mechanical phenomena are better described using infinite dimensional Hilbert spaces. Also, a practical quantum computer must be able to handle a large number of qubits so that the underlying Hilbert space must have a very large dimension. We will also consider the joint rank k-numerical range of an m-tuple A = (A 1,..., A m of infinite dimensional operators A 1,..., A m for positive integers k and k =. It is interesting to note that Λ (A has intimate connection with the joint essential numerical range of A defined as W e (A = {cl (W(A + F : F F(H m },

3 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 103 where F(H denotes the set of finite rank operators in B(H and cl (S denotes the closure of the set S. Clearly, the joint essential numerical range is useful for the study of the joint behaviors of operators under perturbations of finite rank (or compact operators. The purpose of this paper is to study the joint rank k-numerical range of A = (A 1,..., A m B(H m. Understanding the properties of Λ k (A is useful for constructing quantum error correcting codes and studying their properties such as their stability under perturbation. Our paper is organized as follows. In Section 2, we present some basic properties of Λ k (A. Section 3 concerns the geometric properties of Λ k (A. We show that if dim H is sufficiently large, then Λ k (A is always star-shaped and contains the convex hull of Λˆk(A for sufficiently large ˆk. In Section 4, we study the connection between W e (A, Λ k (A and its closure cl (Λ k (A. We show that Λ (A is always convex, and is a subset of the set of star centers of Λ k (A for each positive integer k. We also show that W e (A = k 1 cl (Λ k (A. Moreover, we obtain several equivalent formulations of Λ (A, including Λ (A = {Λ k (A + F : F F(H m }. The results extend those in [1, 12]. Let S(H be the real linear space of self-adjoint operators in B(H. Suppose A j = H 2j 1 + ih 2j with H 2j 1, H 2j S(H f or j = 1,..., m. Then Λ k (A C m can be identified with Λ k (H 1,..., H 2m R 2m. Thus, we will focus on the joint rank k-numerical ranges of self-adjoint operators in our discussion. 2. BASIC PROPERTIES OF Λ k (A PROPOSITION 2.1. Suppose A = (A 1,..., A m S(H m, and T = ( t ij is an m n real matrix. If B j = m i=1 t ija i for j = 1,..., n, then {at : a Λ k (A} Λ k (B. Equality holds if {A 1,..., A m } is linearly independent and span {A 1,..., A m } = span {B 1,..., B n }. Proof. The set inclusion follows readily from definitions. Evidently, the equality holds if n = m and T is invertible. Suppose {A 1,..., A m } is linearly independent and span {A 1,..., A m } = span {B 1,..., B n }.

4 104 CHI-KWONG LI AND YIU-TUNG POON First consider the special case when A i = B i for 1 i m. Then T = [I m T 1 ] for some m (n m matrix T 1. Let (b 1,..., b n Λ k (B. Then there exists a rank k orthogonal projection P such that PB i P = b i P for i = 1,..., n. Therefore, we have (b 1,..., b m Λ k (A and for 1 j n, b j P = PB j P = P ( m t ij A i P = P i=1 ( m t ij B i P = i=1 ( m t ij b i P i=1 implying that b j = ( m i=1 t ijb i. Therefore, (b1,..., b n = (b 1,..., b m T. For the general case, by applying a permutation, if necessary, we may assume that {B 1,..., B m } is a basis of span {B 1,..., B n }. Then there exists an m m invertible matrix S = ( s i j such that Aj = m i=1 s ijb i for j = 1,..., m. For 1 j m, we have B j = m t ij A i = i=1 ( m m t ij s ki B k = i=1 k=1 ( m m s ki t ij B k. k=1 i=1 Therefore, m i=1 s kit ij = δ k j and ST = [I m T 1 ] for some m (n m matrix