有限矩陣及有界算子數值域之研究(II)

行政院國家科學委員會專題研究計畫 期中進度報告

有限矩陣及有界算子數值域之研究(2/3)

中 華 民 國 95 年 5 月 17 日

伴隨矩陣之數值域

吳培元

國立交通大學應用數學系

中文摘要：

我們證明任一

n n

×

矩陣

A

的數值域

的邊界上最多有 條

線段且

的邊界上恰好有 條線段之充份必要條件為，當 為奇

數時，

( )

W A

n

( )

W A

n

n

A

是一酉矩陣，且當 為偶數時，

n

A

是酉等價於下列兩個

伴隨矩陣

( / 2) ( / 2)

n

×

n

和

的直和，其中 滿足

a

1

≤ <

| | tan( / ) sec( / )

a

π

n

+

π

n

。

Numerical Ranges of Companion Matrices

Hwa-Long Gau* and Pei Yuan Wu*

Dept. of Mathematics Dept. of Applied Mathematics National Central University National Chiao Tung University Jhongli 32001, Taiwan Hsinchu 300, Taiwan

[email protected] [email protected]

Dedicated to Miroslav Fiedler on his 80th birthday Abstract

We show that an n-by-n companion matrix A can have at most n line segments on the boundary ∂W (A) of its numerical range W (A), and it has exactly n line segments on ∂W (A) if and only if, for n odd, A is unitary, and, for n even, A is unitarily equivalent to the direct sum A1⊕ A2 of two (n/2)-by-(n/2) companion matrices

A1 =         0 1 0 . .. . .. 1 a 0         and A2 =         0 1 0 . .. . .. 1 −1/a 0        

with 1 ≤ |a| < tan(π/n) + sec(π/n).

AMS classification: 15A60

Keywords: Numerical range; Companion matrix

∗_{Part of the results here was presented by the second author in the 12th ILAS} Con-ference at Regina, Canada in June, 2005. The research was partially supported by the National Science Council of the Republic of China under projects NSC 94-2115-M-008-010 and NSC-2115-M-009-007, respectively.

For every complex monic polynomial p(z) = zn+ a1zn−1+ · · · + an−1z + an(n ≥ 2), there is associated an n-by-n matrix

                 0 1 0 · · · · · · · 0 1 −an −an−1 · · · −a2 −a1

                 , (1)

called its companion matrix. In this paper, we consider properties of the numerical ranges of such matrices. To be more precise, we study the number of line segments on the boundary of such a numerical range. We show that for an n-by-n companion matrix, this number is at most n, and we also completely determine all the companion matrices which attain this number “n”. In the case of an odd n, this happens exactly when the companion matrix is unitary, while, for even n, the condition is that the matrix be unitarily equivalent to the direct sum of the two (n/2)-by-(n/2) companion matrices         0 1 0 . .. . .. 1 a 0         and         0 1 0 . .. . .. 1 −1/a 0         for some complex number a satisfying 1 ≤ |a| < tan(π/n) + sec(π/n).

Recall that the numerical range W (A) of an n-by-n complex matrix A is by def-inition the subset {hAx, xi : x ∈ Cn, kxk = 1} of the complex plane, where h·, ·i and k · k denote the standard inner product and norm in Cn_{. The numerical radius w(A)} of A is max {|z| : z ∈ W (A)}. It is known that the numerical range is always convex.

For other properties, the reader can consult [6, Chapter 1].

The study of the numerical ranges of the companion matrices was started in [4]. Among other things, it was shown therein that an n-by-n companion matrix A whose numerical range W (A) is a closed circular disc centered at the origin must be equal to the Jordan block of size n:

Jn =         0 1 0 . .. . .. 1 0        

(cf. [4, Theorem 2.9]). We start with an improvement of this result by weakening the assumption on A to “W (A) contains a closed circular disc D centered at the origin with the boundary ∂W (A) intersecting ∂D at more than n points”. For any matrix A, Re A denotes its real part (A + A∗)/2.

Theorem 1. If A is an n-by-n companion matrix with W (A) containing a closed circular disc D centered at the origin and with ∂W (A) ∩ ∂D having more than n points, then A = Jn.

