Kippenhahn curves of some tridiagonal matrices

arXiv:2011.00849v1 [math.FA] 2 Nov 2020 Kippenhahn curves of some tridiagonal matrices ⋆ Natália Bebianoa , Joáo da Providénciab , Ilya M. Spitkovskyc , Kenya Vazquezc a Bepartamento de Matemática, Universidade da Coimbra, Portugal b Departamento de Fı́sica, Universidade da Coimbra, Portugal c Division of Science and Mathematics, New York University Abu Dhabi (NYUAD), Saadiyat Island, P.O. Box 129188 Abu Dhabi, United Arab Emirates Abstract Tridiagonal matrices with constant main diagonal and reciprocal pairs of offdiagonal entries are considered. Conditions for such matrices with sizes up to 6-by-6 to have elliptical numerical ranges are obtained. 1. Introduction Let Mn stand for the algebra of all n-by-n matrices A with the entries aij ∈ C, i, j = 1, . . . n. We will identify A ∈ Mn with a linear operator acting on Cn , the latter being equipped with the standard scalar product h·, ·i and the associated norm kxk := hx, xi1/2 . The numerical range of A is defined as W (A) = {hAx, xi : kxk = 1}, (1.1) see e.g. [10, Chapter 1] or more recent [6, Chapter 6] for the basic properties of W (A), in particular its convexity (the Toeplitz-Hausdorff theorem) and invariance under unitary similarities. The results are partially based on the Capstone project of [KV] under the supervision of [IMS]. The latter was also supported in part by Faculty Research funding from the Division of Science and Mathematics, New York University Abu Dhabi. The work of the first author [NB] was supported by the Centre for Mathematics of the University of Coimbra — UIDB/00324/2020, funded by the Portuguese Government through FCT/MCTES. Email addresses: bebiano@mat.uc.pt (Natália Bebiano), providencia@uc.pt (Joáo da Providéncia), ims2@nyu.edu, ilya@math.wm.edu, imspitkovsky@gmail.com (Ilya M. Spitkovsky), kvn222@nyu.edu (Kenya Vazquez) ⋆ Preprint submitted to arXiv November 3, 2020 For our purposes it is important that W (A) is the convex hull of a certain curve C(A) associated with the matrix A, described as follows. Using the standard notation Re A = A + A∗ , 2 Im A = A − A∗ , 2i let λ1 (θ), . . . , λn (θ) be the spectrum of Re(eiθ A), counting the multiplicities. Then the tangent lines of C(A) with the slope cot θ are {e−iθ (λk (θ) + it) : t ∈ R}, k = 1, . . . , n. We will thus call the characteristic polynomial of Re(eiθ A),
PA (λ, θ) = det Re(eiθ A) − λI , (1.2) the NR generating polynomial of A. The connection between C(A) and W (A) was established by Kippenhahn [11] (see also the English translation [12]) and following [6, Chapter 13] we will call C(A) the Kippenhahn curve of the matrix A. In algebraic-geometrical terms, C(A) is the dual of a curve of degree n, which makes it a curve of class n. Class two curves are the same as curves of degree two, thus implying that for A ∈ M2 the Kippenhahn curve is an ellipse, in the case of normal A degenerating into the doubleton of its foci. A matrix A ∈ Mn is tridiagonal if aij = 0 whenever |i − j| > 1. We will be making use of the well known (and easy to prove) recursive relation for the determinants ∆n of such matrices, ∆n = ann ∆n−1 − an−1,n an,n−1 ∆n−2 . (1.3) In this paper, we are interested mostly in tridiagonal matrices with an additional property that their main diagonal is constant: ajj := a, j = 1, . . . , n. Let us standardize the notation as follows:   a b1 0 ... 0 c1 a b2 ... 0     ..  . .. (1.4) A(n; a, b, c) =  ... . . . . . . . .     0 0 cn−2 a bn−1  0 . . . 