QUANTUM ERROR CORRECTION AND HIGHER RANK ...

arXiv:0902.4869v1 [math.FA] 27 Feb 2009

QUANTUM ERROR CORRECTION AND HIGHER RANK NUMERICAL RANGES OF NORMAL MATRICES HWA-LONG GAU, CHI-KWONG LI, YIU-TUNG POON, AND NUNG-SING SZE∗ Abstract. The higher rank numerical range is useful for constructing quantum error correction code for a noisy quantum channel. It is known that if a normal matrix A ∈ Mn has eigenvalues a1 , . . . , an , then its rank-k numerical range Λk (A) is the intersection of convex polygons with vertices aj1 , . . . , ajn−k+1 , where 1 ≤ j1 < · · · < jn−k+1 ≤ n. In this paper, it is shown that the higher rank numerical range of a normal matrix with m distinct vertices can be written as the intersection of no more than m closed half planes. In addition, given a convex polygon P a construction is given for a normal matrix A ∈ Mn with minimum n such that Λk (A) = P. In particular, if P has p vertices, with p ≥ 3, there is a normal matrix A ∈ Mn with n ≤ max {p + k − 1, 2k + 2} such that Λk (A) = P.

1. Introduction Let Mn be the algebra of n × n complex matrices regarded as linear operators acting on the n-dimensional Hilbert space Cn . In the context of quantum information theory, if the quantum states are represented as matrices in Mn , then a quantum channel is a trace preserving completely positive map L : Mn → Mn with the following operator sum representation L(A) =

r X

Ej∗ AEj ,

j=1

Pr

where E1 , . . . , Er ∈ Mn satisfy j=1 Ej Ej∗ = In . The matrices E1 , . . . , Er are known as the error operators of the quantum channel L. A subspace V of Cn is a quantum error correction code for the channel L if and only if the orthogonal projection P ∈ Mn with range space V satisfies P Ei∗ Ej P = γij P for all i, j ∈ {1, . . . , r}; for example, see [7, 8, 9]. In this connection, for 2000 Mathematics Subject Classification. 15A60, 15A90, 47N50, 81P68. Key words and phrases. Quantum error correction, higher rank numerical range, normal matrices, convex polygon. Research of Gau was supported by National Science Council of the Republic of China. Research of Li was supported by USA NSF and the William and Mary Plumeri Award. ∗ Corresponding author. 1

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HWA-LONG GAU, CHI-KWONG LI, YIU-TUNG POON, AND NUNG-SING SZE∗

1 ≤ k < n researchers define the rank-k numerical range of A ∈ Mn by Λk (A) = {λ ∈ C : P AP = λP for some rank-k orthogonal projection P }, and the joint rank-k numerical range of A1 , . . . , Am ∈ Mn by Λk (A1 , . . . , Am ) to be the collection of complex vectors (a1 , . . . , am ) ∈ C1×m such that P Aj P = aj P for a rank-k orthogonal projection P ∈ Mn . Evidently, there is a quantum error correction code V of dimension k for the quantum channel L described above if and only if Λk (A1 , . . . , Am ) is non-empty for (A1 , . . . , Am ) = (E1∗ E1 , E1∗ E2 , . . . , Er∗ Er ). Also, it is easy to see that if (a1 , . . . , am ) ∈ Λk (A1 , . . . , Am ) then aj ∈ Λk (Aj ) for j = 1, . . . , m. When k = 1, Λk (A) reduces to the classical numerical range defined and denoted by W (A) = {x∗ Ax ∈ C : x ∈ Cn with x∗ x = 1}, which is a useful concept in studying matrices and operators; see [6]. Recently, interesting results have been obtained for the rank-k numerical range and the joint rank-k numerical range; see [1, 2, 3, 4, 5, 11, 12, 13, 14, 16]. In particular, an explicit description of the rank-k numerical range of A ∈ Mn is given in [14], namely, \ (1.1) Λk (A) = {µ ∈ C : e−iξ µ + eiξ µ ≤ λk (e−iξ A + eiξ A∗ )}, ξ∈[0,2π)

where λk (X) is the kth largest eigenvalue of a Hermitian matrix X. In the study of quantum error correction, there are channels such as the randomized unitary channels and Pauli channels whose error operators are commuting normal matrices. Thus, it is of interest to study the rank-k numerical ranges of normal matrices. Given S ⊆ C, let conv S denote the smallest convex subset of C containing S. For a normal matrix A ∈ Mn with eigenvalues a1 , . . . , an , it was conjectured in [3, 4] that \ (1.2) Λk (A) = conv {aj1 , . . . , ajn−k+1 } 1≤j1 0}, R2 = {j : Im (aj ) = 0}, and R3 = {j : Im (aj ) < 0}. Then each of R1 and R3 has at most k − 1 elements and at least one of them is non-empty. By (1.2), Λk (A) ⊆ conv {ar : r ∈ R2 ∪ R3 }. Therefore, R2 has at least 2 elements r, s such that ar and as are distinct points in C. Let a = max{L(ri , rj ) ∩ R : 1 ≤ i < j ≤ 3, ri ∈ Ri , rj ∈ Rj , (ri , rj ) ∈ S}, b = min{L(rj , ri ) ∩ R : 1 ≤ i < j ≤ 3, ri ∈ Ri , rj ∈ Rj , (ri , rj ) ∈ S}. Choose rij ∈ Rij for j = 1, . . . , 4 such that a = L (ri1 , ri2 ) ∩ R

and

b = L (ri3 , ri4 ) ∩ R.

Then we have Λk (A) = [a, b] = H(r, s) ∩ H(s, r) ∩ H (ri1 , ri2 ) ∩ H (ri3 , ri4 ) . Clearly, a non-degenerate line segment cannot be written as an intersection of less than four half spaces H(r, s). (b) Suppose Λk (A) is a singleton. We may replace A by A − νI for some suitable ν and assume that \ (2.4) {0} = Λk (A) = H(r, s), (r,s)∈S

where S is defined as in the proof of Theorem 2.2. Evidently, in the intersection (2.4), we only need half spaces of the form H(rj , sj ) = {z ∈ C : Re (e−iξj z) ≥ 0} = Hj for j = 1, . . . , p. Since the intersection is a singleton, we have p ≥ 3. We may assume that 0 ≤ ξ1 < ξ2 < · · · < ξp < 2π. If p ≥ 5, then either ξ3 − ξ1 or ξ5 − ξ3 is less than π. If ξ3 − ξ1 < π, then we have H1 ∩ H2 ∩ H3 = H1 ∩ H3 and H2 can be omitted. Similarly, H4 can be omitted if ξ5 − ξ3 < π. Hence, we have p ≤ 4. From the proof, we can see that p = 4 if and only if {H(r, s) : (r, s) ∈ S and 0 ∈ L(r, s)} = {Hj : 1 ≤ j ≤ 4} and 0 ≤ ξ1 < ξ2 < π ≤ ξ3 = ξ1 + π < ξ4 = ξ2 + π. (c) If Λk (A) = ∅, then by Helly’s Theorem [10, Theorem 24.9], there are three half spaces H(rj , sj ) for j = 1, 2, 3, such that ∩3j=1 H(rj , sj ) = ∅ = Λk (A). If the boundary of two of these half spaces are parallel lines, then

