Nonnegative block shift matrix

We start by reviewing a couple of basic facts on a block shift matrix. Recall that a block shift matrix A is one of the form



Then it is easy to see that the numerical range W (A) is an m-symmetric compact convex region since U^∗AU = e^iϕA, where U is a unitary matrix of the form

where the diagonal identity matrix Ij is of the same size as the corresponding 0j

(j = 1, ..., m). Let hAx, xi be a boundary point of W (A), where x = ⊕^mk=1xj is a unit vector. We define x0φ = x and xjϕ = ⊕^mk=1e^i(k−1)jϕxk for j = 1, ..., m − 1. With these notations, we can give a lower bound for k(A).

Proposition 3.2.1. Let A be a block shift of the above form with the corresponding notations as above. Then kxkk is equal to 1/√

m for all k = 1, ..., m if and only if the vectors xpϕ, 0 ≤ p ≤ m − 1, are orthonormal. In this case, we have k(A) ≥ m .

Proof. Assume that hx^pϕ, xqϕi = 0 for 0 ≤ p 6= q ≤ m − 1. This is equivalent to the above polynomial with those of kx^mk²Qm−1

j=1 (t −e^ijϕ). Conversely, if kx^kk is equal to 1/√

m for all k = 1, ..., m, then it is a routine matter to check that xpϕ and xqϕ

are orthonormal for 0 ≤ p 6= q ≤ m − 1. Clearly, in this case, k(A) has a lower bound

m. 

Recall that the numerical radius ω(A) of a matrix A is the quantity max {|z| : z ∈ W (A)}. For a nonnegative matrix with irreducible real part, [16, Lemma 1] says that, for ω(A)e^iθ in W (A), where θ is a real number with e^iθ 6= 1, (a) if θ is an irrational multiple of 2π, then A is permutationally similar to a matrix of the form

(1)

where the diagonal zeros are zero square matrices, and, in particular, W (A) is a circu-lar disc centered at the origin, and (b) if θ is a rational multiple of 2π, say, θ = 2πp/q, where p and q are relatively prime integers and q ≥ 2, then A is permutationally sim-ilar to

and, in particular, W (A) = e^2πi/qW (A).

The following lemma is a generalization of [19, Lemma 3.6], which is useful for the proof of Proposition 3.2.3. Recall that a vector x with positive components, denoted by x ≻ 0, is called positive.

Lemma 3.2.2. Let A be an n-by-n (n ≥ 2) nonnegative matrix of the form (1) with irreducible real part and m ≥ 2. Then the following hold:

(a) W (A) = {z ∈ C : |z| ≤ ω(A)}.

(b) There is a unique positive vector x = x1⊕ · · · ⊕ x^m ∈ Cⁿ such that hAx, xi = ω(A).

(c) For any a = ω(A)e^iθ, θ ∈ [0, 2π), in ∂W (A), if x^θ = x1⊕e^iθx2⊕· · ·⊕e^i(m−1)θxm, then a = hAxθ, x_θi and Ha is generated by x_θ.

(d) Let aj = ω(A)e^iθj (θj ∈ [0, 2π)), j = 1, 2, be two points in ∂W (A) with the corresponding unit vector x_θj. Then x_θ1 and x_θ2 are orthogonal to each other if and only if e^i(θ1−θ2) is a zero of the polynomial kx¹k²+ kx²k²t + · · · + kx^mk²t^m−1.

Proof. Since U_θ^∗AUθ = e^iθA for any θ, where Uθ = ⊕^mk=1e^i(k−1)θIk, that is, A is unitarily similar to e^iθA for any θ, (a) follows immediately. (b) is a consequence of [11, Proposition 3.3]. To prove (c), note that

a = ω(A)e^iθ = he^iθAx, xi = hUθ^∗AUθx, xi = hA(U^θx), (Uθx)i = hAx^θ, xθi, which shows that xθ is in Ha. That dim Ha = 1 is by [11, Corollary 3.10]. Thus Ha

is generated by xθ. (d) follows from the fact that hx^θ1, xθ2i = Pm

k=1e^{i(k−1)(θ1−θ2)}x²_k.