Proof. This is done by modifying the proof of [4, Theorem 2.9]. Let A be as in (1) and let r be the radius of D. For |z| = 1, consider the expansion of det(rIn− Re (zA)) as a trigonometric polynomial p(z) in z. Since zJn−1 is unitarily equivalent to Jn−1 for all z, |z| = 1, the numerical range W (zJn−1) is a circular disc with center the origin and radius w(Re (zJn−1)). On the other hand, since Re (zJn−1) is an (n − 1)-by-(n − 1) compression of Re (zA), we infer from our assumption on W (A) that w(Re (zJn−1)) ≤ r ≤ w(Re (zA)) for all z, |z| = 1, and r = w(Re (zA)) for more than n values of z. Also, w(Re (zJn−1)) lies between w(Re (zA)) and the second largest

eigenvalue of Re (zA). Thus the same is true for r. Therefore, p(z) ≤ 0 for all z, |z| = 1, and p(z) = 0 for n values of z. By a classical result of Fej´er [7, p. 77, Problem 40], there is a polynomial q of degree n such that |q(z)|2 = −p(z) for all z. Since |q(z)|2 _{= −p(z) = 0 for more than n values of z, we conclude that q ≡ 0 and} thus p ≡ 0. In particular, the coefficients of zj _{in p for j = 0, ±1, . . . , ±n are all zero.} Using the arguments for the second half of the proof of [4, Theorem 2.9], we can show that the ak’s in A are all zero. Thus A = Jn as asserted. 

The preceding theorem is analogous to a result of Anderson’s: if A is an n-by-n matrix whose numerical range W (A) is contained in a closed circular disc D such that ∂W (A) ∩ ∂D has more than n points, then W (A) = D. A proof of this which makes use of Fej´er’s result on nonnegative trigonometric polynomials can be found in [8, Lemma 6].

An immediate corollary of Theorem 1 is the following:

Theorem 2. For any n-by-n companion matrix A, there can be at most n points in ∂W (A) ∩ ∂W (Jn−1).

In this case, Theorem 1 is applicable since Jn−1 is a compression of A and hence W (A) contains the circular disc W (Jn−1) = {z ∈ C : |z| ≤ cos(π/n)} (cf. [5, Propo-sition 1]).

Next we give an alternative proof of Theorem 2 based on the following Lemma 3. It is simpler and more direct. Moreover, the techniques involved are useful in the determining of when ∂W (A) ∩ ∂W (Jn−1) contains exactly n points for an n-by-n companion matrix A.

Lemma 3. Let A be the companion matrix given by (1). If z0cos(π/n) is a point in ∂W (A) ∩ ∂W (Jn−1), where |z0| = 1, then z0 is a zero of the polynomial

p(z) = znsinπ n − n X j=2 zn−jajsin (n − j + 1)π n .

Proof. It is easily seen that cos(π/n) is an eigenvalue of Re (z0Jn−1) with the corresponding unit eigenvector

x0 = r 2 n  z0sin π n, z 2 0sin 2π n , . . . , z n−1 0 sin (n − 1)π n T in Cn−1 _{(cf. [5, Proposition 1]). Let y} 0 = [x0, 0]T in Cn. Then hRe (z0A)y0, y0i = hRe (z0Jn−1)x0, x0i = cos

π n.

From this we deduce that cos(π/n) is an eigenvalue of Re (z0A) with the corresponding eigenvector y0, that is, it satisfies (Re (z0A) − cos(π/n)In)y0 = 0. Carrying out the computations, we obtain from the equality of the nth components the equation

z₀nsin(n − 1)π n − n X j=2 z₀n−jajsin (n − j + 1)π n = 0.

Hence z0 is a zero of p as asserted. 

Proof of Theorem 2. If ∂W (A) ∩ ∂W (Jn−1) has more than n points, then the degree-n polynomial in Lemma 3 has more than n zeros. The fundamental theorem of algebra dictates that, in particular, the leading coefficient sin(π/n) be zero, which

is a contradiction. _

We next consider the number of line segments on the boundary of the numerical range of a companion matrix. The following theorem says that this number is at most

the size of the matrix.

Theorem 4. An n-by-n companion matrix can have at most n line segments on the boundary of its numerical range.

This is the consequence of the next lemma and Theorem 2.

Lemma 5. Let A be an n-by-n matrix and let B be the (n − 1)-by-(n − 1) subma-trix of A obtained by deleting the last row and last column from A. Then every line segment of ∂W (A) intersects ∂W (B).

Proof. Let [a, b] be a line segment in ∂W (A) and let K = {x ∈ Cn _{: hAx, xi =} λkxk2 _{for some λ in [a, b]}. It is known that K is a subspace of C}n _{with dimension} at least two (cf. [2, Lemma 2]). If L = Cn−1_{⊕ {0}, then}

dim(K ∩ L) = dim K + dim L − dim(K + L) ≥ 2 + (n − 1) − n = 1.

Hence there is in K a unit vector x = x1 ⊕ 0 with x1 in Cn−1. Thus hBx1, x1i = hAx, xi ∈ [a, b], showing that [a, b] ∩ ∂W (B) 6= ∅. _

Proof of Theorem 4. Let A be an n-by-n companion matrix and let B = Jn−1. Lemma 5 says that every line segment of ∂W (A) intersects the circle ∂W (B). Our assertion then follows from Theorem 2. _

As the preceding proof shows, for an n-by-n companion matrix A every line seg-ment on ∂W (A) intersects ∂W (Jn−1). The converse is in general false, namely, not every point in ∂W (A) ∩ ∂W (Jn−1) arises as the intersection of a line segment on

∂W (A) with ∂W (Jn−1). This is illustrated by the following example.