QUANTUM ERROR CORRECTION AND HIGHER RANK NUMERICAL RANGES

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the two half spaces must be disjoint so that intersecting the two half spaces is the empty set. Clearly, we need to intersect at least two half spaces H(r, s) to get Λk (A). Therefore, p ∈ {2, 3}. By Theorems 2.3 and 2.6, we have the following. Corollary 2.7. Suppose A ∈ Mn is normal with m distinct eigenvalues. If Λk (A) 6= ∅, then Λk (A) is a convex polygon with at most m vertices. Corollary 2.8. Suppose A ∈ Mn is normal such that W (A) is an n-sided polygon containing the origin as its interior point. Let v1 , . . . , vn be the vertices of W (A) having arguments 0 ≤ ξ1 < · · · < ξn < 2π. If k < n/2, then Λk (A) is an n-sided convex polygon obtained by joining vj and vj+k , where vj+k = vj+k−n if j + k > n. By Theorem 2.3, it is easy to see that the boundary of Λk (A) are subsets of the union of line segments of the form conv {ar , as } such that ar and as satisfy the H(r, s) condition. However, it is not easy to determine which part of the line segment actually belong to Λk (A). Here are some examples.

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

imaginary axis

imaginary axis

Example 2.9. Let A = diag (a1 , . . . , an ) = diag (1, w, w2 , . . . , wn−1 ) with w = e2πi/n . Then for k ≤ n/2, we have Λk (A) = ∩nj=1 H(j, j + k), where aj+k = aj+k−n if j + k > n, and only a small part of conv {wj−1 , wj−1+k } lies in Λk (A).

0 −0.2

0 −0.2

−0.4

−0.4

−0.6

−0.6

−0.8

−0.8

−1

−1

−1

−0.8

−0.6

−0.4

−0.2

0 0.2 real axis

0.4

0.6

Λ2 (A) with n = 9

0.8

1

−1

−0.8

−0.6

−0.4

−0.2

0 0.2 real axis

0.4

0.6

0.8

1

Λ3 (A) with n = 9

Example 2.10. Suppose B = diag (1, i, −1, −i, 2, 2i, −2, −2i, 3, 3i, −3, −3i). One can see from the figures that the eigenvalues 1, i, −1, −i are interior points of Λ2 (B) while these eigenvalues are the vertices of Λ3 (B).

HWA-LONG GAU, CHI-KWONG LI, YIU-TUNG POON, AND NUNG-SING SZE∗ 4

4

3

3

2

2

1

1 imaginary axis

imaginary axis

8

0

0

−1

−1

−2

−2

−3

−3

−4 −4

−3

−2

−1

0 real axis

1

2

3

4

−4 −4

−3

−2

Λ2 (B)

−1

0 real axis

1

2

3

4

Λ3 (B)

Remark 2.11. Note that if the eigenvalues a1 , . . . , an of a normal matrix A is given, one can easily construct Λk (A) by the following algorithm from the discussion in this section. Case 1. If the eigenvalues lie on a line L, then arrange the points with real parts in descending order if L is not perpendicular to the real axis; otherwise, arrange the points with imaginary parts in descending order. Relabel the subscripts as a1 , . . . , an according to this ordering. If n ≥ 2k − 1, then Λk (A) is the line segment joining ak and an−k+1 , which may degenerate to a point if ak = an−k+1 . If n < 2k − 1, then it is the singleton {ak } if ak = an−k+1 ; otherwise, it is the empty set. Case 2. Suppose the eigenvalues do not lie on a straight line. For each distinct eigenvalue ar , identify the set Sr and the point asr . Determine the intersection of the half spaces H(r, sr ), which is Λk (A). 3. Matrices with prescribed higher rank numerical ranges We study the following problem in this section. Problem 3.1. Let k > 1 be a positive integer, and let P be a p-sided polygon in C. Construct a normal matrix A with smallest size (dimension) such that Λk (A) = P. If P degenerates to a line segment joining two points a1 and a2 . Then the smallest n to get a normal matrix with Λk (A) = P is n = 2k, if a1 and a2 are distinct and n = k if a1 = a2 . So we focus on the case when the polygon P is non-degenerate. A natural approach to Problem 3.1 is to reverse the construction of Λk (A) in Example 2.2. Suppose we have a non-degenerate p-sided polygon P, with vertices, v1 , . . . , vp . Without loss of generality, we may assume that 0 is in the interior of P and vj = rj eiθj , where rj > 0 and 0 ≤ θ1 < θ2 < · · · < θp < 2π. Let Lj be the line passing through vj and vj+1 . Suppose the following conditions hold.

QUANTUM ERROR CORRECTION AND HIGHER RANK NUMERICAL RANGES

9

(1) k < p/2. (2) for every 1 ≤ j ≤ p, Lj and Lj+k intersects at aj . (3) a1 , . . . , ap are vertices of a p-sided convex polygon arranged in the counterclockwise direction around 0. Let A be a normal matrix with eigenvalues a1 , a2 , . . . , ap . Then Λk (A) = P. The above construction fails when one or more of conditions (1) – (3) are not satisfied. Conditions (2) and (3) motivate the following definition. Definition 3.2. Let Ω = {z ∈ C : |z| = 1}. Given distinct α1 , α2 ∈ Ω, α2 /α1 = eiθ for a unique 0 < θ < 2π. Then [α1 , α2 ] = {eit α1 : 0 ≤ t ≤ θ} is the arc on Ω from α1 to α2 in the counterclockwise direction. Let (α1 , α2 ) = [α1 , α2 ]\{α1 , α2 }. The value θ is called the length of these intervals. Suppose 1 ≤ k ≤ n and Π ⊆ Ω. Then Π is said to be k-regular if for each α ∈ Π, (α, −α) ∩ Π contains at least k elements. If Π = {eiξj : 1 ≤ j ≤ n} with 0 ≤ ξ1 < · · · < ξn < 2π, then the above definition is equivalent to (3.1)

sin(ξr+k − ξr ) > 0 for all r = 1, . . . , n.