This completes the proof. 

Thus, for a nonnegative matrix A of the form (1) with irreducible real part, k(A) equals the maximum number of θ1, ..., θk in [0, 2π) for which e^i(θj−θl) is a zero of p(t) ≡ kx1k² + kx2k²t + · · · + kxmk²t^m−1 for all j and l, 1 ≤ j 6= l ≤ k. If m = 2, then the polynomial p(t) has degree one. Hence k(A) = 2 if m = 2. The following

proposition says that k(A) ≤ m − 1 if m ≥ 3.

Proposition 3.2.3. Let A be an n-by-n (n ≥ 2) nonnegative matrix of the form (1), where m ≥ 3, with irreducible real part and ω(A) = 1. Then k(A) ≤ m − 1.

Proof. From our assumptions on A and Lemma 3.2.2 (b), there is a unique vector x = x₁ ⊕ · · · ⊕ xm ∈ Cⁿ with positive x_j for all j such that hAx, xi = 1. Letting k(A) = k, we may assume, by the proof of [19, Theorem 3.10] that θ0 = 0, θ1 = 2π/k, ..., θ_k−1 = 2(k − 1)π/k, so that xθj and x_θl are orthogonal to each other for all j and l, 0 ≤ j 6= l ≤ k − 1. From Lemma 3.2.2 (d), this yields that k(A) ≤ m since the degree of p(t) is m − 1. Assume that k(A) = m. If m is odd, then the degree of the polynomial p(t) is equal to the even m − 1. We note that −1 is a zero of p(t) and the zeros of the real polynomial p(t) appear in conjugate pairs, which is a contradiction. Hence k(A) ≤ m − 1 for odd m. On the other hand, if m is even, then kx1k = · · · = kxmk = 1/√m by examining the coefficients of p(t). From the assumption that ω(A) = 1, we have (Re A)x = x by [11, Proposition 3.3]. That is,

(i) (A1/2)x2 = x1,

(ii) (A^T_j/2)x_j + (A_j+1/2)x_j+2 = x_j+1 for 1 ≤ j ≤ m − 2, and (iii) (A^T_m−1/2)xm−1 = xm.

Taking the transpose of (i) and then multiplying x1 from right on both sides, we obtain x^T₂(A^T₁/2)x₁ = kx1k². Next, multiplying (ii) on both sides by x^T_j, we have x^T_j(Aj/2)xj+1 = 0 if j is even and x^T_j(Aj/2)xj+1= kx^jk²if j is odd, where 2 ≤ j ≤ m.

Similarly, we multiply (iii) on both sides by x^T_m to get x^T_m−1(A_m−1/2)x_m = kxm−1k². These are the same as

hAjx_j+1, x_ji = 0 for j = 2, 4, ..., m − 2, and hA^jxj+1, xji = kx^jk² for j = 1, 3, ..., m − 1.

This implies that A2 = A4 = · · · = A^m−2 = 0, which contradicts the assumption of the irreducibility of the real part of A. Hence k(A) ≤ m − 1 for even m. We complete

the proof. 

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Our next result is a characterization of n-by-n (n ≥ 2) nonnegative matrices A of the form (1) with irreducible real part for which k(A) = m − 1, where m ≥ 3.

Theorem 3.2.4. Let A be an n-by-n (n ≥ 2) nonnegative matrix of the form (1), where m ≥ 3, with irreducible real part and ω(A) = 1. Then the following are equivalent:

(a) k(A) = m − 1.

(b) there is an α > 0 satisfying p(−α) = 0, where

p(t) =







1

2(m−1) +_m−1¹ t + · · · +m−1¹ t^m−2+_2(m−1)¹ t^m−1 if m is even, or

1

2(1+α)(m−1) + _m−1¹ t + · · · + _m−1¹ t^m−2+2(1+α)(m−1)^α t^m−1 if m is odd.