Example 6. Let A be the 3-by-3 companion matrix associated with the polynomial p(z) = (z − (1/2))(z − 2ω)(z − 2ω2_{), where ω = (−1 +}√_{3i)/2. It can be checked that} A is unitarily equivalent to [1/2] ⊕



2ω 3 0 2ω2



. Thus W (A) is the elliptic disc with foci 2ω and 2ω2 _{and minor axis of length 3. Hence ∂W (A) ∩ ∂W (J}

2) consists of the singleton 1/2 and there is no line segment on the ellipse ∂W (A).

We remark that via Kippenhahn’s result we can show that the number of line segments on ∂W (A) for an n-by-n matrix A is at most n(n − 1)/2. It was asked in [1, p. 108] whether this number can be further reduced to 2(n − 2). As of now, nobody knows.

In the remaining part of this paper, we determine when the boundary of the nu-merical range of an n-by-n companion matrix has exactly n line segments. This is given by the following theorem.

Theorem 7. The following conditions are equivalent for an n-by-n (n ≥ 3) com-panion matrix A:

(a) ∂W (A) has n line segments on it;

(b) ∂W (A) ∩ ∂W (Jn−1) consists of n points; (c) for n odd, A =         0 1 0 . .. . .. 1 a 0        

for some a, |a| = 1, and, for n even, A =             0 1 0 1 0 · · · · · · · · 1 a 0 · · · 0 b 0 · · · 0             ,

where |a| = 1 and b is in the (n, (n/2) + 1)-position with |b| < 2 tan(π/n) and, if b 6= 0, arg b = (arg a ± π)/2;

(d) for n odd, A is unitary, and, for n even, A is unitarily equivalent to the direct sum of the two (n/2)-by-(n/2) matrices

A1 =         0 1 0 . .. . .. 1 c 0         and A2 =         0 1 0 . .. . .. 1 −1/c 0         ,

where c is a complex number satisfying 1 ≤ |c| < tan(π/n) + sec(π/n).

The implication (a)⇒(b) follows from Lemma 5 and Theorem 2. The proofs for the remaining implications (b)⇒(c), (c)⇒(d) and (d)⇒(a) are more laborious. We start with the following lemma on an expression for some determinants associated with the real part of the Jordan block. This is useful in proving the subsequent lem-mas.

Lemma 8. For any k, 1 ≤ k ≤ n − 1, we have

det((cosπ n)Ik− Re Jk) = 1 2k · sin(k+1)π_n sinπ_n .

In particular, if dk, 1 ≤ k ≤ n − 1, denotes the above determinant and d0 = 1, then dk = 2n−2k−2dn−k−2 for 0 ≤ k ≤ n − 3.

Proof. For k = n − 1, the asserted equality is obviously true since cos(π/n) is an eigenvalue of Re Jn−1 and thus both sides are equal to zero. We next consider k = n − 2. Since               cosπ_n −1 2 −1 2 cos π n · · · · · · · · · −1 2 −1 2 cos π n                             sinπ_n sin2π_n · · · sin(n−2)π_n               =               0 · · · 0 1 2sin (n−1)π n               , (2)

applying Cramer’s rule to solve for sin(π/n), we obtain

sinπ n = (−1)n−1 1₂sin(n−1)π_n (−1₂)n−3 dn−2 = ( 1 2) n−2_sin(n−1)π n dn−2 .

It follows that dn−2 = 1/2n−2 as asserted.

Assume now 1 ≤ k ≤ n − 3. In this case, we solve sin((k + 1)π/n) by Cramer’s rule to obtain sin(k + 1)π n = (−1)n−1+k 1 2sin (n−1)π n · dk(− 1 2) n−3−k dn−2 = ( 1 2) n−2−k_sin(n−1)π n · dk (1₂)n−2 = 2 k_sinπ n · dk.

Our asserted expression for dk follows immediately. 

unitarily equivalent to the companion matrix            0 1 0 . .. . .. 1 0 1

−aneinθ −an−1ei(n−1)θ · · · −a2ei2θ −a1eiθ            .

This will be used in the proofs below.

Lemma 9. Let A be the n-by-n companion matrix (1). If ∂W (A) ∩ ∂W (Jn−1) consists of n points, then aj = 0 for all j, 1 ≤ j ≤ n − 1, except possibly, when n is even, for j = n/2.