0 cn−1 a Since W (A − λI) = W (A) − λ for any A ∈ Mn and λ ∈ C, it suffices to concentrate our attention on matrices A(n; 0, b, c) with the zero main diagonal. For n > 3, we will restrict our attention further to what we will call reciprocal 2 matrices. Namely, in addition to A being of the form A(n; 0, b, c) we will also suppose that bj cj = 1 for all j = 1, . . . , n − 1, writing this symbolically as A = A(n; 0, b, b−1 ). (1.5) Section 2 contains some properties of the NR generating polynomials and curves of tridiagonal, in particular reciprocal, matrices. A necessary condition for ellipticity of the numerical range is also established in this section, while its concrete implementations (along with the proof of sufficiency) for 4-by-4 and 5-by-5 reciprocal matrices are provided in Section 3. The last two sections are devoted to 6-by-6 reciprocal matrices. In Section 4 criteria for the Kippenhahn curve of such matrices to contain an elliptical component are derived. By way of example it is also shown there that this component may be the exterior one thus guaranteeing the ellipticity of W (A) without all the components of C(A) being ellipses. Section 5 in turn is devoted to the case when C(A) consists of three ellipses and contains an example of a non-Toeplitz reciprocal matrix with this property. 2. Preliminary results We start with some basic observations concerning NR generating polynomials of matrices (1.4). Proposition 1. The NR generating polynomial of A(n; 0, b, c) is an even/odd function of λ if n is even (resp., odd). Proof. Matrices Re(eiθ A) − λI are tridiagonal along with A = A(n; 0, b, c). More specifically,
Re(eiθ A(n; 0, b, c)) − λI = A n; −λ, (eiθ b + e−iθ c)/2, (e−iθ b + e−iθ c)/2 and so (1.3) implies

det Re(eiθ A(n; 0, b, c)) − λI = −λ det Re(eiθ A(n − 1; 0, b, c)) − λI 2
eiθ bn−1 + e−iθ cn−1 det Re(eiθ A(n − 2; 0, b, c)) − λI . − 4
iθ Since det Re(e A(1; 0, b, c) = −λ and
2 det Re(eiθ A(2; 0, b, c) = λ2 − eiθ b1 + e−iθ c1 /4, the result follows by induction. 3 Corollary 1. The Kippenhahn curve of A = A(n; 0, b, c) is central symmetric, for n odd having the origin as one of its components. In particular, the numerical range of A(n; 0, b, c) is central symmetric — a fact observed for the first time (to the best of our knowledge) in [4, Theorem 1]. If n = 3, Corollary 1 implies that C(A) consists of an ellipse centered at the origin and the origin itself. Thus, W (A) is an elliptical disk (degenerating into a line segment if A is normal) centered at the origin. This fact also follows from [2, Theorem 4.2]. If A is a reciprocal matrix, then eiθ bj + e−iθ cj 2 = |bj |2 + |bj |−2 + 2 cos(2θ) = 2(Aj + τ ), j = 1, . . . , n − 1, (2.1) where for notational convenience we relabeled |bj |2 + |cj |2 := 2Aj and cos(2θ) = τ. (2.2) Note that Aj ≥ 1, and the extremal case A1 = . . . = An−1 = 1 (2.3) is easy, due to the following: Proposition 2. An n-by-n reciprocal matrix A is normal if and only if (2.3) holds. If this is the case, then A is in fact hermitian, and W (A) is the real π . line segment with the endpoints ±2 cos n+1 Proof. If a tridiagonal matrix (1.4) is normal, the equalities |bj | = |cj | can be obtained via a direct verification, or by applying the normality criterion from [2, Lemma 5.1]. For a reciprocal A from |bj | = |cj | it then follows that bj = cj , thus making A hermitian. Moreover, such A is unitarily similar to a tridiagonal Toeplitz matrix (1.4) with a = 0 and bj = cj = 1. The description of W (A) then follows from the well known formula λj = 2 cos πj , n+1 j = 1, . . . , n for the eigenvalues of A (see e.g. [1, Theorem 2.4]). In this case C(A) = σ(A). 