T 1. Hence, we have Λ k (B = Λ k (B 1,..., B m [I T 1 ] = Λ k (B 1,..., B m ST = Λ k (A 1,..., A m T. In view of the above proposition, in the study of the geometric properties of Λ k (A, we may always assume that A 1,..., A m are linearly independent. PROPOSITION 2.2. Let A = (A 1,..., A m S(H m, and let k < dim H. (a For any real vector µ = (µ 1,..., µ m, Λ k (A 1 µ 1 I,..., A m µ m I = Λ k (A µ. (b If (a 1,..., a m Λ k (A then (a 1,..., a m 1 Λ k (A 1,..., A m 1. (c Λ k+1 (A Λ k (A. REMARK 2.3. By Proposition 2.2 (a, we can replace A j by A j µ j I for j = 1,..., m, without affecting the geometric properties of Λ k (A 1,..., A m. Suppose dim H = n < 2k 1 and A 1 = diag (1, 2,..., n. Then Λ k (A 1 =. By Proposition 2.2 (c, we see that Λ k (A = for any A 2,..., A m. Thus, Λ k (A can be empty if dim H is small. However, a result of Knill, Laflamme and Viola [10] shows that Λ k (A is non-empty if dim H is sufficiently large. By modifying the proof of Theorem 3 in [10], we can get a slightly better bound in the following proposition. The proof given here is essentially the same as that of Theorem 3 and 4 in [10], except for the choice of x 1. We include the details here for completeness. PROPOSITION 2.4. Let A S(H m. For m 1 and k > 1. If then Λ k (A =. dim H = n (k 1(m + 1 2,

5 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 105 Proof. We may assume that dim H = n = (k 1(m Otherwise, replace each A j by U A j U for some U such that U U = I n. Let q = (m + 1(k Choose an eigenvector x 1 of A 1 with x 1 = 1. Then choose a unit vector x 2 orthogonal to x 1, A 2 x 1,..., A m x 1. By the assumption on n, we can choose an orthonormal set {x 1, x 2,..., x q } of q vectors in C n such that for 1 < r q, x r is orthogonal to A j x i for all 1 i < r and 1 j m. Let X be the n q matrix with x i as the i-th column. Then X A j X is a diagonal matrix for 1 j m. By Tverberg s Theorem [17], we can partition the set {i : 1 i q} into k disjoint subset R j, 1 j k such that R = k j=1 conv { Ax i, x i : i R j } =. Suppose a R. Then there exist non-negative numbers t i j, 1 j k, i R j such that for all 1 j k, i Rj t i j = 1 and i Rj t i j Ax i, x i = a. Let y j = i Rj ti j x i for 1 j k. Then {y 1,..., y k } is orthonormal and Ay j, y j = a for all 1 j k. PROPOSITION 2.5. Suppose A S(H m and 1 r < k dim H. Let V r be the set of operator X : H 1 H such that X X = I H 1 for an r-dimensional subspace H 1 of H. Then (2.1 Λ k (A {Λ k r (X A 1 X,..., X A m X : X V r } and (2.2 conv Λ k (A conv ( {Λ k r (X A 1 X,..., X A m X : X V r }. Proof. Suppose (a 1,..., a m Λ k (A. Let H 2 be a k-dimensional subspace and V : H 2 H such that V V = I H2 and V A j V = a j I k for j = 1,..., m. Let X V r and X X = I H for an r-dimensional subspace H 1 of H. Then 1 ( ( H 0 = X V (H 2 X H1 has dimension at least s = k r. Let U : H 0 H be given by Ux = x for all x H 0. Then we have U U = I H0 and U ( X A j X U = a j I H0 for j = 1,..., m. Thus, (2.1 holds, and the inclusion (2.2 follows. Proposition 2.5 extends [14, Proposition 4.8] corresponding to the case when m = 2. In such a case, the set inclusion (2.1 becomes a set equality if dim H < or if (A 1, A 2 is a commuting pair, i.e., A 1 + ia 2 is normal; see [14, Corollary 4.9]. The following example shows that the set equality in (2.1 may not hold even in the finite dimensional case if m 3. ( ( ( i EXAMPLE 2.6. Let B 1 = B = B =. For i 0 k > 1, let A j = B j I k for j = 1, 2, 3. (a We have Λ k (A 1,..., A m = Λ 1 (B 1, B 2, B 3 = {a R 3 : a = 1}, which is not convex.