Here, for notational convenience, we set ξm = ξm−n if m > n. For this reason, a set {ξ1 , . . . , ξn } ⊆ [0, 2π) is also called k-regular if {eiξj : 1 ≤ j ≤ n} is k-regular as defined in Definition 3.2. For ξ, ξ ′ ∈ [0, 2π), [ξ, ξ ′ ] will denote ′ the subset {t ∈ [0, 2π) : eit ∈ [eiξ , eiξ ]}; the intervals [ξ, ξ ′ ), (ξ, ξ ′ ] and (ξ, ξ ′ ) will also be defined similarly. In Example 2.9, a direct computation shows that for 1 ≤ r, k ≤ n, ( 2kπ/n if r + k ≤ n, ξr+k − ξr = 2kπ/n − 2π if r + k > n. Therefore, the set {ξ1 , . . . , ξn } is k-regular and Λk (A) is nonempty for 1 ≤ k < n/2. Otherwise, the set {ξ1 , . . . , ξn } is not k-regular and Λk (A) is either empty or a singleton. In the following, we need an alternate formulation of (1.1). For any d ∈ R and ξ ∈ [0, 2π), consider the closed half plane H(d, ξ) = {µ ∈ C : Re (e−iξ µ) ≤ d}, and its boundary, which is the straight line L(d, ξ) = ∂H(d, ξ) = {µ ∈ C : Re (e−iξ µ) = d}. For A ∈ Mn , let ReA = (A + A∗ )/2. Then (1.1) is equivalent to \ Λk (A) = H(λk (Re (e−iξ A)), ξ). ξ∈[0,2π)

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Theorem 3.3. Suppose P is a non-degenerate polygon on C. Then there is a normal matrix A ∈ Mn such that Λk (A) = P if and only if there are d1 , . . . , dn ∈ R and a k-regular set {ξ1 , . . . , ξn } such that P=

n \

H(dj , ξj ).

j=1

Notice that a necessary condition for the set degenerate polygon is that

Tp

j=1

H(dj , ξj ) to be a non-

{eiξ1 , . . . , eiξp } is 1-regular.

(3.2)

To prove Theorem 3.3, we need some lemmas. Lemma 3.4. Given A = diag (a1 , . . . , an ) and 1 ≤ m < n. Suppose the eigenvalues am+1 , . . . , an are in Λk (A) but not extreme points of Λk (A). Then Λk (diag (a1 , . . . , am )) = Λk (A). Proof. It suffices to show that if an is in Λk (A) but not an extreme point of Λk (A), then Λk (diag (a1 , . . . , an−1 )) = Λk (A). Suppose an satisfy the above assumption. Clearly, Λk (diag (a1 , . . . , an−1 )) is a subset of Λk (A). On the other hand, for any 1 ≤ j1 < · · · < jn−k ≤ n − 1, Λk (A) ⊆ conv {aj1 , . . . , ajn−k , an }. Since an is not an extreme point of Λk (A), it follows that an lies in conv {aj1 , . . . , ajn−k , an } but is not its extreme point. Therefore, conv {aj1 , . . . , ajn−k } = conv {aj1 , . . . , ajn−k , an }. Thus, \ {conv {aj1 , . . . , ajn−k+1 } : 1 ≤ j1 < · · · < jn−k < jn−k+1 ≤ n} \ ⊆ {conv {aj1 , . . . , ajn−k , an } : 1 ≤ j1 < · · · < jn−k ≤ n − 1} \ = {conv {aj1 , . . . , ajn−k } : 1 ≤ j1 < · · · < jn−k ≤ n − 1} = Λk (diag (a1 , . . . , an−1 )).

Λk (A) =

Lemma 3.5. Suppose P = ∩m j=1 H(dj , ξj ) such that 0 ≤ ξ1 ≤ · · · ≤ ξm < 2π and for each r = 1, . . . , m, there are 1 ≤ j1 < · · · < jk ≤ m such that ξj1 , . . . , ξjk ∈ (ξr , ξr + π). For every n ≥ m, there are real numbers dˆ1 , . . . , dˆn and a k-regular set {ξˆ1 , . . . , ξˆn } such that P = ∩nj=1 H(dˆj , ξˆj ) and P ∩ L(dˆj , ξˆj ) 6= ∅ for each j = 1, . . . , n. Proof. Set ξˆ1 = ξ1 , and for s ∈ {2, . . . , m}, let ξˆs = ξs if ξs−1 < ξs . For the remaining values, we have ξs−1 = ξs and we can write that ξˆs−t1 = ξs−t1 +1 = · · · = ξs = · · · = ξs+t2 < ξˆs+t2 +1

QUANTUM ERROR CORRECTION AND HIGHER RANK NUMERICAL RANGES 11

for some t1 ≥ 1 and t2 ≥ 0. Let ℓ = min{j : ξˆj > ξˆs−t1 + π}, then we can replace ξs+j by ξˆs+j = ξs+j + ǫj for sufficient small ǫj > 0 for j = −t1 + 1, −t1 + 2, . . . , 0, . . . , t2 such that ξˆs−t < ξˆs−t +1 < · · · < ξˆs < · · · < ξˆs+t < min{ξˆℓ − π, ξˆs+t +1 }. 1

1

2

2

After this modification, ξˆ1 , . . . , ξˆm are distinct and {ξˆ1 , . . . , ξˆm } is k-regular. If n > m, pick distinct ξˆm+1 , . . . , ξˆn ∈ [0, 2π)\{ξˆ1 , . . . , ξˆm }. Then {ξˆ1 , . . . , ξˆn } ˆ also forms a k-regular set. Finally, let dˆj = maxµ∈P Re e−iξj µ for j = 1, . . . , n. Clearly, we have P ∩ L(dˆj , ξˆj ) 6= ∅ and P ⊆ H(dˆj , ξˆj ) for all j. Since {ξ1 , . . . , ξm } is a subset of {ξˆ1 , . . . , ξˆn }, hence P = ∩m j=1 H(dj , ξj ) = n ˆ ˆ ∩ H(dj , ξj ). j=1

We can now present the proof of Theorem 3.3. T Proof of Theorem 3.3. Suppose P = nj=1 H(dj , ξj ) is a non-degenerate polygon, where {ξ1 , . . . , ξn } is k-regular. By Lemma 3.5, we may assume that P ∩ L(dj , ξj ) 6= ∅ for all j = 1, . . . , n, and 0 ≤ ξ1 < · · · < ξn < 2π such that condition (3.1) holds. For each r = 1, . . . , n, let i eiξr dr+k − eiξr+k dr ar = sin(ξr+k − ξr ) and A = diag (a1 , . . . , an ). Then Re (e−iξr ar ) = dr

and

Re (e−iξr+k ar ) = dr+k .