4 On the other hand, the case A1 = . . . = An−1 := A0 > 1 (2.4) is covered by [2]. According to [2, Theorem 3.3], W (A) is then an elliptical disk. Following the proof of this theorem more closely, as in [3, Section 5], reveals that in fact the “hidden” components of C(A) are also all elliptical. ⌈n/2⌉ Theorem 3. Let (2.4) hold. Then C(A) = ∪j=1 σj E, where E is the ellipse p with the foci ±1 and the axes having lengths 2(A0 ± 1), while σj = cos jπ , n+1 j = 1, . . . , ⌈n/2⌉. (2.5) Note that the number of non-degenerate ellipses constituting C(A) is ⌊n/2⌋. Indeed, if n is odd, then σ⌈n/2⌉ = 0, and the respective component degenerates into the point {0}. The only difference in the statement of Theorem 3 from the material already contained in [3] is the explicit formula (2.5) for the s-numbers σj of the ⌈n/2⌉ × ⌊n/2⌋ matrix X with the entries ( 1 if i − j = 0, 1 xij = 0 otherwise. This covers in particular tridiagonal Toeplitz matrices, for which the ellipticity of their numerical ranges was established much earlier in [7]. In order to move beyond cases (2.3) and (2.4), we need to further analyze properties of NR generating polynomials (1.2). Tracking the coefficients of PA (λ, θ) in the proof of Proposition 1 when matrices under consideration are reciprocal yields the following, more precise, statement. For convenience of notation, in addition to (2.2) we also introduce ζ = λ2 , k = ⌊n/2⌋. Proposition 4. Let A ∈ Mn be a reciprocal matrix. Then its NR generating polynomial has the form Pn (ζ, τ ) = ζ k + k−1 X pj (τ )ζ j , (2.6) j=0 premultiplied by −λ in case n is odd. Here pj are polynomials in τ of degree k − j and coefficients depending only on A1 , . . . An−1 . 5 In particular, NR generating polynomials of reciprocal matrices are invariant under the change θ 7→ −θ. Combined with the central symmetry of C(A), this observation implies Corollary 2. Let A be a reciprocal matrix. Then its Kippenhahn curve C(A), and thus also the numerical range W (A), is symmetric about both coordinate axes. We also need the following simple technical observation, which is a reformulation of a well known fact repeatedly used in the numerical range related literature (see, e.g., [5] or [8]). It holds for all A ∈ Mn , not just tridiagonal matrices. Proposition 5. The Kippenhahn curve C(A) of A ∈ Mn contains an ellipse centered at the origin if and only if the NR generating polynomial of A is divisible by λ2 − (x cos(2θ) + y sin(2θ) + z) (2.7) p with x, y, z ∈ R such that z > x2 + y 2 . Combining the results stated above, we arrive at a necessary condition for a reciprocal matrix to have an elliptical numerical range. Theorem 6. Let A be a reciprocal matrix with its numerical range being an elliptical disk. Then polynomial (2.6) is divisible by ζ − (xτ + z), where √ z ± x are the lengths of the half-axes of W (A). Equivalently, Pn (xτ + z, τ ) = 0 for all τ. (2.8) Proof. According to Corollary 2, W (A) has to be axes-aligned and centered at the origin. Since its boundary E is a component of C(A), Proposition 5 implies that Pn is divisible by (2.7) with some triple x, y, z. The geometry of E means that y = 0 and the values of z ±x are as described in the statement. Finally, (2.8) follows from the remainder theorem. Treating Pn (xτ + z, τ ) as a polynomial in τ , (2.8) can be rewritten as the system of k + 1 algebraic equations in x, z, A1 , . . . , An−1 . So, the (n − 1)tuples {A1 , . . . , An−1} for which this system is consistent form an algebraic variety defined by k − 1 polynomial equations. 6 3. Reciprocal 4-by-4 and 5-by-5 matrices In full agreement with Proposition 4, polynomials (2.6) for n = 4 and n = 5 are as follows: 1 1 P4 (ζ, τ ) = ζ 2 − ζ (A1 + A2 + A3 + 3τ ) + (A1 + τ )(A3 + τ ) 2 4 (3.1) and 2 P5 (ζ, τ ) = ζ − ζ +  1 (A1 + A2 + A3 + A4 ) + 2τ 2 