6 106 CHI-KWONG LI AND YIU-TUNG POON (b If r = k 1 and X V r and X X = I H for an r-dimensional subspace 1 H 1, then dim H1 = 2k r = k so that Λ k r (X A 1 X, X A 2 X, X A 3 X = Λ 1 (X A 1 X, X A 2 X, X A 3 X is convex [11]. Consequently, {Λ k r (X A 1 X, X A 2 X, X A 3 X : X V r } is convex and cannot be equal to Λ k (A 1, A 2, A 3. For m > 3, we can take A 1, A 2, A 3 as above and A j = 0 2k for 3 < j m. Then we have Λ k (A 1, A 2, A 3 = {Λ k r (X A 1 X,..., X A m X : X V r }. ( U1 To verify (a, suppose U = is such that U 2 U 1, U 2 M k, U U = U 1 U 1 + U 2 U 2 = I k and U A j U = a j I k. Then U 1 U 1 U 2 U 2 = a 1 I k. It follows that U 1 U 1 = (1 + a 1 I k and U 2 U 2 = (1 a 1 I k. Thus, U 1 U 1 = (1 + a 1I k and U 2 U 2 = (1 a 1I k. As a result, and U i U ju j U i = (1 + a 1 (1 a 1 I = U i U iu j U j for (i, j {(1, 2, (2, 1}, (a a2 2 + a2 3 I k = 3 j=1 (U A j U 2 = (U 1 U 1 U 2 U (U 1 U 2 + U 2 U (iu 1 U 2 iu 2 U 1 2 = (U 1 U 1 + U 2 U 2 2 = I k. Thus, Λ k (A 1, A 2, A 3 {(a 1, a 2, a 3 R 3 : a a2 2 + a2 3 = 1}. Conversely, suppose (a 1, a 2, a 3 R 3 such that a a2 2 + a2 3 = 1. Let (0, 1 i f a 1 = 1, (α, β = (1 + a 1, a 2 ia 3 otherwise. 2(1 + a1 ( αik Let U =. Direct computation shows that U βi A j U = a j I k for 1 j 3. k By a similar argument or putting k = 1, we see that Λ 1 (B 1, B 2, B 3 has the same form. It is natural to ask if the set equality in (2.2 can hold for for m > 2. Also,

7 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 107 conv ( {Λ k r (X A 1 X,..., X A m X : X V r } {conv Λ k r (X A 1 X,..., X A m X : X V r }. It is interesting to determine whether the two sets are equal. 3. GEOMETRIC PROPERTIES OF Λ k (A Let A = (A 1,..., A m S(H m. Much is known about Λ 1 (A; for example see [11] and its references. For instance, Λ 1 (A is always convex if m 3 unless (dim H, m = (2, 3; If (dim H, m = (2, 3 or n > 1, m 4, there are examples A S(H m such that Λ 1 (A is not convex. Furthermore, for any A 1, A 2, A 3 S(H such that span {I, A 1, A 2, A 3 } has dimension 4, there is always an A 4 S(H for which Λ 1 (A 1,..., A 4 is not convex. In the following, we show that Λ k (A is always star-shaped if dim H is sufficiently large. Moreover, it always contains the convex hull of Λˆk(A for ˆk = (m + 2k. If k = 1 and m > 2, we can lower the bound of the dimension of the Hilbert space to get the star-shapedness result. We begin with the following. THEOREM 3.1. Let A = (A 1,..., A m S(H m and k be a positive integer. If Λˆk(A = for some ˆk (m + 2k, then Λ k (A is star-shaped and contains the convex subset conv Λˆk(A so that every element in conv Λˆk(A is a star center of Λ k (A. Note that Λˆk(A may be empty if dim H is small relative to ˆk. Even if Λˆk(A is non-empty, it may be much smaller than its convex hull; for example, see Example 2.6. So, the conclusion in Theorem 3.1 is rather remarkable. Proof. We may assume a = 0 Λˆk(A. Then there exists Y such that Y Y = I (m+2k and Y A j Y = 0 for all 1 j m. Let b Λ k (A. Then there exists X such that X X = I k and (X A 1 X,..., X A m X = (b 1 I k,..., b m I k. Suppose X and Y are the range spaces of X and Y, respectively. Then we have dim (Y (X + A 1 (X + + A m (X dim Y dim (X + A 1 (X + + A m (X k. Let Y 1 be an k-dimensional subspace of Y (X + A 1 (X + + A m (X and Y 2 = Y (X + Y 1. Set Z = [X Y 1 Y 2 ], where Y i has columns forming an orthonormal basis of Y i for i = 1, 2. Then we have Z Z = I (m+2k and for 1 j m, Z A j Z has the form b j I k 0 k 0 k 0 k.