Proof. Let zkcos(π/n), 1 ≤ k ≤ n, be the n points in ∂W (A) ∩ ∂W (Jn−1), where the zk’s all have modulus one. Lemma 3 says that every zkis a zero of the polynomial

p(z) = znsinπ n − n X j=2 zn−jajsin (n − j + 1)π n ,

which, by Lemma 8, is the same as

p(z) = sinπ n(z n₋ n X j=2 zn−jaj2n−jdn−j) ≡ sin π np1(z) (3)

where dm = det((cos(π/n))Im− Re Jm) for 1 ≤ m ≤ n − 2 and d0 = 1. Let σ0 = 1 and let

σj = X k1<···<kj

zk1· · · zkj,

1 ≤ j ≤ n, be the jth elementary symmetric function of the zk’s. Hence we have

p1(z) = n Y k=1 (z − zk) = n X j=0 (−1)jσjzn−j. (4)

Equating the corresponding coefficients of p1(z) in (3) and (4) yields σ1 = 0, σn = (−1)n+1_a

n and

σj = (−1)j+1aj2n−jdn−j, 2 ≤ j ≤ n − 1. (5)

Since |zk| = 1 for all k, we have σj = σn−j/σn and thus

aj2n−jdn−j = −anan−j2jdj, 2 ≤ j ≤ n − 2. (6)

Note that σ1 = 0 implies that σn−1 = 0 and therefore an−1 = 0.

To prove that the remaining aj’s are also zero, we consider the (n − 1)-by-(n − 1) matrices Ak =               cos(π/n) −zk/2 0 −zk/2 · · an−2zk/2 · · · ... · · −zk/2 a3zk/2 −zk/2 cos(π/n) (a2zk− zk)/2 0 an−2zk/2 · · · a3zk/2 (a2zk− zk)/2 cos(π/n) + Re (a1zk)               ,

1 ≤ k ≤ n. Since |zk| = 1, the matrices zkJm and Jm are unitarily equivalent and hence det((cos(π/n))Im− Re (zkJm)) = dm for 1 ≤ m ≤ n − 2. Expanding det Ak by cofactors along its last row and then expanding the latter along their last columns, we obtain det Ak = (cosπ_n + Re (a1zk))dn−2− 1₄|a2zk− zk|2dn−3 −2Re [ n−2 P j=3 (a2zk−zk 2 )(−1) j₍ajzk 2 )(− zk 2) j−2_d n−j−1] − n−2 P j=3 1 4|ajzk| 2_d j−2dn−j−1 +2Re [ n−2 P l=3 (−1)l+1(alzk 2 )( n−2 P j=l+1 (−1)j(ajzk 2 )(− zk 2) j−l_d l−2dn−j−1)] (7) = d1dn−2+ Re (a1zk)dn−2− 1 4(|a2| 2_{− 2Re (a} 2z2k) + 1)dn−3

−1 4 n−2 X j=3 |aj|2dj−2dn−j−1 +2Re [ n−2 X j=3 aj( zk 2) j dn−j−1− 1 4 n−3 X l=2 al( n−2 X j=l+1 aj( zk 2) j−l dl−2dn−j−1)] = (d1dn−2− 1 4dn−3) + Re (a1zk)dn−2+ 1 2Re (a2z 2 k)dn−3− 1 4 n−2 X j=2 |aj|2dj−2dn−j−1 +2Re [ n−2 X j=3 (aj( zk 2 ) j_d n−j−1− 1 4 j−1 X l=2 alaj( zk 2 ) j−l_d l−2dn−j−1)] = Re (a1zk)dn−2+ 1 2Re (a2z 2 k)dn−3− 1 4 n−2 X j=2 |aj|2dj−2dn−j−1 +2Re [ n−2 X j=3 (aj 2j)dn−j−1(z j k+ j−1 X l=2 (−1)lσlzkj−l)],

where in the last equality we used the facts that d1dn−2− (1/4)dn−3 = dn−1= 0, since cos(π/n) is an eigenvalue of Re Jn−1, and

−al2l−2dl−2 = −al2n−ldn−l = (−1)lσl, 2 ≤ l ≤ n − 2, (8)

by Lemma 8 and (5). Since cos(π/n) is the maximum eigenvalue of Re (zkJn−1), we have Ak≥ 0 and thus det Ak ≥ 0 for all k. Hence

0 ≤ n X k=1 det Ak = Re (a1s1)dn−2+ 1 2Re (a2s2)dn−3− n 4 n−2 X j=2 |aj|2dj−2dn−j−1 (9) +2Re [ n−2 X j=3 (aj 2j)dn−j−1(sj + j−1 X l=2 (−1)lσlsj−l)],

where sj =Pn_k=1zkj for 1 ≤ j ≤ n − 1. Note that s1 = σ1 = 0 and the sj’s and σl’s are related by Newton’s identities:

sj = ( j−1 X

l=1

Hence sj + j−1 X l=2 (−1)lσlsj−l = sj+ j−1 X l=1 (−1)lσlsj−l = (−1)j+1jσj = jaj2j−2dj−2, 2 ≤ j ≤ n − 2, by (8). Therefore, (9) becomes 0 ≤ |a2|2dn−3− n 4 n−2 X j=2 |aj|2dj−2dn−j−1+ 2Re [ n−2 X j=3 (aj 2j)dn−j−1jaj2 j−2 dj−2] (10) = n−2 X j=2 2j − n 4 |aj| 2_d j−2dn−j−1.