1 (A1 + τ )(A3 + τ ) + (A1 + τ )(A4 + τ ) + (A2 + τ )(A4 + τ ) . (3.2) 4 Using the explicit formulas (3.1),(3.2), in the next two theorems we restate the necessary condition of Theorem 6 in a constructive way and show that it is also sufficient. The proofs of these results have a similar outline, varying in computational details only. Theorem 7. A reciprocal 4-by-4 matrix A has an elliptical numerical range if and only if √ A2 = φA1 − φ−1 A3 or A2 = φA3 − φ−1 A1 , (3.3) is the golden ratio, and at least one of the inequalities Aj ≥ 1 where φ = 5+1 2 (j = 1, 2, 3) is strict. Moreover, in this case C(A) is the union of two nested axis-aligned ellipses centered at the origin. Proof. Necessity. Case (2.3) is excluded due to Proposition 2. When divided by ζ − (xτ + z), polynomial (3.1) yields the quotient ζ − (x1 τ + z1 ), where x1 = 3 − x, 2 1 z1 = (A1 + A2 + A3 ) − z, 2 while condition (2.8) takes the form of the system: 
1 3  x x − 2 + 4 = 0
z z − 21 (A1 + A2 + A3 ) + 41 A1 A3 = 0 
 (x − 32 )z + x z − 21 (A1 + A2 + A3 ) + 41 (A1 + A3 ) = 0. 7 Solving the first two equations for x and z, respectively: √ p 3± 5 1 x= , z = (A1 + A2 + A3 ± (A1 + A2 + A3 )2 − 4A1 A3 ). (3.4) 4 4 Plugging these values of x and z into the last equation of the system reveals that it is consistent if and only if the signs in (3.4) match, and √ p 5 2 (A1 + A2 + A3 ) − 4A1 A3 = (A1 + 3A2 + A3 ). (3.5) 5 Finally, solving (3.5) for A2 yields (3.3). Sufficiency. Given (3.3) and choosing upper signs in (3.4), we arrive at the factorization of (3.1) as (ζ − (xτ + z)) (ζ − (x1 τ + z1 )). Condition A1 + A2 + A3 > 3 guarantees that (0 <) x < z. From here and the relation 4(xτ − z)(x1 τ − z1 ) = (A1 + τ )(A3 + τ ) we conclude that C(A) consists of a non-degenerate ellipse E and an ellipse E1 , degenerating into the doubleton of its foci if A1 = 1 or A3 = 1. Furthermore, E1 lies completely inside E. Indeed, 1p 1 z − z1 = 2z − (A1 + A2 + A3 ) = (A1 + A2 + A3 )2 − 4A1 A3 2 2 √ √ is bigger than 25 due to (3.5) while x − x1 = 2x − 23 = 25 , implying that xτ + z > x1 τ + z1 for all τ ∈ [−1, 1]. So, W (A) is the elliptical disk bounded by E, both E and E1 are centered at the origin due to the central symmetry of C(A), and their major axes are horizontal because x, x1 > 0. Visualizing reciprocal 4-by-4 matrices as points {A1 , A2 , A3 } in R3+ we see that those with elliptical numerical ranges form a 2-dimensional manifold M4 described by (3.3). These equations show that M4 contains the ray (2.4), as it should according to Theorem 3. The same equations imply the following Corollary 3. The intersection of M4 with any of the bisector planes Ai = Aj (i 6= j) consists only of the ray (2.4). It is worth mentioning that the ray A2 = 1, A1 = A3 > 1 corresponds to matrices with so called bi-elliptical numerical ranges, i.e. convex hulls of two non-concentric ellipses (see [9, Theorem 7]). 8 Theorem 8. A reciprocal 5-by-5 matrix A has an elliptical numerical range if and only if A1 = A4 or A1 − A4 = 2(A3 − A2 ) (3.6) and at least one of the inequalities Aj ≥ 1 (j = 1, . . . , 4) is strict. Moreover, in this case C(A) is the union of the origin and two nested axis-aligned ellipses centered there. So, in contrast with the case n = 4, a reciprocal matrix A ∈ M5 can have an elliptical numerical range while some but not all of its parameters Aj coincide. Proof. The factor −λ of the NR generating polynomial of A corresponds to {0} as a component of C(A). Having duly noted that, we proceed by using (3.2) in place of (3.1) but otherwise basically along the same lines as in the proof of Theorem 7, with the computations surprisingly being even simpler. Necessity. Case (2.3) is excluded due to Proposition 2. When divided by ζ − (xτ + z), polynomial (3.2) yields the quotient ζ − (x1 τ + z1 ), where x1 = 2 − x, 1 z1 = (A1 + A2 + A3 + A4 ) − z, 2 while condition (2.8) takes form of the system  3  x(x − 2) + 4 = 0