8 108 CHI-KWONG LI AND YIU-TUNG POON Let C j = b j I k 0 k. For t [0, 1], we have t(b 1,..., b m Λ k (C 1,..., C m Λ k (Z A 1 Z,..., Z A m Z Λ k (A. Clearly, Λˆk(A Λ k (A. Since every element of Λˆk(A is a star center of Λ k (A and the set of star centers of a star-shaped set is convex, we see that every element in conv Λˆk(A is a star center of Λ k (A. Hence, conv Λˆk(A Λ k (A. If dim H is finite, then Λ k (A is always closed. But this may not be the case if dim H is infinite. Using Theorem 3.1, we can prove the star-shapedness of cl (Λ k (A. COROLLARY 3.2. Let A = (A 1,..., A m S(H m and k be a positive integer. If cl (Λˆk(A = for some ˆk (m + 2k, then cl (Λ k (A is star-shaped and contains the convex subset conv cl (Λˆk(A so that every element in conv cl (Λˆk(A is a star center of cl (Λ k (A. Proof. Suppose a cl (Λˆk(A and b Λ k (A. Then for every ε there is ã = (ã 1,..., ã m Λˆk(A such that l 1 (ã a < ε. By Theorem 3.1, we see that the line segment joining ã and b lies in Λ k (A. Consequently, the line segment joining a and b lies in cl (Λ k (A. The proof of the last assertion is similar to that of Theorem 3.1. It is easy to see that a star center of Λ k (A is also a star center of cl (Λ k (A. However, the converse may not hold. The following example from [12, Example 3.2] illustrates this. EXAMPLE 3.3. Consider H = l 2 with canonical basis {e n : n 1}. Let A = (A 1,..., A 4 with A 1 = diag (1, 0, 1/3, 1/4,..., A 2 = diag (1, 0 0, ( ( i A 3 = 0 and A = 0. i 0 Then (1, 1, 0, 0 W(A and (0, 0, 0, 0 W(A W e (A is a star-center of the closure of W(A. However, (1/2, 1/2, 0, 0 / W(A so that (0, 0, 0, 0 is not a starcenter of W(A. In fact, W(A is not convex even though cl (W(A is convex. By Proposition 2.4, we see that Λˆk(A is non-empty if dim H is sufficiently large. So, Λ k (A is star-shaped and contains a convex set. The same comment also holds for cl (Λ k (A. More specifically, we have the following. THEOREM 3.4. Let A = (A 1,..., A m S(H m. If dim H ((m + 2k 1(m + 1 2, then both Λ k (A and cl (Λ k (A are star-shaped. In Theorem 3.7, we will show that the classical joint numerical range is starshaped with a much milder restriction on dim H comparing with that in Theorem 3.4. To demonstrate this, we need two related results.