(c) If x = x1 ⊕ · · · ⊕ x^m ∈ Cⁿ is the (unique) positive unit vector such that hAx, xi = 1, then its component vectors satisfy

(i) kx¹k = kx^mk = 1/p2(m − 1) and kxjk = 1/√

m − 1, 2 ≤ j ≤ m − 1, if m is even,

(ii) kx¹k = 1/p(1 + α)(m − 1), kx^mk = pα/((1 + α)(m − 1)), and kx^jk = 1/√

m − 1, 2 ≤ j ≤ m − 1 for some α > 0, if m is odd.

Proof. If k(A) = m − 1, then, for even m, zeros of the corresponding polynomial, denoted by pe(t), are exactly −1 and those points which are equally distributed over the unit circle by the proof of Proposition 3.2.3. In other words, the polynomial (t + 1)Qm−2

j=1 (t − ω^j), where ω = e^2πi/(m−1), has the same zeros as pe(t). This implies that p_e(t) = _2(m−1)¹ +_m−1¹ t+· · ·+_m−1¹ t^m−2+_2(m−1)¹ t^m−1 for even m since kxk = kx1k²+

· · · + kx^mk² = 1. Conversely, if pe(t) = _2(m−1)¹ + _m−1¹ t + · · · + _m−1¹ t^m−2+ _2(m−1)¹ t^m−1 for even m, then it is clear that k(A) = m − 1. On the other hand, for odd m zeros of the corresponding polynomial, denoted by po(t), are exactly −α and those points which are equally distributed over the unit circle and −α by the proof of Proposition 3.2.3. This α must be positive since the coefficients of po(t) are nonnegative. That is, the polynomial (t + α)Qm−2

j=1 (t − ω^j), where ω = e^2πi/(m−1), has the same zeros as

po(t). It follows that po(t) = 2(1+α)(m−1)¹ +_m−1¹ t + · · · +_m−1¹ t^m−2+2(1+α)(m−1)^α t^m−1 for odd m since kxk = kx1k² + · · · + kxmk² = 1. The converse is obvious. This proves the equivalence of (a) and (b). The equivalence of (b) and (c) is obvious by the above

arguments. 

An easy consequence of Theorem 3.2.4 is that we can give a necessary condition for k(A) = m − 1 by dealing with the norms of blocks, which are similar to [19, Theorem 3.10]

Corollary 3.2.5. Let A be an n-by-n (n ≥ 2) nonnegative matrix of the form (1), where m ≥ 3, with irreducible real part and ω(A) = 1. If k(A) = m − 1, then either

(a) m is even, kA¹k = kA^m−1k ≥√

2 and kA²k = · · · = kA^m−2k ≥ 1, or (b) m is odd, kA1k ≥ 2/√

1 + α, kA2jk ≥ 2α/(1 + α), kA2j+1k ≥ 2/(1 + α) for 1 ≤ j ≤ (n − 3)/2, and kA^m−1k ≥ 2pα/(1 + α) for some α > 0.

Proof. Let k(A) = m − 1. If ω(A) = 1, then (Re A)x = x by [11, Proposition 3.3]

or, equivalently,

(i) (A1/2)x2 = x1,

(ii) (A^T_j/2)xj + (Aj+1/2)xj+2 = xj+1 for 1 ≤ j ≤ m − 2, and (iii) (A^T_m−1/2)xm−1 = xm.

Assume first that m is even. Then after some computations, we obtain that hA^jxj+1, xji

= 1/(m − 1) for 1 ≤ j ≤ m − 1. This along with Theorem 3.2.4 (c) proves case (a).

Similar arguments apply for odd m. We complete the proof. 

According to the above discussions and [19, Section 3], it is natural to ask whether k(A) ≤ q holds for a matrix A of the form (2). The following gives a counterexample.

36

Example 3.2.6. Let

It is easy to check that A is a nonnegative normal matrix with irreducible real part and σ(A) = {1, ω, · · · , ω⁵}, where ω = e^2πi/6. Hence k(A) = 6  3 by Proposition 2.2.1.

In the next section, we consider the numerical ranges of certain special nonnegative matrices, namely, those of doubly stochastic matrices.