z z − 12 (A1 + A2 + A3 + A4 ) + 41 (A1 A3 + A1 A4 + A2 A4 ) = 0 
 (x − 2)z + x z − 21 (A1 + A2 + A3 + A4 ) + 14 (A1 A3 + A1 A4 + A2 A4 ) = 0. Solving the first two equations for x and z, respectively: 1 x=1± , 2 1 z = (A1 + A2 + A3 + A4 ± D), 4 (3.7) where D= p (A1 + A2 + A3 + A4 )2 − 4(A1 A3 + A1 A4 + A2 A4 ). Plugging these values of x and z into the last equation of the system reveals that it is consistent if and only if the signs in (3.7) match, and D = A2 + A3 . The latter condition is equivalent to (3.6). 9 Sufficiency. Given (3.6) and choosing upper signs in (3.7), we arrive at the factorization of (3.2) as P5 (ζ, τ ) = (ζ − (xτ + z)) (ζ − (x1 τ + z1 )) . (3.8) Conditions D = A2 + A3 ≥ 2 and A1 + A2 + A3 + A4 > 4 guarantee that (0 <) x < z. From here and the relation 4(xτ − z)(x1 τ − z1 ) = (A1 + τ )(A3 + τ ) + (A1 + τ )(A4 + τ ) + (A2 + τ )(A3 + τ ) we conclude that C(A) consists of {0}, a non-degenerate ellipse E and a (possibly degenerating into a doubleton) ellipse E1 . Furthermore, E1 lies inside E, with a non-empty intersection occurring only if A2 = A3 = 1. Indeed, 1 1 1 z − z1 = 2z − (A1 + A2 + A3 + A4 ) = D = (A2 + A3 ) ≥ 1 2 2 2 while x − x1 = 2x − 2 = 1. For convenience of future use, let us provide the explicit form of factorization (3.8) when (3.6) holds: (

3 +3τ ζ − A12+τ ζ − A1 +A2 +A 2

P5 (ζ, τ ) = (3.9) A3 +A4 −A2 +τ 4 +3τ ζ − 2A3 +A ζ − 2 2 if A1 − A4 equals zero or 2(A3 − A2 ), respectively. Example 1. For the 5 × 5 reciprocal matrix (1.5) with the superdiagonal string b = (1.5, 2, 2.5, 1.5) we have A1 = A4 = 1.3472, A2 = 2.125, A3 = 3.205, and according to the upper line of (3.9)    6.67722 + 3τ 1.34722 + τ ζ− . P5 (ζ, τ ) = ζ − 2 2 The Kippenhahn curve of this matrix, consisting of two ellipses and the origin, is shown in Figure 1. 10 1.0 0.5 -2 1 -1 2 -0.5 -1.0 Figure 1: b1 = 1.5, b2 = 2, b3 = 2.5, b4 = 1.5. The curves are exactly elliptical. Now let A be a reciprocal matrix (1.5) with b = (1.5, 2, 2, 3). Then A1 − A4 ≈ −3.2, A3 − A2 = 0, and so (3.6) does not hold. The Kippenhahn curve of this matrix is shown in Figure 2. Its components are non-elliptical; the best fitting ellipses are pictured as dotted curves. 1.5 1.0 0.5 -2 1 -1 2 -0.5 -1.0 -1.5 Figure 2: b1 = 1.5, b2 = 2, b3 = 2, b4 = 3. The curves look elliptical but they are not exactly elliptical. 11 4. Reciprocal 6-by-6 matrices with C(A) containing an ellipse When n increases, things get more complicated. It may become impossible to state the divisibility condition from Proposition 5 in exact arithmetic. Besides, this condition does not any longer guarantee the ellipticity of W (A). We will illustrate these phenomena for n = 6. Formula (2.6) then takes the form: 1 P6 (ζ, τ ) = ζ 3 − ζ 2 (A1 + A2 + A3 + A4 + A5 + 5τ ) 2 1 + ζ (A1 (A3 + A4 + A5 ) + A2 (A4 + A5 ) + A3 A5 + (3(A1 + A5 ) 4
1 + 2 (A2 + A3 + A4 )) τ + 6τ 2 − (A1 + τ ) (A3 + τ ) (A5 + τ ) . (4.1) 8 Consequently,