Note that ar ∈ L(dr , ξr ) ∩ L(dr+k , ξr+k ) is the vertex of the conical region H(dr , ξr ) ∩ H(dr+k , ξr+k ), which contains P. Therefore, Re (e−iξr (ar − µ)) ≥ 0 and Re (e−iξr+k (ar − µ)) ≥ 0, for all µ ∈ P. Since ξr+k ∈ (ξr , ξr + π), we have (3.3)

Re (e−iξ ar ) ≥ max Re (e−iξ µ). µ∈P

for all ξ ∈ [ξr , ξr+k ]. Let µj ∈ L(dj , ξj ) ∩ P for j = r, r + k. As ξr+k ∈ (ξr , ξr + π), we have µr = ar − ieiξr br and µr+k = ar + ieiξr+k cr for some br , cr ≥ 0. Note that Re (e−iξ (µr − ar )) = br sin(ξr − ξ) ≥ 0, for all ξ ∈ [ξr − π, ξr ], and Re (e−iξ (µr+k − ar )) = cr sin(ξ − ξr+k ) ≥ 0, for all ξ ∈ [ξr+k , ξr+k + π]. Since {ξ1 , . . . , ξn } is k-regular, it is easily seen that [0, 2π) \ [ξr , ξr+k ] = [ξr − π, ξr ) ∪ (ξr+k , ξr+k + π]. Therefore, for ξ ∈ [0, 2π) \ [ξr , ξr+k ], we have max{Re (e−iξ (µr − ar )), Re (e−iξ (µr+k − ar ))} ≥ 0.

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Moreover, we have max{Re (e−iξ µr ), Re (e−iξ µr+k )} ≥ Re (e−iξ ar ).

(3.4)

Let ξ ∈ [0, 2π). Then ξ ∈ [ξs , ξs+1 ) for some s ∈ {1, . . . , n}. It follows that ξ ∈ [ξr , ξr+k ] for r = s − k + 1, . . . , s, and ξ ∈ [0, 2π) \ [ξr , ξr+k ] for other r. By (3.3) and (3.4), min

r∈{s−k+1,...,s}

Re (e−iξ ar ) ≥ max Re (e−iξ µ) ≥ µ∈P

max

r ∈{s−k+1,...,s} /

Re (e−iξ ar ).

Thus, λk (Re (e−iξ A)) = minr∈{s−k+1,...,s} Re (e−iξ ar ) and so P ⊆ H λk (Re (e−iξ A)), ξ .

Hence, P ⊆ Λk (A). Furthermore, if ξ = ξs , then Re (e−iξs as ) = ds . Thus λk (Re (e−iξs A)) =

min

r∈{s−k+1,...,s}

Re (e−iξs ar ) ≤ ds .

It follows that \

Λk (A) =

ξ∈[0,2π)

⊆

\

1≤s≤n

H λk (Re (e−iξ A)), ξ

\ H (ds , ξs ) = P. H λk (Re (e−iξs A)), ξs ⊆ 1≤s≤n

Thus, P = Λk (A). Next, we consider the converse. Suppose A = diag (a1 , . . . , an ) and Λk (A) = P. By Lemma 3.4, we can remove the eigenvalues of A in Λk (A) ˜ that are not extreme points of Λk (A) to get A˜ ∈ Mm so that Λk (A) = Λk (A). m ˜ = ∩ H(dj , ξj ) for a k-regular set {ξ1 , . . . , ξm }, If we can show that Λk (A) j=1 ˜ = ∩n H(d˜j , ξ˜j ) for some then Lemma 3.5 will ensure that Λk (A) = Λk (A) j=1

k-regular set {ξ˜1 , . . . , ξ˜n }. For notational convenience, we assume that A = A˜ so that every eigenvalue is either an extreme point of Λk (A) or not in Λk (A). Recall that in the proof of Theorem 2.3, S denotes the set of (r, s) such that H(r, s) contains at least n − k + 1 eigenvalues of A and Sr = {s : (r, s) ∈ S}. We first show that Sr 6= ∅ for all r. As ar is either an extreme point of Λk (A) or not in Λk (A), there is ξ ∈ [0, 2π) such that Re (e−iξ ar ) ≥ λk (Re (e−iξ A)). Thus, the closed half plane H(Re (e−iξ ar ), ξ) contains at least n − k + 1 eigenvalues of A. Let W = conv {at : at ∈ H(Re (e−iξ ar ), ξ)}.


Then W is a convex set containing P and there is a vertex as ∈ W with as 6= ar such that W ⊆ H(r, s). Thus, H(r, s) contains at least n − k + 1 eigenvalues and hence Sr 6= ∅. Now by Theorem 2.3 and its proof, \ \ \ Λk (A) = H(r, sr ) = H(r, sr ) = H(dr , ξr ), Sr 6=∅

1≤r≤n

1≤r≤n

where ξr = arg(asr − ar ) − π/2 and dr = Re (e−iξr ar ). We will prove the following. Claim For each r = 1, . . . , n, there are 1 ≤ j1 < · · · < jk ≤ n such that ξj1 , . . . , ξjk ∈ (ξr , ξr + π). Once the claim is proved, Lemma 3.5 will ensure that Λk (A) = ∩nj=1 H(d˜j , ξ˜j ) for some k-regular set {ξ˜1 , . . . , ξ˜n }. To prove our claim, we first establish the following two assertions. (1) for any ξ ∈ [0, 2π), if H(Re (e−iξ ar ), ξ) contains at least n − k + 1 eigenvalues of A, then either ξr = ξ or sin(ξr − ξ) > 0, and (2) the closed half plane {z ∈ C : Re (e−iξr z) ≥ dr } contains at least k eigenvalues aj1 , . . . , ajk satisfying ξjt 6= ξr . To prove the two assertions, without loss of generality, we may assume that ar = 0 and asr lies in the positive imaginary axis. Then dr = 0, ξr = 0 and H(r, sr ) = H = {z ∈ C : Re (z) ≤ 0}. Thus, the closed left half plane contains at least n − k + 1 eigenvalues. First we consider Assertion (1). Note that in terms of the set Sr associate with ar in the proof of Theorem 2.3, Assertion (1) simply says that Sr cannot have any element in the interior of H(r, sr ) = H(dr , ξr ). Suppose H(Re (e−iξ ar ), ξ) contains at least n − k + 1 eigenvalues. By our assumption, H(Re (e−iξ ar ), ξ) = eiξ H. Notice that the closed right half plane −H can contain at most n − k eigenvalues. Otherwise, Λk (A) has to be a subset in H ∩ (−H) = iR, which contradicts the assumption that Λk (A) is a non-degenerate polygon. From this argument, we first see that ξ 6= π. Now suppose that ξ ∈ (0, π). If the intersection of H ∩ (eiξ H) = {z ∈ C : arg(z) ∈ [ξ + π/2, 3π/2]} does not contain any nonzero eigenvalue of A, then the closed half plane −H will contain all the eigenvalues that are in eiξ H and hence −H has at least n−k+1 eigenvalues. But this is impossible by the previous argument. Thus, the intersection H ∩ (eiξ H) contains at least one nonzero eigenvalue of A. Take the nonzero eigenvalue as in H ∩ (eiξ H) such that as has the smallest argument arg(as ) ∈ [ξ + π/2, 3π/2] among all the nonzero eigenvalues of A

