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Linear and Multilinear Algebra
Vol. 57, No. 4, June 2009, 365–368
Condition for the higher rank numerical
range to be non-empty
Chi-Kwong Lia*, Yiu-Tung Poonb and Nung-Sing Szec
a
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Department of Mathematics, The College of William and Mary, Williamsburg,
VA, USA; bDepartment of Mathematics, Iowa State University,
Ames, IA, USA; cDepartment of Mathematics, University of Connecticut,
Storrs, CT, USA
Communicated by S. Kirkland
(Received 8 April 2007; final version received 3 November 2007)
It is shown that the rank-k numerical range of every n-by-n complex matrix is
non-empty if k 5 n/3 þ 1. The proof is based on a recent characterization of the
rank-k numerical range by Li and Sze, the Helly’s theorem on compact convex
sets, and some eigenvalue inequalities. In particular, the result implies that rank-2
numerical range is non-empty if n 4. This confirms a conjecture of Choi et al. If
k n/3 þ 1, an n-by-n complex matrix is given for which the rank-k numerical
range is empty. An extension of the result of bounded linear operators acting on
an infinite dimensional Hilbert space is also discussed.
Keywords: higher rank numerical range; eigenvalue inequalities; Helly’s theorem
AMS Subject Classification: 15A21; 15A24; 15A60; 15A90; 81P68
1. Introduction
Let Mn be the algebra of n n complex matrices. In [3], the authors introduced the notion
of the ‘rank-k numerical range’ of A 2 Mn defined and denoted by
k ðAÞ ¼ 2 C : X AX ¼ Ik ; X is n k such that X X ¼ Ik
in connection with the study of quantum error correction [4]. Evidently, 2 k(A) if and
only if there is a unitary matrix U 2 Mn such that U*AU has Ik as the leading principal
submatrix. When k ¼ 1, this concept reduces to the classical numerical range, which is
well-known to be convex by the Toeplitz–Hausdorff theorem; for example, see [7] for
a simple proof. In [1], the authors conjectured that k(A) is convex, and reduced the
convexity problem to the problem of showing that 0 2 k(A) for
Ik X
A¼
Y Ik
*Corresponding author. Email: [email protected]
ISSN 0308–1087 print/ISSN 1563–5139 online
ß 2009 Taylor & Francis
DOI: 10.1080/03081080701786384
http://www.informaworld.com
366
C.-K. Li et al.
for arbitrary X, Y 2 Mk. They further reduced this problem to the existence of a Hermitian
matrix H satisfying the matrix equation
Ik þ MH þ HM HPH ¼ H
ð1Þ
for arbitrary M 2 Mk and positive definite P 2 Mk. In [10], the author observed that
Equation (1) can be rewritten as the continuous Riccati equation
HPH HðM Ik =2Þ ðM Ik =2ÞH Ik ¼ 0k ,
ð2Þ
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and existing results on Riccati equation will ensure its solvability; for example, see
[5, Theorem 3.1]. This establishes the convexity of k(A).
For a Hermitian X 2 Mn, let 1(X) n(X) be the eigenvalues of X. In [8], it was
shown that
ð3Þ
k ðAÞ ¼ 2 C : eit þ eit k ðeit A þ eit A Þ for all t 2 ½0; 2Þ :
In particular, k(A) is the intersection of closed half planes on C, and therefore is always
convex. Moreover, if A 2 Mn is normal with eigenvalues 1, . . . , n, then
\
k ðAÞ ¼
convfj1 , . . . , jnkþ1 g:
1 j1 5 5 jnkþ1 n
This confirms a conjecture in [2].
While many interesting results have been obtained for k(A), see [1–4] for example,
there are some basic questions whose answers are unknown. The purpose of this article is
to answer the following.
Problem
Determine n and k such that k(A) is non-empty for every A 2 Mn.
It is well-known that the classical numerical range 1(A) is non-empty. For k 4 n/2,
k(A) has at most one element and one can easily construct A 2 Mn such that k(A) ¼ ;,
see Proposition 2.2 and Corollary 2.3 in [3]. The situation for k(A) with n/2 k 4 1 is
not so clear. In [2], the authors conjectured that 2(A) 6¼ ; if A 2 Mn with n 4.
In the next section, we show that k(A) is non-empty for every A 2 Mn if and
only if k 5 n/3 þ 1. In particular, it confirms the conjecture in [2] that 2(A) 6¼ ; if A 2 Mn
with n 4. We also consider extension of the result to infinite dimensional bounded linear
operators.