1 P6 (xτ +z, τ ) = τ 3 8x3 − 20x2 + 12x − 1 +τ 2 Q1 (x, z)+τ Q2 (x, z)+Q3 (x, z). 8 Here Q1 = q11 z + q10 , Q2 = q22 z 2 + q21 z + q20 , Q3 = z 3 + q32 z 2 + q31 z + q30 (4.2) with qij given by 3 5 q11 =3x2 − 5x + , q22 = 3x − , 2 2 1 q10 = − (A1 + A2 + A3 + A4 + A5 ) x2 2 1 1 + (3A1 + 2A2 + 2A3 + 2A4 + 3A5 ) x − (A1 + A3 + A5 ) , 4 8 1 q21 = − (A1 + A2 + A3 + A4 + A5 ) x + (3A1 + 2A2 + 2A3 + 2A4 + 3A5 ) , 4 1 q20 = (A1 A3 + A5 A3 + A1 A4 + A2 A4 + A1 A5 + A2 A5 ) x 4 1 − (A1 A3 + A5 A3 + A1 A5 ) , 8 1 q32 = − (A1 + A2 + A3 + A4 + A5 ) , 2 1 q31 = (A1 A3 + A5 A3 + A1 A4 + A2 A4 + A1 A5 + A2 A5 ) , 4 1 q30 = − A1 A3 A5 . 8 12 Condition (2.8) therefore is nothing but the requirement for all three Qj from (4.2) to vanish at the same root of the cubic equation 8x3 − 20x2 + 12x − 1 = 0. (4.3) In turn, this happens if and only if the resultant R1 of Q1 and Q2 , as well as the resultant R2 of Q1 and Q3 , are equal to zero. A somewhat lengthy but straightforward computation, with the repeated use of (4.3), reveals that R1 , R2 can be treated as quadratic polynomials in x. Namely, Rj (x) = rj2 x2 + rj1x + rj0 , j = 1, 2, (4.4) where r12 = − 8(A21 + A2 A3 + A3 A4 + A25 ) − 28(A1 A2 − A1 A4 − A2 A5 + A4 A5 ) + 4(A1 A3 + A3 A5 + A23 ) − 12(A22 + A24 − A1 A5 ) + 32A2 A4 , r11 =12(A22 + A24 ) + 18(A1 A2 − A2 A4 + A3 A5 ) − 24(A1 A4 + A2 A5 ) + 6(A1 A3 − A23 + A3 A5 ), r10 =4(A21 − A22 − A24 + A25 + A2 A3 + A3 A4 ) − (A1 A2 + A2 A4 + A4 A5 ) − 7(A1 A3 + A3 A5 ) + 6(A1 A4 − A1 A5 + A2 A5 ) + 3A23 and r22 = − 12(A31 + A35 ) − 48(A21 A2 + A4 A25 ) − 60(A1 A22 + A24 A5 ) − 16(A32 + A34 ) + 44(A1 A23 − A1 A2 A3 + A1 A2 A5 + A23 A5 + A1 A4 A5 − A3 A4 A5 ) − 36(A22 A3 − A22 A4 − A2 A24 + A3 A24 ) − 24(A2 A23 + A23 A4 − A1 A24 − A22 A5 ) + 8(A21 A4 + A2 A25 ) + 20(−A1 A2 A4 + A21 A5 + A2 A4 A5 + A1 A25 ) + 68(A1 A3 A4 + A2 A3 A5 ) − 4A33 + 40A2 A3 A4 − 84(A1 A3 A5 ), r21 =14(A31 A35 − A1 A24 − A22 A5 ) + 56(A1 A22 + A24 A5 ) + 16(A32 + A34 ) − 12(A21 A3 + A3 A25 ) + 30(A22 A3 + A3 A24 ) − 50(A1 A23 + A23 A5 ) + 20(A2 A23 + A23 A4 ) − 22(A21 + A22 A4 + A2 A24 + A2 A25 ) − 28(A1 A2 A4 + A21 A5 + A2 A4 A5 + A1 A25 ) − 66(A1 A3 A4 + A2 A3 A5 ) − 44(A1 A2 A5 + A1 A4 A5 ) + 48(A21 A2 + A4 A25 ) + 46(A1 A2 A3 + A3 A4 A5 ) + 6A33 + 158A1 A3 A5 − 52A2 A3 A4 , 13 r20 = − 2(A31 + A32 + A33 + A34 + A35 + A2 A23 + A23 A4 ) − 4(A21 A2 + A1 A22 + A1 A24 + A22 A5 + A24 A5 + A4 A25 ) + 6(A21 A3 + A1 A2 A4 − A22 A4 − A2 A24 + A1 A2 A5 + A1 A4 A5 + A2 A4 A5 + A3 A25 ) − 10(A1 A2 A3 + A3 A4 A5 − A21 A4 − A2 A25 ) − (A22 A3 + A3 A24 ) + 12(A1 A23 + A23 A5 ) + 11(A1 A3 A4 + A2 A3 A5 ) + 8(A21 A5 + A1 A25 ) + 19A2 A3 A4 − 65A1 A3 A5 . We thus arrive at the following Theorem 9. A reciprocal 6-by-6 matrix A has an elliptical component in its Kippenhahn curve C(A) if and only if both polynomials (4.4) share a common root with equation (4.3). Note that the requirement on the elliptical component of C(A) to be centered at the origin is not included in the statement of Theorem 9. It holds automatically for the following reason: if C(A) contains an ellipse E not centered at the origin, then due to the symmetry it also contains −E which is different from E. But them the third component of C(A) also has to be elliptical and, since there cannot be four, this one is then centered at the origin. Plugging in the approximate values x1 ≈ 0.0990311, x2 ≈ 0.777479, x3 ≈ 1.62349 for the roots of (4.3) into (4.4) thus delivers the numerical tests for C(A) to contain an elliptical component. Due to the structure of rij , these are systems of two homogeneous polynomial equations (one quadratic and one cubic) in five unknowns A1 , . . . , A5 . Fixing any three of Aj (or otherwise reducing the number of free parameters to two) and solving for the other two yields the specific examples of matrices satisfying Theorem 9. Example 2. Under additional constraints A1 = A5 := uA3 , A2 = A4 := vA3 the polynomials (4.4) both have the root x = x3 if and only if the pairs (u, v) are as follows: (1, 1), (8.84369, −2.49077), (−1.80414, 2.24796), and (1.7724359313231006, 0.6562336702811362). (4.5) The first solution corresponds to the situation (2.4) in which all three components of C(A) are elliptical; the next two are irrelevant because of nonpositivity. The remaining pair (4.5) delivers a non-trivial example in which 14 one component of C(A) is an ellipse while two others are not (the latter fact follows from Theorem 11 below). The respective C(A) is shown in Figure 3. Only the outer component is an exact ellipse, though deviations of two others from being elliptical are almost negligent. 