9 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 109 PROPOSITION 3.5. Suppose dim H = n and A = (A 1,..., A m S(H m is such that {A 1,..., A m } is linearly independent. ( Assume that 0 Λ n 1 (A, i.e., there is a basis such that A j has operator matrix for j = 1,..., m. 0 n 1 (a If m = 2n 1, then there is an invertible S M m (R such that { } Λ 1 (A = (1 + u 1, u 2,..., u m S : u 1,..., u m R, m j=1 u 2 j = 1 so that Λ 1 (A is not star-shaped. (b If m < 2n 1, then Λ 1 (A is star-shaped with 0 as a star center. Proof. (a If m = 2n 1, there is an invertible m m real matrix T = (t ij such that for B j = i=1 t ij A i for j = 1,..., m with B = (B 1,..., B m = (E 11, E 12 + E 21, ie 12 + ie 21,..., ie 1n + ie n1. Let x = µ(cos t, sin t(v 2 + iv 3,..., sin t(v m 1 + iv m t be a unit vector in C n such that µ = 1, t [0, π/2], and v 2,..., v m R with m j=2 v2 j = 1. Then Bx, x = ((1 + cos(2t/2, v 2 sin(2t, v 2 sin(2t,..., v m sin(2t = (1 + u 1, u 2,..., u m D, where D = [1/2] I n 1, u 1 = cos(2t and u j = v j sin(2t for j = 2,..., m. It follows that { } Λ 1 (B = (1 + u 1, u 2,..., u m D : u 1,..., u m R, m j=1 u 2 j = 1 By Proposition 2.1, Λ 1 (A = {bt 1 : b Λ 1 (B}. The result follows. (b Now, suppose m < 2n 1. By adding more A j, if necessary, we only need to consider the case when m = 2n 2. Let v j be the row vector obtained by removing the first entry of the first row of A j for j = 1,..., m. Case 1 Suppose span {v 1,..., v m } has real dimension m 1 = 2n 3. Then there is a unitary matrix of the form U = [1] U 0 M n such that the (1, n entry of U A j U is real for j = 1,..., m. Hence, there is an invertible m m real matrix T = (t ij such that for B j = i=1 t ij A i for j = 1,..., m with B = (B 1,..., B m = (E 11, E 12 + E 21, ie 12 + ie 21,..., ie 1,n 1 + ie n 1,1, E 1n + E n1. Suppose b Λ 1 (B, b = 0. Then there exists a unit vector x = µ(u 0, u 1 + iu 2,..., u m 1 + iu m t such that µ = 1 and u 0 > 0, with b = Bx, x = u 0 (u 0, 2u 1,..., 2u m 1..

10 110 CHI-KWONG LI AND YIU-TUNG POON For any t (0, 1, we can choose a unit vector of the form x t = t(u 0, u 1 + iu 2,..., u m 1 + iũ m t with tũ 2 m = 1 m 1 j=0 tu2 j so that Bx t, x t = t Bx, x. Case 2. Suppose {v 1,..., v m } has real dimension m = 2n 2. Then there is an invertible m m real matrix T = (t ij such that B = (B 1,..., B m = t i1 A i,..., t im A i ( m i=1 m i=1 = (a 1 E 11,..., a m E 11 + (E 12 + E 21, ie 12 + ie 21,..., E 1n + E n1, ie 1n + ie n1 with a 1,..., a m R. Suppose b Λ 1 (B, b = 0. Then there exists a unit vector x = µ(u 0, u 1 + iu 2,..., u m 1 + iu m t such that µ = 1 and u 0 > 0, with b = Bx, x = u 0 (a 1 u 0 + 2u 1, a 2 u 0 + 2u 2,..., a m u 0 + 2u m. For any t (0, 1, consider a vector of the form x ξ = (ξu 0, w 1 + iw 2, w 3 + iw 4,..., w m 1 + iw m t, where ξ t and w j = a j u 0 (t ξ 2 /(2ξ + tu j /ξ for j = 1,..., m. Then so that ξu 0 (a j ξu 0 + 2w j = tu 0 (a j u 0 + 2u j, j = 1,..., m, Bx ξ, x ξ = t Bx, x. If ξ = t, then x ξ = t(u 0, u 1 + iu 2,..., u m 1 + iu m t has norm less than 1; if ξ, then x ξ ξu 0. Thus, there is ξ > t such that x ξ is a unit vector satisfying Bx ξ, x ξ = t Bx, x. So, Λ 1 (B is star-shaped with 0 as a star center. By Proposition 2.1, Λ 1 (A = {bt 1 : b Λ 1 (B}. The result follows. THEOREM 3.6. Let A = (A 1,..., A m S(H m. If Λˆk(A = for some ˆk > (m + 1/2, then Λ 1 (A is star-shaped and contains conv Λˆk(A such that every element in conv Λˆk(A is a star center of Λ 1 (A. Proof. We may assume a = 0 Λˆk(A with ˆk > (m + 1/2. Suppose x H is a unit vector and b = Ax, x Λ 1 (A. Suppose X is such that X X = Iˆk and X A j X = 0ˆk for j = 1,..., m. Let Y be such that Y Y = Iˆk+1, and the range space of Y contains the range space of X and x. Suppose B = (B 1 (,..., B m = (Y A 1 Y,..., Y A m Y. Then we may assume that B j has the form for 0ˆk j = 1,..., m. Clearly, span {B 1,..., B m } has dimension at most m < 2ˆk 1. By Proposition 3.5, the line segment joining 0 and b lies entirely in Λ 1 (B Λ 1 (A. Thus, 0 is a star center of Λ 1 (A. Since the set of star centers of Λ 1 (A is convex, we see that every element in conv Λˆk(A is a star center of Λ 1 (A.