14


in H ∩ (eiξ H). Let τ = arg(as ) − π/2. Then H(r, s) = eiτ H contains all the eigenvalues that are in eiξ H and hence H(r, s) contains at least n − k + 1 eigenvalues. Then the property of asr implies asr ∈ H(r, s) and so H(r, s) contains the positive imaginary axis. As τ = arg(as ) − π/2 ∈ [ξ, π], this is possible only when τ = π. But then eiτ H = −H, which contradicts our previous argument. Therefore, we must have ξ ∈ (π, 2π] and hence Assertion (1) holds. Next, turn to Assertion (2). Suppose that the closed half plane −H contains eigenvalues aj1 , . . . , ajg+h with ξjt = 0 for t = 1, . . . , g, and ξjt 6= 0 for t = g + 1, . . . , g + h. Then Assertion (2) holds if h ≥ k. Fix a sufficiently small ǫ > 0. We choose 1 ≤ ℓ ≤ g so that Re (e−iǫ ajℓ ) = max Re (e−iǫ ajt ). 1≤t≤g

Then {aj1 , . . . , ajg } ⊆ H(Re (e−iǫ ajℓ ), ǫ). On the other hand, this closed half plane H(Re (e−iǫ ajℓ ), ǫ) also contains all eigenvalues of A that are in the left open half plane. Thus, this closed half plane H(Re (e−iǫ ajℓ ), ǫ) has at least n−h eigenvalues of A. By Assertion (1), H(Re (e−iǫ ajℓ ), ǫ) has at most n−k eigenvalues. Thus, we have h ≥ k and the assertion holds. Now we prove our claim. By Assertion (2), for any r = 1, . . . , n, the closed half plane {z ∈ C : Re (e−iξr z) ≥ dr } has at least k eigenvalues aj1 , . . . , ajk with ξjt 6= ξr . For each t = 1, . . . , k, let ht = Re (e−iξr ajt ), then ht ≥ dr and H(dr , ξr ) ⊆ H(ht , ξr ). Thus, the closed half plane H(ht , ξr ) contains at least n − k + 1 eigenvalues. Now by Assertion (1), we see that (3.5)

sin(ξjt − ξr ) > 0

for t = 1, . . . , k.

Thus, our claim is proved, and the result follows. By Theorem 3.3, we can deduce the following. T Theorem 3.6. Let P = pj=1 H(dj , ξj ) be a non-degenerate p-sided polygon, where d1 , . . . , dp ∈ R and ξ1 , . . . , ξp ∈ [0, 2π). The following two statements are equivalent. (I) There is a (p + q) × (p + q) normal matrix A such that Λk (A) = P. (II) There are ξp+1 , . . . , ξp+q ∈ [0, 2π) such that {ξ1 , . . . , ξp+q } is k-regular. T Proof. Let P = pj=1 H(dj , ξj ) be a non-degenerate p-sided polygon, where d1 , . . . , dp ∈ R and ξ1 , . . . , ξp ∈ [0, 2π). Suppose (I) holds. By Theorem 3.3, there are h1 , . . . , hp+q ∈ R and ζ1 , . . . , ζp+q ∈ [0, 2π) such that {ζ1 , . . . , ζp+q } is k-regular and p \

j=1

H(dj , ξj ) = P = Λk (A) =

p+q \

j=1

H(hj , ζj ).


Then {ξ1 , . . . , ξp } ⊆ {ζ1 , . . . , ζp+q }. Now we can take ξp+1 , . . . , ξp+q ∈ [0, 2π) so that {ξ1 , . . . , ξp+q } = {ζ1 , . . . , ζp+q }. Therefore, (II) holds. Suppose (II) holds, namely, there are ξp+1 , . . . , ξp+q such that {ξ1 , . . . , ξp+q } is k-regular. For j = p + 1, . . . , p + q, define dj = max Re (e−iξj µ). µ∈P

Then P ⊆ H(dj , ξj ) and so P=

p \

j=1

H(dj , ξj ) =

p+q \

H(dj , ξj ).

j=1

Then the result follows from Theorem 3.3.

By Theorem 3.6, Problem 3.1 is equivalent to the following combinatorial problem. Problem 3.7. Suppose {ξ1 , . . . , ξp } is 1-regular. For k > 1, determine the smallest q so that {ξ1 , . . . , ξp+q } is k-regular for some ξp+1 , . . . , ξp+q ∈ [0, 2π). 4. Solutions for Problems 3.1 and 3.7 In this section, we give the solutions for Problems 3.1 and 3.7. Given a set of 1-regular complex units Π = {α1 , . . . , αp } ⊆ Ω, Problem 3.7 is equivalent to the study of smallest q so that {α1 , . . . , αp+q } is k-regular for some αp+1 , . . . , αp+q ∈ Ω. As shown in Assertion 1 later in this section, the existence of αi , αj ∈ Π with αj = −αi has implication on the size of a k-regular set Π. So, to deal with our problems, we always partition a set of complex units Π = {α1 , . . . , αp } ⊆ Ω into Π = Π1 ∪ Π2 with (4.1)

Π1 = {αj ∈ Π : −αj ∈ / Π} and Π2 = {αj ∈ Π : −αj ∈ Π}.

Here either Π2 is empty or Π2 contains an even number of elements. For any finite set S, denote the number of elements of S by n(S). Suppose n(Π1 ) = r and n(Π2 ) = 2s. Then 0 ≤ s ≤ p/2 and p = r + 2s. To see the impact of the size of Π1 and Π2 on the solution of our optimization problem, let us consider the following example. Example 4.1. Suppose S1 = {1, w, w2 , w3 } with w = e2iπ/5 . Then α 6= −β for any two elements α, β ∈ S1 and adding w4 to S1 results in a 2-regular set. Suppose S2 = {1, −1, i, −i}. Then we need to add at least two points, say, z, −z ∈ Ω \ S2 , to get a 2-regular set. In general, we have the following result allowing us to determine the minimum number of elements to be added to a finite set Π ⊆ Ω to get a k-regular set. We use the convention that a non-empty subset set of Ω is 0-regular in Theorem 4.2.