2 1 -3 -2 1 -1 2 3 -1 -2 Figure 3: C(A) for A1 = A5 = 3.28117, A2 = A4 = 6.5623367, A3 = 5. 5. Reciprocal 6-by-6 matrices with C(A) consisting of three ellipses From Theorem 9 it follows that a reciprocal matrix A ∈ M6 has C(A) consisting of three concentric ellipses (and thus an elliptical numerical range) if and only if R1 (xj ) = R2 (xj ) = 0 for all three roots of (4.3). This means exactly that rjk = 0 for all j = 1, 2; k = 0, 1, 2. We have arrived at the system of six polynomial equations in five variables A1 , . . . , A5 , which therefore seems overdetermined. As was already observed in Section 4, the ellipticity of two components of C(A) automatically implies that the third one is an ellipse, centered at the origin. So, the number of equations can be easily reduced to four. An alternative approach shows, however, that in fact reciprocal 6-by-6 matrices with C(A) consisting of three ellipses are characterized by just three equations, and so form a two-parameter set. 15 Q To describe this approach, observe that (4.1) factors as 3j=1 (ζ − (xj τ + zj )) if and only if   z1 + z2 + z3 = 21 (A1 + A2 + A3 + A4 + A5 )      z1 z2 z3 = 81 A1 A3 A5      z1 z2 + z1 z3 + z2 z3 = 14 (A1 A3 + A1 A4 + A1 A5 + A2 A4 + A2 A5 + A3 A5 )    5   x1 + x2 + x3 = 2 x1 x2 x3 = 81    x1 x2 + x1 x3 + x2 x3 = 23     z1 x2 x3 + z2 x1 x3 + z3 x1 x2 = 1 (A1 + A3 + A5 )  8    1  (A1 A5 + A1 A3 + A3 A5 ) z z x + z z x + z z x = 1 2 3 1 3 2 2 3 1  8   z (x + x ) + z (x + x ) + z (x + x ) = 3 (A + A ) + 1 (A + A + A ). 1 5 2 3 4 1 2 3 2 1 3 3 1 2 4 2 The three equations of this system not containing variables zj mean exactly that x1 , x2 , x3 are the roots of (4.3). The three equations linear in zj can be rewritten as    1  1 1 1 z1 (A1 + A2 + A3 + A4 + A5 ) 2 x2 + x3 x1 + x3 x1 + x2  z2  =  3 (A1 + A5 ) + 1 (A2 + A3 + A4 ) . 4 2 1 z3 x2 x3 x1 x3 x1 x2 (A + A3 + A5 ) 1 8  Solving this system: zj = R(xj ) , 8(xj − xi )(xj − xk ) j = 1, 2, 3, (5.1) where {i, k} = {1, 2, 3} \ {j} and R(x) = r2 x2 + r1 x + r0 with r2 = 4(A1 + A2 + A3 + A4 + A5 ), r1 = 4(A2 + A3 + A4 ) − 6(A1 + A5 ), r0 = A1 + A3 + A5 . (5.2) Finally, the remaining three nonlinear equations in zj yield the following: 16r02 + 24 (r1 + r2 ) r0 + 12r12 + 15r22 + 28r1 r2 + 112 (A1 A3 + A5 A3 + A1 A4 + A2 A4 + A1 A5 + A2 A5 ) = 0; 32r02 + 8 (3r1 + r2 ) r0 + 4r12 + 3r22 + 6r1 r2 + 112 (A1 A3 + A5 A3 + A1 A5 ) = 0; 1 3 64r0 + 16(10r1 + 13r2 )r02 + 8(12r12 + 27r2 r1 + 13r22 )r0 + 8r13 + r23 64  2 2 + 12r1 r2 + 20r1 r2 + 49A1 A3 A5 = 0 16 or, plugging in the values of rj from (5.2): − (A22 + A24 ) − 2(A21 + A25 ) + 2A3 (A1 − A2 − A4 + A5 ) + 3(A1 A4 + A1 A5 + A2 A5 ) − 4(A1 A2 + A4 A5 ) + 5A2 A4 = 0, and (5.3) A22 − A23 + A24 + 3(A1 A3 + A1 A5 + A3 A5 ) (5.4) − 2(A21 + A25 + A2 A3 − A2 A4 + A3 A4 ) − (A1 + A5 )(A2 + A4 ) = 0, − A31 + A32 + A33 + A34 − A35 + (A1 − A3 + A5 )(A22 + A24 ) − 2(A2 + A4 )(A21 + A23 + A25 ) + 2A2 A4 (A1 − A3 + A5 ) − 3(A2 + A4 ) (A3 (A1 + A5 ) − A2 A4 ) − 3(A1 + A5 )(A23 + A1 A5 ) − 4(A21 A3 + A3 A25 + A1 A2 A5 + A1 A4 A5 ) + 41A1 A3 A5 = 0, (5.5) respectively. We have thus reached the conclusion. Theorem 10. A reciprocal 6-by-6 matrix A has its Kippenhahn curve consisting of three concentric ellipses if and only if