11 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 111 THEOREM 3.7. Let A = (A 1,..., A m S(H m. If [ ] m + 1 dim H (m + 1 2, 2 then Λ 1 (A is star-shaped. [ ] Proof. Let ˆk m + 1 = + 1 > m + 1. Then dim H (ˆk 1(m and 2 2 Λˆk(A = by Proposition 2.4. The result then follows from Theorem RESULTS ON Λ (A In this section, we always assume that H has infinite dimension. Denote by P the set of infinite rank orthogonal projections in S(H. For A S(H m let Λ (A = {(γ 1,..., γ m R m : there is P P By the result in Section 3, we have the following. such that PA i P = γ i P for all 1 i m}. PROPOSITION 4.1. Suppose A S(H m, where H is infinite-dimensional. Then Λ k (A is star-shaped for each positive integer k. Moreover, if a Λ (A, then a is a star center for Λ k (A for every positive integer k. When m = 2, it was conjectured in [16] and confirmed in [14] that Λ (A 1, A 2 = Λ k (A 1, A 2 ; in [1, Theorem 4], it was proven that k 1 Λ (A 1, A 2 = {W(A 1 + F 1, A 2 + F 2 : F 1, F 2 S(H F(H}. In the following, we extend the above results to Λ (A 1,..., A m for m > 2. Moreover, we show that Λ (A = k 1 S k (A, where S k (A is the set of star centers of Λ k (A. Hence, Λ (A is always convex. THEOREM 4.2. Suppose A S(H m, where H is infinite-dimensional. For each k 1, let S k (A be the set of star-center of Λ k (A. Then (4.1 Λ (A = k S k (A = k Λ k (A = {W(A + F : F S(H m F(H m }. Consequently, Λ (A is convex. Proof. It follows from definitions and Theorem 3.1 that Λ (A k S k (A k Λ k (A. We are going to prove that k Λ k (A {W(A + F : F S(H m F(H m }. Suppose F = (F 1,..., F m S(H m F(H m and K = m i=1 rank (F i + 1. Let

12 112 CHI-KWONG LI AND YIU-TUNG POON µ = (µ 1,..., µ m k Λ k (A. Then there exists a rank K orthogonal projection P such that PA j P = µ j P for 1 j m. Let H 0 = range P ker F 1 ker F 2 ker F m = range P (range F 1 + range F range F m. Then dim H 0 1. Let x be a unit vector in H 0. Then we have (A + F x, x = Ax, x = µ. Therefore, µ W(A + F. Hence, we have k Λ k (A {W(A + F : F S(H m F(H m }. Next, we prove that {W(A + F : F S(H m F(H m } Λ (A. Suppose µ {W(A + F : F S(H m F(H m }. By Remark 2.3, we may assume that µ = 0. Then 0 W(A and there exists a unit vector x 1 such that Ax 1, x 1 = 0. Suppose we have chosen an orthonormal set of vectors {x 1,..., x n } such that Ax i, x j = 0 for all 1 i, j n. Let H 0 be the subspace spanned by {x i : 1 i n} {A j