16


Theorem 4.2. Given Π = Π1 ∪ Π2 as defined in (4.1) with n(Π1 ) = r and n(Π2 ) = 2s. Suppose Π is 1-regular. Then there is a set Π3 ⊆ Ω with n(Π3 ) = q such that Π1 ∪ Π2 ∪ Π3 is k-regular if and only if one of the following holds. (a) k ≥ r + s and ( 2k + 1 − p if s = 0, ∗ q ≥ ℓ := 2k + 2 − p if s 6= 0. (b) k < r + s and q ≥ t∗ , ˆ 1 with where t∗ is the minimum t such that Π1 contains a subset Π ˆ ˆ n(Π1 ) = r − t and Π1 ∪ Π2 being (k − t)-regular. Proof. Let P = {z ∈ C : Im (z) > 0} be the open upper half plane. We divide the proof into five assertions. Assertion 1. Suppose S = {α1 , . . . , αn } is k-regular. Then n ≥ 2k + 1. Furthermore, if there are distinct i and j such that αj = −αi , then n ≥ 2k + 2. Proof. For any r ∈ {1, . . . , n}, each of the intervals (αr , −αr ) and (−αr , αr ) contains k elements of S. Thus, n ≥ 2k + 1. For the last statement, if we take r = j, then there are 2k elements in the intervals. Together with αi and αj , we see that n ≥ 2k + 2. The proof of the assertion is complete. Assertion 2. Suppose k ≥ r+s. Then there is a set Π3 ⊆ Ω with n(Π3 ) = ℓ∗ such that Π1 ∪ Π2 ∪ Π3 is k-regular. Proof. Suppose k ≥ r + s. Let Π′ be a set containing (k − r − s + 1) pairs of opposite points of the form {α, −α} such that Π′ ∩ Π is empty. Take Π3 = (−Π1 ) ∪ Π′ . Then n(Π3 ) = n(−Π1 ) + n(Π′ ) = r + 2(k − r − s + 1) = 2k + 2 − p. Furthermore, the set Π1 ∪ Π2 ∪ Π3 contains k + 1 pairs of opposite points of the form {α, −α} and hence it is k-regular. Thus, the result follows if s 6= 0. Now suppose s = 0, i.e., Π2 = ∅. Without loss of generality, we may assume that 1 ∈ Π1 and −1 ∈ Π3 . We now modify Π3 . We first delete the point −1 in Π3 . Then for all other points α ∈ Π3 , we replace α by eiξ α with sufficiently small positive ξ > 0 if α lies in P , and with sufficiently small negative ξ < 0 if α lies in −P . Then we see that for every α ∈ Π1 ∪ Π3 , αP still contains exactly k elements. Thus, Π1 ∪ Π3 is k-regular. Furthermore, the modified set Π3 has one fewer point, i.e., n(Π3 ) = 2k + 1 − p. The proof of the assertion is complete. Combining Assertions 1 and 2, we complete the proof of the case (a).


ˆ 1 of Π1 with n(Π1 ) = Assertion 3. Suppose k < r+s and there is a subset Π ˆ 1 ∪ Π2 is (k − t)-regular. Then there is a set Π3 ⊆ Ω with r − t such that Π n(Π3 ) = t such that Π1 ∪ Π2 ∪ Π3 forms a k-regular set. ˆ 1 ∪ Π2 is (k − t)-regular, for every α ∈ Ω, the open half plane Proof. As Π ˆ 1 ∪ Π2 . αP contains at least k − t elements of Π ˆ 1 = {αt+1 , . . . , αr } ⊆ Π1 . Take Without loss of generality, we assume Π αp+j = −αj for j = 1, . . . , t and let Π3 = {αp+1 , . . . , αp+t }. Then (Π1 \ ˆ 1 ) ∪ Π3 contains t pairs of opposite points of the form {α, −α}. Now for Π ˆ 1 ) ∪ Π3 , the open half plane αP contains at least k − t every α 6∈ (Π1 \ Π ˆ 1 ∪ Π2 and at least t elements of (Π1 \ Π ˆ 1 ) ∪ Π3 . Hence, αP elements of Π contains at least k elements of Π1 ∪ Π2 ∪ Π3 . On the other hand, when ˆ 1 ) ∪ Π3 , the open half plane αP contains at least k − 1 elements α ∈ (Π1 \ Π of Π1 ∪ Π2 ∪ Π3 . ˆ 1 ∪ Π2 has at least 2(k − t) + 1 elements and hence By Assertion 1, Π Π1 ∪ Π2 ∪ Π3 contains at least 2k + 1 elements. Then for each j = 1, . . . , t, at least one of the open half plane, αj P or αp+j P (= −αj P ), contains at least k elements while the other contains at least k − 1 elements. As αp+j lies in the boundary of both αj P and αp+j P , we can perturb αp+j so that both the open half planes αj P and αp+j P contain at least k elements of Π1 ∪ Π2 ∪ Π3 . Thus the assertion follows. Assertion 4. Suppose there is a set Π3 ⊆ Ω with n(Π3 ) = q such that ˆ 1 of Π1 and a set Π1 ∪ Π2 ∪ Π3 is k-regular. Then there exist a subset Π ˜ ˆ ˜ ˆ 1 ∪ Π2 ∪ Π ˜ 3 is Π3 ⊆ Ω with n(Π1 ) = r − 1 and n(Π3 ) = q − 1 such that Π (k − 1)-regular. Proof. We claim that there are β ∈ Π3 and γ ∈ Π1 ∪ Π3 such that the set (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ} is (k − 1)-regular. Suppose our claim is proved. If γ ∈ Π1 , then the result follows with ˆ 1 = Π1 \ {γ} and Π ˜ 3 = Π3 \ {β}. If γ ∈ Π3 , then the result follows with Π ˆ ˜ 3 = (Π3 ∪ {α}) \ {β, γ}. Π1 = Π1 \ {α} for any α ∈ Π1 and Π It remains to prove our claim. If there is β ∈ Π3 such that −β ∈ Π1 ∪ Π3 , then it is easy to see that the set (Π1 ∪ Π2 ∪ Π3 ) \ {β, −β} is (k − 1)-regular. Thus, the claim follows by taking γ = −β. We can assume that for each β ∈ Π3 , −β ∈ / Π1 ∪ Π3 . Notice that we may assume that Π3 ∩ Π = ∅. So we have −β ∈ / Π2 . Fixed a point β ∈ Π3 . Without loss of generality, we may assume that β = −1. Furthermore, we have the following extra assumption when Π2 6= ∅. By Assertion 1, Π1 ∪ Π2 ∪ Π3 contains at least 2k + 2 elements. Thus, either the upper open half plane or the lower open half plane contains at least k + 1 elements. By the fact that a set S is k-regular if and only if the set ¯1 ∪ Π ¯2 ∪ Π ¯ 3 if S¯ = {¯ z : z ∈ S} is k-regular, replacing Π1 ∪ Π2 ∪ Π3 with Π