For any real number x, we use bxc to denote the largest integer which is less than or equal to x. The second half of the above summation, namely,

n−2 X j=bn/2c+1 2j − n 4 |aj| 2_d j−2dn−j−1, equals b(n−1)/2c X j=2 2(n − j) − n 4 |an−j| 2 dn−j−2dj−1, (11)

which we want to express as a linear combination of the |aj|2dj−2dn−j−1’s as in the first half. For this purpose, note that |aj|2n−jdn−j = |an−j|2jdj for 2 ≤ j ≤ n − 2 from (6). Therefore, |an−j|2dn−j−2dj−1 = |aj|222n−4j d2_n−j d2 j dn−j−2dj−1 = |aj|222n−4j (22j−n−2_d j−2)2 d2 j (22j−n−2dj)(2n−2jdn−j−1) = |aj|2dj−2dn−j−1· 1 4 dj−2 dj = |aj|2dj−2dn−j−1· sin(j−1)π_n sin(j+1)π_n

with the aid of Lemma 8. Plugging this into (11), we obtain from (10) the nonnega-tivity of − b(n−1)/2c X j=2 (n − 2j 4 )(1 − sin(j−1)π_n sin(j+1)π_n )|aj| 2_d j−2dn−j−1.

Since all the terms except |aj|2 in the above summation are strictly positive, we con-clude that aj = 0 for all j, 2 ≤ j ≤ b(n − 1)/2c. By (6), we also have aj = 0 for bn/2c + 1 ≤ j ≤ n − 2. To complete the proof, we need only show that a1 = 0. Since |an| = 1, we may assume, by the remark in the paragraph preceding Lemma 9, that an = −1. Consider the cases of odd and even n separately.

Assume first that n is odd. Then, from (3),

p1(z) = zn− 2an−1d1z − an= zn+ 1.

We assume that the zeros of p1 are given by zk = e(2k−1)πi/n, 1 ≤ k ≤ n. Now we obtain from (7) that det Ak = Re (a1zk)dn−2. Hence

0 ≤ Re (a1zk) = cos

(2k − 1)π

n Re a1+ sin

(2k − 1)π n Im a1 for all k, 1 ≤ k ≤ n. Replacing k by n − k + 1 in the above, we also have

cos(2k − 1)π

n Re a1− sin

(2k − 1)π

n Im a1 ≥ 0.

Thus cos((2k − 1)π/n)Re a1 ≥ 0 for all k. Since cos((2k − 1)π/n) can be positive or negative for different values of k, we infer that Re a1 = 0. Then, from above, ± sin((2k − 1)π/n)Im a1 ≥ 0 for all k, which implies that Im a1 = 0. Hence, as as-serted, a1 = 0 for odd n.

Finally, assume that n is even. In this case, we deduce from (6) that an/2 = −anan/2 = an/2, that is, an/2 is real, and from (3) that

= (zn/2− z+)(zn/2− z−),

where z± = (2n/2an/2dn/2 ± (2na2n/2d2n/2 − 4)1/2)/2. Since the zeros zk’s of p1 have modulus one, we have |z±| = 1, which is equivalent to |2n/2an/2dn/2| ≤ 2. Hence, in particular, Re z±= 2(n/2)−1an/2dn/2. On the other hand, from (7) we have

det Ak = Re (a1zk)dn−2− 1 4a 2 n/2d(n/2)−2d(n/2)−1 +2Re (an/2 2n/2d(n/2)−1z n/2 k ), where, since z_kn/2 = z±, the last term can be simplified as

2Re (an/2 2n/2d(n/2)−1z n/2 k ) = 2 an/2 2n/2d(n/2)−1Re z± = 2an/2 2n/2d(n/2)−12 (n/2)−1_a n/2dn/2 = a2_n/2d(n/2)−1dn/2. Hence 0 ≤ det Ak = Re (a1zk)dn−2− a2n/2d(n/2)−1( 1 4d(n/2)−2− dn/2) = Re (a1zk)dn−2

by Lemma 8. Because dn−2> 0, we have Re (a1zk) ≥ 0 for all k, 1 ≤ k ≤ n. If z+= eiθ0 _{for some real θ}