x i : 1 i n, 1 j m} and P the orthogonal projection of H onto H 0. Suppose ( B = (I PA 1 (I P H,..., (I PA m (I P 0 H. 0 Let b = (b 1,..., b m be a star-center of W(B. Then bi H0 B = (b 1 P + (I PA 1 (I P,..., b m P + (I PA m (I P = A + F for some F S(H m F(H m. Therefore, 0 W(bI H0 B. Hence, there exists a unit vector x H such that 0 = (A + Fx, x. Let x = y + z, where y H 0 and z H0. Then y 2 + z 2 = x 2 = 1. If z = 0, then 0 = b W(B. If z = 0, then by Proposition 4.1, we have ( ( z z 0 = (A + Fx, x = y 2 b + z 2 B, W(B z z So there exists a unit vector x n+1 H 0 such that 0 = (A + Fx n+1, x n+1 = Bx n+1, x n+1 = Ax n+1, x n+1 Hence, inductively, we can choose an orthonormal sequence of vectors {x n } n=1 such that Thus, we have Ax i, x j = 0 for all i, j. Λ (A k S k (A k Λ k (A {W(A + F : F S(H m F(H m } Λ (A. Since S k (A is convex for all k 1, the last statement follows.

13 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 113 The last equality in (4.1 establishes a relationship between Λ (A and the joint numerical ranges of finite rank perturbation of A. The following result gives an extension. THEOREM 4.3. Suppose A = (A 1,..., A m S(H m. Let n 1 and F 0 S(H m F(H m. Then the following sets are equal. (a Λ (A. (b Λ (A + F 0. (c {Λ n (A + F : F F(H m S(H m }. (d k 1 ( {Λk (A + F : F F(H m S(H m }. Proof. By Theorem 4.2, we have Λ (A + F 0 = {W(A + F 0 + F : F S(H m F(H m } = {W(A + F : F S(H m F(H m } = Λ (A. This proves the equality of the sets in (a and (b. For the equality of the sets of (a and (c, let F F(H m S(H m. Then we have It follows that Λ (A = Λ (A + F Λ n (A + F W(A + F. Λ (A {Λ n (A + F : F S(H m F(H m } {W(A + F : F S(H m F(H m } = Λ (A. The equivalence of (a and (d follows immediately. Recall that V r is the set of X : H1 H such that dim H 1 = r and X X = I H. Then X AX is a compression of A to H 1 1. The next result is an analog to Theorem 4.3 for Λ k (X AX. THEOREM 4.4. Suppose A = (A 1,..., A m S(H m. Let n, r 0 1 and X 0 V r0. Then for X AX = (X A 1 X,..., X A m X, the following sets are equal. (a Λ (A. (b Λ (X 0 AX 0. (c {Λ n (X AX : X r 1 V r }. (d k 1 ( {Λ k (X AX : X r 1 V r }. Proof. By (4.1 and (2.1, we have Λ (A = k Λ k (A = k Λ k+r (A k Λ k (X 0 AX 0 = Λ (X 0 AX 0 Λ (A. This proves the equality of the sets in (a and (b. For the equality of the sets of (a and (c, we will first show that (4.2 {Λ 1 (X AX : X r 1 V r } Λ (A.