2. Results and proofs
THEOREM 1 Let A 2 Mn, and let k be a positive integer such that k 5 n/3 þ 1. Then k(A)
is non-empty.
Proof Evidently, k(A) 1(A). Given A 2 Mn and t 2 [0, 2), let A(t) ¼ eitA þ eitA*.
Consider the compact convex sets
SðtÞ ¼ 2 1 ðAÞ : eit þ eit k ðAðtÞÞ , t 2 ½0, 2Þ:
By (3),
k ðAÞ ¼
\
t2½0, 2Þ
SðtÞ:
Linear and Multilinear Algebra
367
By Helly’s Theorem [6, Theorem 24.9], it suffices to show that S(t1) \ S(t2) \ S(t3) 6¼ ; for
all choices of t1, t2, t3 with 0 t1 5 t2 5 t3 5 2.
For 1 j 3, let Vj be the subspace spanned by the eigenvectors of A(tj) corresponding
to the eigenvalues k(A(tj)), . . . , n(A(tj)). Then dim Vj n k þ 1. Hence, we have
dimðV1 \ V2 \ V3 Þ
¼ dimðV1 \ V2 Þ þ dim V3 dimððV1 \ V2 Þ þ V3 Þ
¼ dim V1 þ dim V2 dimðV1 þ V2 Þ þ dim V3 dimððV1 \ V2 Þ þ V3 Þ
3ðn k þ 1Þ 2n
¼ n 3k þ 3
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1:
Let v be a unit (column) vector in V1 \ V2 \ V3. Then ¼ v* Av 2 1(A) and for t1, t2, t3, we
have
eit þ eit ¼ v ðAðtj ÞÞv k ðAðtj ÞÞ:
g
Hence, 2 S(t1) \ S(t2) \ S(t3).
The following answers a question in [2].
COROLLARY 2
Let A 2 Mn with n 4. Then 2(A) 6¼ ;.
Without additional information on A 2 Mn, the bound on k in Theorem 1 is best
possible as shown by the following result.
THEOREM 3 Suppose k is a positive integer such that k n/3 þ 1. There exists A 2 Mn
such that k(A) ¼ ;.
Proof
We first consider the case when 3k ¼ n þ 3. Let w ¼ ei2/3, and
A ¼ Ik1 wIk1 w2 Ik1 :
Write A ¼ H þ iG with H ¼ H* and G ¼ G*. Then H ¼ Ik1 (1/2)I2k2. Thus,
k(H) ¼ {1/2}; see also [3, Theorem 2.4]. So,
k ðAÞ L ¼ fz : Re z ¼ 1=2g:
By rotating k(A) through an angle of 2/3 and 4/3, one can show that k(A) wL and
k ðAÞ w2 L. So,
k ðAÞ L \ wL \ w2 L ¼ ;:
Now, suppose 3k 4 n þ 3. Then we can consider a principal submatrix B 2 Mn of the
matrix A 2 M3k3 constructed in the preceding paragraph. Then k(B) k(A) ¼ ;. g
Note that we can perturb the example in the above proof to get a non-normal matrix
A 2 Mn such that k(A) ¼ ; if k n/3 þ 1. Also, Theorem 3 can be obtained from parts (1),
(2), (3) of [2, Theorem 4.7] and the fact that k(A) is a subset of
\
conv j1 , . . . , jnkþ1
1 j1 5 5 jnkþ1 n
368
C.-K. Li et al.
if A 2 Mn is normal with eigenvalues 1, . . . , n.
Let BðHÞ be the algebra of bounded linear operator acting on an infinite dimensional
Hilbert space H. One can extend the definition of k(A) for a bounded linear operator
A 2 BðHÞ by
k ðAÞ ¼ 2 C : X AX ¼ Ik , X : Ck ! H, X X ¼ Ik :
By Theorem 1, we have the following.
COROLLARY 4 Suppose k is a positive integer and A 2 BðHÞ for an infinite dimensional
Hilbert space H. Then
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k ðAÞ 6¼ ;:
Acknowledgements
We would like to thank the authors of [1] and [10] for sending us their preprints. We also thank
Professor John Holbrook for some helpful comments, and the referee for a careful reading of this
article. Research of Li was partially supported by a USA NSF grant and a HK RGC grant.
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