0, then z− = e−iθ0 and the zk’s are equal to uj ≡ e(2θ0+4jπ)/n and vj ≡ e(−2θ0+4jπ)/n, 0 ≤ j ≤ (n/2)−1. Since uj = v(n/2)−j, both Re (a1uj) and Re (a1uj) (= Re (a1v(n/2)−j)) are nonnegative. Hence (Re a1) cos((2θ0 + 4jπ)/n) ≥ 0 for all j. Since different values of j yield positive and negative values of cos((2θ0+ 4jπ)/n), we infer that Re a1 = 0. Then

Re (a1uj) = (Im a1) sin 2θ0+ 4jπ n ≥ 0 and Re (a1uj) = −(Im a1) sin 2θ0+ 4jπ n ≥ 0

for all j. Hence Im a1 = 0 and, therefore, a1 = 0. This completes the proof. 

We now resume the proof of Theorem 7.

Proof of Theorem 7, (b)⇒(c). If n is odd, then, as proved in Lemma 9,

A =         0 1 0 . .. . .. 1 −an 0         with |an| = 1 as required.

Now assume that n is even. From Lemma 9, we have

A =             0 1 0 · · · · · · · · · · 1 −an 0 · · · 0 −an/2 0 · · · 0            

with |an| = 1. Let an= eiθ0 with θ0 real and let θ = (π − θ0)/n. Then eiθA is unitarily equivalent to A0 =             0 1 0 · · · · · · · · · · 1 1 0 · · · 0 −ian/2e−iθ0/2 0 · · · 0            

(cf. the paragraph before Lemma 9). If b0 = −ian/2e−iθ0/2, then Lemma 3 as applied to A0 yields that the zeros of the polynomial p1(z) = zn + zn/2b0cot(π/n) + 1 are

distinct and have modulus one. However, the zeros of p1 are the (n/2)th roots of (−b0cot(π/n) ± (b02cot2_{(π/n) − 4)}1/2_{)/2. Thus we must have |b}0_{cot(π/n)| < 2 or} |b0_{| < 2 tan(π/n). On the other hand, (6) as applied to A}0 _{with j = n/2 yields that} b0 (= −ian/2e−iθ0/2) is real. Hence for nonzero b0 we have arg an/2 = (θ0 ± π)/2. Let-ting a = −an and b = −an/2, we conclude that |a| = 1, |b| < 2 tan(π/n) and, if b 6= 0,

arg b = (θ0± π)/2. 

We next prove the implication (c)⇒(d) of Theorem 7.

Proof of Theorem 7, (c)⇒(d). We need only prove the case for even n. Considering eiθ_{A with θ = (π − arg a)/n instead of A, we may assume that a = 1 and b is real} (cf. the paragraph before Lemma 9). Let c = (b ± (b2_{+ 4)}1/2_{)/2 with the “+” sign if} b ≥ 0 and “−” sign if b < 0. Then

1 ≤ |c| = 1 2|b ± (b 2_{+ 4)}1/2_| ≤ 1 2(|b| + |b 2_{+ 4|}1/2_{) < tan}π n + sec π n and b = c − (1/c). Let d = 1/(1 + c2)1/2 and

U = d   In/2 cIn/2 cIn/2 −In/2  .

Then U is unitary and U A = (A1⊕ A2)U , completing the proof. 

Lemma 10. Let A1 =         0 1 0 . .. . .. 1 c 0         and A2 =         0 1 0 . .. . .. 1 −1/c 0        

be (n/2)-by-(n/2) matrices, where n (≥ 4) is even and c is real satisfying 1 ≤ c < tan(π/n) + sec(π/n). Let z0 be a zero of p1(z) = zn+ zn/2(c − (1/c)) cot(π/n) + 1 and let x =  z0sin π n, z 2 0sin 2π n , . . . , z n/2 0 sin n 2π n T , y =  z₀(n/2)+1cosπ n, z (n/2)+2 0 cos 2π n , . . . , z n−1 0 cos (n₂ − 1)π n , 0 T ,

u = (x + cy)/kx + cyk and v = (cx − y)/kcx − yk be vectors in Cn/2. Then

hz0A1u, ui = cos π n − i ncIm (z₀n/2) sinπ_n n 2(1 + c2) + (1 − c2) csc2( π n) and hz0A2v, vi = cos π n + i ncIm (z₀n/2) sin π_n n 2(1 + c2) + (c2− 1) csc2( π n) .