Aj defined by (2.2) satisfy (5.3)–(5.5), and in addition A1 + · · · + A5 > 1. Note that the difference of (5.3) and (5.4), A23 − A3 (A1 + A5 ) − 2(A22 + A24 ) − 3(A1 A2 + A4 A5 ) + 3A2 A4 + 4(A1 A4 + A2 A5 ) = 0, (5.6) is somewhat simpler than either of these conditions, and for computational purposes it thus might be useful to replace (5.3) or (5.4) (but not both) with (5.6). The semialgebraic subset M6 of R5+ defined by the system (5.3)–(5.5) is of a more complicated structure than the pairs of hyperplanes (3.3),(3.6) corresponding to n = 4, 5 cases. Even though we do not have an explicit description of M6 , a direct verification confirms that it contains the ray A1 = . . . = A5 > 0 — as it should, in order to agree with Theorem 3. Our next result, somewhat analogous to Corollary 3, shows that this ray is in fact the intersection of M6 with either of the hyperplanes A2 = A4 and A1 = A5 . Theorem 11. If (A1 , . . . , A5 ) ∈ M6 and A2 = A4 or A1 = A5 , then all Aj coincide. 17 Proof. Equations (5.3)–(5.6) are homogeneous, so by scaling without loss of generality we may set A3 = 1. Case 1. A2 = A4 := s. Relabeling also for convenience of notation A1 = x, A5 = y, we may rewrite (5.6) as (s − 1)(x + y − (s + 1)) = 0, and conclude from there that s = 1 or x + y = s + 1. In turn, (5.4) in our abbreviated notation amounts to 4s2 − 4s − 1 + (3 − 2s)(x + y) − 2(x2 + y 2 ) + 3xy = 0. (5.7) If s = 1, (5.7) simplifies further to 2(x2 + y 2) − 3xy − (x + y) + 1 = 0, which for x, y > 0 is possible only when x = y = 1. On the other hand, plugging x + y = s + 1 into (5.7) yields xy = s, so that x = s, y = 1, or x = 1, y = s. (5.8) Since (5.5) in our setting is nothing but 8s3 −4s2 −4s+1−(x+y)3 −4s(x+y)2 +(4s2 −6s−3)(x+y)−4(x2 +y 2 )+41xy = 0, in view of (5.8) it amounts to 7(s − 1)3 = 0, thus implying s = 1. According to (5.8), then also x = y = 1. Case 2. A1 = A5 := s. Letting now A2 = x, A4 = y, rewrite (5.4) as (x + y − (s + 1))2 − 2(s − 1)2 = 0. So, x+y = s+1± √ 2(s − 1). (5.9) On the other hand, in our abbreviated notation, (5.5) is nothing but (x + y)3 + (2s − 1)(x + y)2 − (8s2 + 6s + 2)(x + y) − 8s3 + 33s2 − 6s + 1 = 0 or, plugging in the expression for x + y from (5.9): √ √ √ −1−10s+25s2 −8s3 −(s± 2(−1+s))−2s(s± 2(−1+s))−8s2 (s± 2(−1+s)) √ √ √ + 2(s ± 2(−1 + s))2 + 2s(s ± 2(−1 + s))2 + (s ± 2(−1 + s))3 = 0. 18 The left hand side of the latter equation is simply (3 ± therefore s = 1. Equation(5.6) therefore takes the form √ 2)(s − 1)3 , and −1 − 2(x + y)2 + x + y + 7xy = 0. Observing that x + y = 2 due to (5.9) and s = 1, we conclude that xy = 1, and so in fact x = y = 1. Non-trivial elements of M6 can be constructed as follows. Fix two of the parameters A1 , A2 , A4 , A5 (making sure to avoid the A1 = A5 or A2 = A4 situations), and solve (5.3)–(5.4) for the other two, expressing the solutions as function of A3 . Then find A3 as a solution to (5.5). Example 3. Set A1 = 20, A5 = 40. Then A2 = 64.9396, A3 = 36.9387548, A4 = 28.9008 deliver a solution to (5.3)–(5.5) The respective C(A) curve is plotted in Figure 4. 3 2 1 -4 2 -2 4 -1 -2 -3 Figure 4: C(A) for A1 = 10, A2 = 32.4698, A3 = 18.4693774, A4 = 14.4004, A5 = 20. 19 References [1] A. Böttcher and S. M. Grudsky, Spectral properties of banded Toeplitz matrices, SIAM, Philadelphia, 2005. [2] E. Brown and I. Spitkovsky, On matrices with elliptical numerical ranges, Linear Multilinear Algebra 52 (2004), 177–193. [3] K. A. Camenga, L. Deaett, P. X. Rault, T. Sendova, I. M. Spitkovsky, and R. B. J. Yates, Singularities of base polynomials and Gau-Wu numbers, Linear Algebra Appl. 581 (2019), 112–127. 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