18


necessary, we may further assume that the upper open half plane contains at least k + 1 elements if Π2 is non-empty. Let γ be the point in Π1 ∪Π3 such that arg(γ) ≤ arg(α) for all α ∈ Π1 ∪Π3 . We show that (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ} is (k − 1)-regular. Take any α ∈ (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ}. Suppose α ∈ βP ∪ γP . Then the open half plane αP can contain at most one of points β and γ. As the open half plane αP contains at least k elements of Π1 ∪ Π2 ∪ Π3 , αP contains at least k − 1 elements of (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ}. Now let ω1 , . . . , ωt be the points in (Π1 ∪ Π2 ∪ Π3 ) \ (βP ∪ γP ). By the choice of γ, {ω1 , . . . , ωt } ⊆ Π2 . Furthermore, all of them lie in the upper open half plane P . Therefore, we may assume that 0 < arg(ω1 ) < · · · < arg(ωt ) < arg(γ) < π. Clearly, each open half plane ωj P contains at least k elements of Π1 ∪ Π2 ∪ Π3 . Suppose ωj P contains exactly k elements. Notice that ωj P contains −ω1 , . . . , −ωj−1 , and β, which do not lie in the upper open half plane P . Thus, the intersection P ∩ (ωj P ) contains at most k − j elements and hence the upper open half plane P contains at most k elements only. But this contradicts to our assumption that the upper open half plane contains at least k + 1 elements when Π2 6= ∅. Thus, every open half plane wj P contains at least k + 1 elements of Π1 ∪ Π2 ∪ Π3 and hence every open half plane wj P contains at least k − 1 elements of (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ}. Therefore, (Π1 ∪ Π2 ∪ Π3 ) \ {β, γ} is a (k − 1)-regular set and the assertion follows. Assertion 5. Suppose there is a set Π3 ⊆ Ω with n(Π3 ) = q such that Π1 ∪ Π2 ∪ Π3 is k-regular. Let t = min{k, q, r}. Then there are subsets ˆ 1 of Π1 and Π ˜ 3 of Ω with n(Π ˆ 1 ) = r − t and n(Π ˜ 3 ) = q − t such that Π ˆ ˜ ˆ 1 ∪ Π2 Π1 ∪ Π2 ∪ Π3 is (k − t)-regular. Furthermore, if k < r + s, the set Π is also (k − t)-regular. Proof. The first part is clear by Assertion 4. Assume k < r + s. The last ˜ 3 = ∅ in this case. assertion is also clear when t = q as Π Suppose t = k. As k < r + s, either r > k or s > 0. In both cases, ˆ Π1 ∪ Π2 is 0-regular. Finally, suppose t = r. Then s > 0. Notice that ˆ 1 ∪ Π2 contains s pairs of opposite points of the form {α, −α}. Thus, the Π ˆ 1 ∪ Π2 is (s − 1)-regular and hence (k − r)-regular as k − r ≤ s − 1. set Π Then the result follows. Combining the five assertions, we get the conclusion in Theorem 4.2. Remark 4.3. Note that if condition (a) of the theorem holds, then the minimum number ℓ∗ can be computed immediately. If condition (b) holds, ˆ 1 of Π1 to see whether Π ˆ 1 ∪ Π2 is one can check each r − t element subset Π k − t regular, for t = 0, 1, . . . . Thus, in a finite number of steps, one can determine the minimum number t∗ .


t∗

The following proposition gives more information about the value ℓ∗ and in Theorem 4.2.

Proposition 4.4. Using the notation in Theorem 4.2. If k < r + s, then the value t∗ exists and satisfies ( k if (r, s) = (k + 1, 0) or (k, 1), ℓ∗ ≤ t ∗ ≤ k − 1 otherwise. Furthermore, t∗ = ℓ∗ if p ≤ k + 2 with s = 0 or p ≤ k + 3 with s 6= 0. Proof. We first consider the case when (r, s) = (k + 1, 0) or (k, 1). Take ˆ 1 of Π1 with n(Π1 ) = r − t. In both cases, t = k and any arbitrary subset Π ˆ 1 ∪ Π2 is non-empty and hence is 0-regular. Thus, t∗ ≤ t = k. Π Now we assume that (r, s) ∈ / {(k + 1, 0), (k, 1)}. Consider the case when ˆ 1 of Π1 with s ≥ 2. Take t = min{k − 1, r} and an arbitrary subset Π ˆ 1 ) = r − t. Clearly, Π ˆ 1 ∪ Π2 is (s − 1)-regular and hence (k − t)-regular n(Π as k − t = max{1, k − r} ≤ s − 1. Thus, we have t∗ ≤ t ≤ k − 1. Next we consider the case when s = 1 and r ≥ k + 1. Let Π2 = {α, −α}. Since Π is 1-regular, there exists α1 ∈ (α, −α) ∩ Π1 and α2 ∈ (−α, α) ∩ Π1 . ˆ 1 of Π1 with n(Π ˆ 1 ) = r − t ≥ 2 such Take t = k − 1. Choose any subset Π ˆ 1 . Then Π ˆ 1 ∪ Π2 is 1-regular. Then the result follows. that α1 , α2 ∈ Π Finally consider the case when s = 0 and r ≥ k + 2. We may assume that Π = {eiξj : 1 ≤ j ≤ p} with 0 = ξ1 < · · · < ξp < 2π. Since Π is 1-regular, we can choose ℓ such that ξℓ = max{ξj : 0 < ξj < π}. Then S = {ξ1 , ξℓ , ξℓ+1 } is ˆ 1 ⊆ Π1 with n(Π ˆ 1 ) = r − t ≥ 3. Then 1-regular. Take t = k − 1. Let S ⊆ Π ˆ 1 ∪ Π2 is 1-regular. Π By Theorem 4.2 and Proposition 4.4, we can answer Problems 3.1 and 3.7 and obtain some additional information on the solutions. Theorem 4.5. Let Π = {eiξj : 1 ≤ j ≤ p}. The optimal solution (minimum value) q˜ in Problem 3.7 is equal to ℓ∗ or t∗ , depending on case (a) or (b) of Theorem 4.2. Furthermore, (4.2)

q˜ ≤ max{k − 1, 2k + 2 − p}.

For Problem 3.1, if a p-sided polygon P is expressed as P = ∩pj=1 H(dj , ξj ) for some d1 , . . . , dp > 0, then the minimum dimension n ˜ for the existence of a normal matrix A ∈ Mn˜ such that Λk (A) = P is equal to p + ℓ∗ or p + t∗ depending on case (a) or (b) of Theorem 4.2. Furthermore, (4.3)

n ˜ ≤ max{p + k − 1, 2k + 2}.

Moreover, the inequalities in (4.2) and (4.3) become equalities if we let Π = {α1 , . . . , αp }, where (α1 , α2 , α3 ) = (1, i, −1) and α4 , . . . , αp lie in the open lower half plane.