Proof. Since 1 ≤ c < tan(π/n) + sec(π/n), we have 0 ≤ c − tan(π/n) < sec(π/n) and therefore c2−2c tan(π/n)+tan2_{(π/n) < sec}2_{(π/n) or c}2_{−2c tan(π/n) < 1. Hence} (c − (1/c)) cot(π/n) < 2. Thus z₀n/2 = −1 2(c − 1 c) cot π n ± 1 2i(4 − (c − 1 c) 2 cot2π n) 1/2

and, in particular, z0 has modulus one. Since

hz0A1u, ui = 1

kx + cyk2(hz0A1x, xi + chz0A1x, yi + chz0A1y, xi + c 2_hz

we need compute the values of kx + cyk and the four inner products above. To obtain the former, note that

kxk2 ₌ n/2 X j=1 |z0|2jsin2( jπ n ) = 1 2 n/2 X j=1 (1 − cos2jπ n ) = n 4 − 1 2Re ( 1 − e(1+(2/n))πi 1 − e2πi/n − 1) = n 4 − 1 2(−1) = 1 4(n + 2), kyk2 ₌ (n/2)−1 X j=1 |z0|n+2jcos2( jπ n ) = 1 2 (n/2)−1 X j=1 (1 + cos2jπ n ) = 1 4(n − 2), hx, yi = zn/2₀ (n/2)−1 X j=1 sin(jπ n ) cos( jπ n ) = 1 2z n/2 0 (n/2)−1 X j=1 sin2jπ n = 1 2z n/2 0 Im ( 1 − eπi 1 − e2πi/n − 1) = 1 2z n/2 0 cot π n, and

kx + cyk2 _{= kxk}2_{+ 2cRe hx, yi + c}2_kyk2 = 1 4(n + 2) + c cot π n · Re (z n/2 0 ) + 1 4(n − 2)c 2 = n 4(1 + c 2_{) +} 1 2(1 − c 2_{) + c cot}π n(− 1 2(c − 1 c) cot π n) = n 4(1 + c 2_{) +} 1 2(1 − c 2_{) csc}2₍π n). Moreover, we have hz0A1x, xi = ( (n/2)−1 X j=1 sinjπ n sin (j + 1)π n ) + cz n/2 0 sin π nsin π 2

= czn/2₀ sinπ n − 1 2 (n/2)−1 X j=1 (cos(2j + 1)π n − cos π n) = czn/2₀ sinπ n − 1 2Re (e 3πi/n_· 1 − e (2πi/n)(n−2)/2 1 − e2πi/n ) + 1 2( n 2 − 1) cos π n = czn/2₀ sinπ n + n 4 cos π n, hz0A1x, yi = z n/2 0 (n/2)−1 X j=1 sin(j + 1)π n cos jπ n = 1 2z n/2 0 (n/2)−1 X j=1 (sin (2j + 1)π n + sin π n) = 1 2z n/2 0 (Im (e 3πi/n_· 1 − e(2πi/n)(n−2)/2 1 − e2πi/n ) + ( n 2 − 1) sin π n) = 1 2z n/2 0 (csc π n + ( n 2 − 2) sin π n), hz0A1y, xi = zn/20 ( (n/2)−2 X j=1 cos(j + 1)π n sin jπ n ) + c cos π n = c cosπ n + 1 2z n/2 0 (n/2)−2 X j=1 (sin (2j + 1)π n − sin π n) = c cosπ n + 1 2z n/2 0 ((csc π n − sin π n − sin (n − 1)π n ) − ( n 2 − 2) sin π n) = c cosπ n + 1 2z n/2 0 (csc π n − n 2sin π n), and hz0A1y, yi = (n/2)−2 X j=1 cos(j + 1)π n cos jπ n = 1 2 (n/2)−2 X j=1 (cos(2j + 1)π n + cos π n) = (n 4 − 1) cos π n. Hence hz0A1u, ui = 1 kx + cyk2[(cz n/2 0 sin π n + n 4cos π n) + 1 2cz n/2 0 (csc π n + ( n 2 − 2) sin π n)

+c(c cosjπ n + 1 2z n/2 0 (csc π n − n 2 sin π n)) + c 2₍n 4 − 1) cos π n] = 1 kx + cyk2( n 4(1 + c 2_{) cos} π n + 1 2c(z n/2 0 + z n/2 0 ) csc π n + n 4c(z n/2 0 − z n/2 0 ) sin π n) = 1 kx + cyk2( n 4(1 + c 2_{) cos} π n + cRe (z n/2 0 ) csc π n − 1 2nciIm (z n/2 0 ) sin π n) = 1 kx + cyk2( n 4(1 + c 2 ) cos π n − 1 2c(c − 1 c) cot π ncsc π n − 1 2nciIm (z n/2 0 ) sin π n) = 1 kx + cyk2(( n 4(1 + c 2_{) +} 1 2(1 − c 2_{) csc}2₍π n)) cos π n − 1 2nciIm (z n/2 0 ) sin π n) = cosπ n − i ncIm (z₀n/2) sinπ_n n 2(1 + c 2_{) + (1 − c}2_{) csc}2₍π n) as asserted.