14 114 CHI-KWONG LI AND YIU-TUNG POON Let µ {Λ 1 (X AX : X r 1 V r }. By Remark 2.3, we may assume that µ = 0. Then there exists a unit vector x 1 such that Ax 1, x 1 = 0. Suppose we have chosen an orthonormal set of vectors {x 1,..., x N } such that Ax i, x j = 0 for all 1 i, j N. Let H 1 be the subspace spanned by {x i : 1 i N} {A j x i : 1 i N, 1 j m} and X : H1 H be given by X(v = v for all v H 1. Then 0 Λ 1(X AX. So there exists a unit vector x N+1 H1 such that 0 = (X AXx N+1, x N+1 = Ax N+1, x N+1. Inductively, we can find an orthonormal sequence {x i } in H such that Ax i, x j = 0 for all i, j. Hence, 0 Λ (A. To continue the proof of the equality of the sets of (a and (c. Let X r 1 V r. Then we have It follows that Λ (A = Λ (X AX Λ n (X AX Λ 1 (X AX. Λ (A {Λ (X AX : X r 1 V r } {Λ 1 (X AX : X r 1 V r } Λ (A. The equality of the sets of (a and (d follows immediately. Recall that the joint essential numerical range of A S(H m is defined by W e (A = {cl (W(A + F : F S(H m F(H m }. Using the last two theorems, we have the following. COROLLARY 4.5. Let A S(H m, where H is infinite dimensional. Denote by S k (A the set of star center of cl (Λ k (A. Then W e (A = cl (Λ k (A = S k (A. k 1 In addition, let n 1 and F 0 S(H m F(H m. Moreover, let V be a finite dimensional subspace of H and X 0 : V H such that X 0 X 0 = I V. Then the following sets are equal. k 1 (a W e (A. (b W e (A + F 0. (c W e (X 0 AX 0. (d {cl (Λ n (A + F : F F(H m S(H m }. (e {cl (Λ n (X AX : X r 1 V r }. (f k 1 ( {cl (Λk (A + F : F F(H m S(H m }. (g k 1 ( {cl ((Λ k (X AX : X r 1 V r }.

15 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 115 Let A S(H and k be a positive integer. Denote by U k the set of X : C k H such that X X = I k and λ k (A = sup{min σ(x AX : X U k }, where σ(b is the spectrum of the operator B. For c = (c 1,..., c m R m and A S(H m, let c A = i=1 m c i A i. Define Ω (A = {a R m : c a λ k (c A for all k 1}. c R m The following extends [14, Theorem 2.1]. THEOREM 4.6. Let A S(H m, where H is infinite dimensional. Then Λ (A Ω (A = W e (A. Proof. For k 1, let Ω k (A = {a R m : c a λ k (c A}. c R m Clearly, Ω (A = k 1 Ω k (A. Suppose k 1 and a = (a 1,..., a m Λ k (A. Then there exists P P k such that PA j P = a j P for all 1 j m. For every c R m, we have P(c AP = (c ap. Hence, c a = λ k (c PAP = λ k (P(c AP λ k (c A. Therefore, Λ k (A Ω k (A. Since Ω k (A is closed, we have cl (Λ k (A Ω k (A. Hence, Λ (A = k 1 Λ k (A k 1 cl (Λ k (A = W e (A k 1 Ω k (A = Ω (A. To show that Ω (A W e (A. Suppose a Ω k (A. By Remark 2.3, we may assume that a = 0. So, λ k (c A 0 for all k 1 and c R m. For F = (F 1,..., F m S(H m F(H m, let K = m i=1 rank (F i + 1. Then λ 1 (c (A + F λ K+1 (c A 0 and λ 1 ( (c (A + F λ K+1 ( c A 0. Therefore, c 0 = 0 cl (W(c (A + F = c cl (W(A + F. Hence, c 0 c W e (A for all c R m. By the convexity of W e (A, we have 0 W e (A. [ 1n ] [ 0 1n ] 0 EXAMPLE 4.7. For n 1, let B n = 0 n 1, C n =, A = n=1 B n, A 2 = n=1 C n and A = (A 1, A 2. Then (0, 0 Ω (A but Λ (A =. Acknowledgements. Li is an honorary professor of the University of Hong Kong and an honorary professor of the Taiyuan Unversity of Technology. His research was partially supported by an NSF grant and the William and Mary Pulmeri Award. Poon s research was partially supported by an NSF grant.

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17 GENERALIZED NUMERICAL RANGES AND QUANTUM ERROR CORRECTION 117 CHI-KWONG LI, DEPARTMENT OF MATHEMATICS, COLLEGE OF WILLIAM AND MARY, WILLIAMSBURG, VIRGINIA 23185, USA address: YIU-TUNG POON, DEPARTMENT OF MATHEMATICS, IOWA STATE UNIVERSITY, AMES, IOWA 50011, USA address: Received Month dd, yyyy; revised Month dd, yyyy.