20


Proof. The assertions on q˜ and n ˜ are clear. To see the last assertion, we see that in order to get a k-regular set by adding q points to Π, we need to add at least k − 1 points eiξ , with 0 < ξ < π. If 2k + 2 − p > k − 1, then p − 3 < k and we need to add an extra k − (p − 3) points eiξ , with π < ξ < 2π, giving a total of k − 1 + k − (p − 3) = 2k + 2 − p points. This proves the equality in (4.2), from which the equality in (4.3) follows. To close our paper, let us illustrate our results by the following example. Example 4.6. Let the polygon P = conv {1, w, w2 , w3 , w4 , w5 , w6 , w9 } with w = e2πi/12 , see the following. 1.5

1

imaginary axis

0.5

0

−0.5

−1

−1.5 −1.5

−1

−0.5

0 real axis

0.5

1

1.5

The polygon P Then P =

T8

π with d1 = · · · = d6 = cos 12 , d7 = d8 = cos π4 , and π 3π 5π 7π 9π 11π 15π 21π (ξ1 , . . . , ξ8 ) = , , , , , , , . 12 12 12 12 12 12 12 12 j=1 H(dj , ξj )

Thus, n πi 3πi 5πi 7πi 9πi 11πi 15πi 21πi o Π = {α1 , . . . , α8 } = e 12 , e 12 , e 12 , e 12 , e 12 , e 12 , e 12 , e 12 .

In particular, n πi 5πi 7πi 11πi o Π1 = e 12 , e 12 , e 12 , e 12

n 3πi 9πi 15πi 21πi o and Π2 = e 12 , e 12 , e 12 , e 12 ,

i.e., r = 4 and s = 2. By Theorem 4.5 and Proposition 4.4, for k ≥ 5, a (2k + 2) × (2k + 2) normal matrix A can be constructed so that Λk (A) = P. It remains to consider the cases for k ≤ 4. Clearly, Π is 2-regular. Thus, a 8 × 8 normal matrix A2 can be constructed so that Λ2 (A2 ) = P. However, Π is not k-regular for k ≥ 3. ˆ 1 ∪ Π2 is 2-regular if Π ˆ1 = Now we consider the case k = 3. Then Π 5πi Π1 \ {e 12 }. Theorem 4.5 shows that there is a 9 × 9 normal matrix A3 such that Λ3 (A3 ) = P. Indeed, following the proof of Assertion 3, we see that if 18πi Π3 = {e 12 }, Π1 ∪ Π2 ∪ Π3 is 3-regular.


3

3

2

2

1

1 imaginary axis

imaginary axis

ˆ 1 ∪ Π2 is 2-regular Finally, we turn to the case when k = 4. Notice that Π 5πi 7πi ˆ 1 = Π1 \ {e 12 , e 12 }. Thus, Theorem 4.5 shows that there is a 10 × 10 if Π normal matrix A4 such that Λ4 (A4 ) = P. In the following, we display the higher rank numerical ranges of A2 , A3 , and A4 . In the figures, the points “o” correspond to the vertices of the polygon while the points “∗” correspond to the eigenvalues of the normal matrices.

0

0

−1

−1

−2

−2

−3 −3

−2

−1

0 real axis

1

2

−3 −3

3

−2

−1

Λ2 (A2 ) = P

0 real axis

1

2

3

Λ3 (A3 ) = P

3

2

imaginary axis

1

0

−1

−2

−3 −4

−3

−2

−1

0 real axis

1

2

3

4

Λ4 (A4 ) = P Acknowledgement This research began at the 2008 IMA PI Summer Program for Graduate Students, where the second author is a lecturer and the third author is a co-organizer. The support of IMA and NSF for the program is graciously acknowledged. The hospitality of the colleagues at Iowa State University is deeply appreciated.

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References [1] M.D. Choi, M. Giesinger, J. A. Holbrook, and D.W. Kribs, Geometry of higherrank numerical ranges, Linear and Multilinear Algebra 56 (2008), 53-64. ˙ [2] M.D. Choi, J.A. Holbrook, D. W. Kribs, and K. Zyczkowski, Higher-rank numerical ranges of unitary and normal matrices, Operators and Matrices 1 (2007), 409-426. ˙ [3] M.D. Choi, D. W. Kribs, and K. Zyczkowski, Higher-rank numerical ranges and compression problems, Linear Algebra Appl. 418 (2006), 828-839. ˙ [4] M.D. Choi, D. W. Kribs, and K. Zyczkowski, Quantum error correcting codes from the compression formalism, Rep. Math. Phys., 58 (2006), 77–91. [5] H.L. Gau, C.K. Li, and P.Y. Wu, Higher-Rank Numerical Ranges and Dilations, J. Operator Theory, to appear. [6] R.A. Horn and C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, Cambridge, 1991. [7] E. Knill and R. Laflamme, Theory of quantum error-correcting codes, Phys. Rev. A 55 (1997), 900-911. [8] E. Knill, R. Laflamme, and L. Viola, Theory of quantum error correction for general noise, Phys. Rev. Lett. 84 (2000), 2525. [9] D.W. Kribs, R. Laflamme, D. Poulin, and M. Lesosky, Operator quantum error correction, e-preprint. arXiv:quant-ph/0504189v3. [10] S.R. Lay, Convex Sets and Their Applications, Pure and Applied Mathematics, John Wiley & Sons, Inc., New York, 1982. [11] C.K. Li and Y.T. Poon, Quantum error correction and generalized numerical ranges, submitted. e-preprint http://arxiv.org/abs/0812.4772. [12] C.K. Li, Y.T. Poon, and N.S. Sze, Higher rank numerical ranges and low rank perturbation of quantum channels, J. Math. Anal. Appl. 348 (2008), 843-855. [13] C.K. Li, Y.T. Poon, and N.S. Sze, Condition for the higher rank numerical range to be non-empty, Linear and Multilinear Algebra, to appear. e-preprint http://arxiv.org/abs/0706.1540. [14] C.K. Li and N.S. Sze, Canonical forms, higher rank numerical ranges, totally isotropic subspaces, and matrix equations, Proc. Amer. Math. Soc., 136 (2008), 3013-3023. [15] M.A. Nielsen and I.L. Chuang, Quantum computation and quantum information, Cambridge, New York, 2000. [16] H. Woerdeman, The higher rank numerical range is convex, Linear and Multilinear Algebra 56 (2008), 65-67. Department of Mathematics, National Central University, Chung-Li 320, Taiwan E-mail address: [email protected] Department of Mathematics, College of William & Mary, Williamsburg, VA 23185 E-mail address: [email protected] Department of Mathematics, Iowa State University, Ames, IA 50051 E-mail address: [email protected] Department of Mathematics, University of Connecticut, Storrs, CT 06269 E-mail address: [email protected]