In a similar fashion, we derive that

kcx − yk2 ₌ n 4(1 + c 2_{) +} 1 2(c 2_{− 1) csc}2₍π n), hz0A2x, xi = − 1 cz n/2 0 sin π n + n 4cos π n, hz0A2x, yi = 1 2z n/2 0 (csc π n + ( n 2 − 2) sin π n), hz0A2y, xi = − 1 ccos π n + 1 2z n/2 0 (csc π n − n 2sin π n), and hz0A2y, yi = ( n 4 − 1) cos π n.

The asserted expression for hz0A2v, vi can be proved analogously as before. 

Proof of Theorem 7, (d)⇒(a). If A is unitary, then A =         0 1 0 . .. . .. 1 −an 0        

with |an| = 1 and ∂W (A) is a regular n-gon (cf. [4, Corollary 1.2]). For the re-maining part of the proof, we assume that n is even and A = A1 ⊕ A2, where A1 and A2 are as in (d). Multiplying A by an eiθ with θ = − arg c, we may further assume that c is positive. If c = 1, then A1 and A2, and hence A, are all unitary, in which case ∂W (A) has n line segments. Under the hypotheses that n ≥ 4 and 1 < c < tan(π/n) + sec(π/n), we have 0 < (c − (1/c)) cot(π/n) < 2. Since the zeros of the polynomial p1(z) = zn+ zn/2(c − (1/c)) cot(π/n) + 1 are the (n/2)th roots of (−(c − (1/c)) cot(π/n) ± ((c − (1/c))2_cot2_{(π/n) − 4)}1/2_{)/2, we infer that they are all} distinct and have modulus one. These we denote by zk, 1 ≤ k ≤ n.

We now show that cos(π/n) is a multiple eigenvalue of Re (zkA) for any k. Indeed, if xk =  zksin π n, z 2 ksin 2π n , . . . , z n/2 k sin n 2π n T , yk=  z_k(n/2)+1cosπ n, z (n/2)+2 k cos 2π n , . . . , z n−1 k cos (n₂ − 1)π n , 0 T ,

uk = (xk+ cyk)/kxk+ cykk and vk = (cxk− yk)/kcxk− ykk, then it is easily checked that Re (zkA1)uk = cos(π/n)uk and Re (zkA2)vk = cos(π/n)vk, where for the equal-ity of the (n/2)th components we need that zk be a zero of p1. Hence cos(π/n) is a multiple eigenvalue of Re (zkA).

Next note that cos(π/n) is the maximum eigenvalue of Re (zkA). To prove this, let c1 ≥ c2 ≥ · · · ≥ cn and d1 ≥ d2 ≥ · · · ≥ dn−1 be the eigenvalues of Re (zkA) and

Re (zkJn−1), respectively. Since Re (zkJn−1) is unitarily equivalent to Re Jn−1, the dj’s are all distinct and d1 = cos(π/n) (cf. [3, Corollary 2.7]). On the other hand, we proved in the preceding paragraph that cos(π/n) = cj0 = cj0+1 for some j0. If j0 > 1,

then from the interlacing of the cj’s and the dj’s: c1 ≥ d1 ≥ c2 ≥ d2 ≥ · · · ≥ cn−1 ≥ dn−1 ≥ cn, we obtain d1 = c2 = d2 = · · · = cj0+1 = cos(π/n), which contradicts the

distinctness of the dj’s. Hence j0 ≤ 1 and therefore c1 = cos(π/n) as required. In particular, we have cos(π/n) = max W (Re (zkA)) = max Re W (zkA).

Finally, we check that W (A) has n line segments on its boundary. For this, consider u0_k = uk⊕ 0 and v0k= 0 ⊕ vk as vectors in Cn. Then

hzkAu0k, u 0 ki = hzkA1uk, uki = cos π n − i ncIm (z_kn/2) sinπ_n n 2(1 + c2) + (1 − c2) csc2( π n) and hzkAvk0, v 0 ki = hzkA2vk, vki = cos π n + i ncIm (z_kn/2) sinπ_n n 2(1 + c2) + (c2− 1) csc2( π n) by Lemma 10. Hence Re hzkAu0k, u 0 ki = Re hzkAvk0, v 0 ki = cos π n = max Re W (zkA) and Im hzkAu0k, u 0 ki 6= Im hzkAvk0, v 0 ki.

Therefore, the vertical line Re z = cos(π/n) yields a line segment on ∂W (zkA). Thus ∂W (A) has n line segments given by Re (zkz) = cos(π/n), 1 ≤ k ≤ n. This completes

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