On Numerical Radius of a Matrix and Estimation of Bounds for Zeros of a Polynomial

Abstract

We obtain inequalities involving numerical radius of a matrix 𝐴∈𝑀_𝑛(ℂ). Using this result, we find upper bounds for zeros of a given polynomial. We also give a method to estimate the spectral radius of a given matrix 𝐴∈𝑀_𝑛(ℂ) up to the desired degree of accuracy.

1. Introduction

Suppose 𝐴∈𝑀_𝑛(ℂ). Let  𝑊(𝐴), 𝜎(𝐴) denote respectively the numerical range, spectrum of  𝐴  and 𝑤(𝐴), 𝑟_𝜎(𝐴) denote respectively the numerical radius, spectral radius of 𝐴, that is, 𝑊(𝐴)=⟩∶‖𝑥‖},𝑤(𝐴)𝜆|(𝐴,𝜎(𝐴)=,𝑟{⟨𝐴𝑥,𝑥=1=sup{|∶𝜆∈𝑊)}𝜆∶𝜆isaneigenvalueof𝐴_𝜎(𝐴)=sup{|𝜆|∶𝜆∈𝜎(𝐴)}. (1.1) It is well known that(i)‖𝐴‖/2≤𝑤(𝐴)≤‖𝐴‖.

Kittaneh [1] improved on the second inequality to prove that.(ii)𝑤(𝐴)≤1/2‖𝐴‖+1/2‖𝐴²‖^1/2.

Clearly, (1/2)‖𝐴‖+(1/2)‖𝐴²‖^1/2≤‖𝐴‖ so that inequality (ii) is sharper than the second inequality of (i).

Let 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀ be a monic polynomial where 𝑎₀,𝑎₁,…,𝑎_𝑛−1 are complex numbers and let 𝐶𝑝=−𝑎_𝑛−1−𝑎_𝑛−2⋅⋅−𝑎₁−𝑎₀10⋅⋅0001⋅⋅0000⋅⋅00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅10 (1.2) be the Frobenius companion matrix of the polynomial 𝑝(𝑧). Then, it is well known that zeros of  𝑝  are exactly the eigenvalues of 𝐶(𝑝). Considering 𝐶(𝑝) as an element of 𝑀_𝑛(ℂ), we see that if 𝑧 is root of the polynomial equation 𝑝(𝑧)=0, then |𝑧|≤𝑤𝐶𝑝,|𝑧|≤𝑟_𝜎𝐶𝑝. (1.3) Based on inequality (ii), Kittaneh [1] obtained an estimation for 𝑤(𝐶(𝑝)) which gives an upper bound for zeros of the polynomial 𝑝(𝑧).

In Section 1 we find numerical radius of some special class of matrices and use the results obtained to give a better estimation of bounds for zeros of a polynomial.

2. On Numerical Radius of a Matrix

We first obtain bounds for numerical radius of a matrix in 𝑀_𝑛(ℂ) and use it to obtain numerical radius for some special class of matrices.

Theorem 2.1. Suppose 𝑇∈𝑀_𝑛(ℂ) and 𝑇=𝐴𝐵𝐶𝐷, (2.1) where 𝐴∈𝑀_𝑟(ℂ),  𝐵∈𝑀_{𝑟,𝑛−𝑟}(ℂ),  𝐶∈𝑀_{𝑛−𝑟,𝑟}(ℂ)  and  𝐷∈𝑀_𝑛−𝑟(ℂ). Then,(i)𝑤(𝑇)≤(1/2)[𝑤(𝐴)+𝑤(𝐷)+(𝑤(𝐴)−𝑤(𝐷))²+(‖𝐵‖+‖𝐶‖)²]  and (ii)‖𝑇‖²≤(1/2)(‖𝐴‖²+‖𝐵‖²+‖𝐶‖²+‖𝐷‖²)  +  (1/2)(‖𝐴‖²+‖𝐶‖²−‖𝐵‖²−‖𝐷‖²)²+4(‖𝐴‖‖𝐵‖+‖𝐶‖‖𝐷‖)².

Proof. (i) Let 𝑍∈ℂ^𝑛 and 𝑋𝑌𝑍=, (2.2) where 𝑋∈ℂ^𝑟 and 𝑌∈ℂ^𝑛−𝑟 with ‖𝑍‖=1.
Then, ⟨⟩=,𝑋𝑌𝑇𝑍,𝑍=⟨⟩+⟨⟩+⟨⟩+⟨⟩𝐴𝑋+𝐵𝑌𝐶𝑋+𝐷𝑌𝐴𝑋,𝑋𝐵𝑌,𝑋𝐶𝑋,𝑌𝐷𝑌,𝑌 (2.3) and so |⟨𝑇𝑍,𝑍⟩|≤|⟨𝐴𝑋,𝑋⟩|+|⟨𝐷𝑌,𝑌⟩|+‖𝐵‖‖𝑋‖‖𝑌‖+‖𝐶‖‖𝑋‖‖𝑌‖. (2.4) Therefore, we have 𝑤(𝑇)≤sup_{‖𝑋‖2+‖𝑌‖2=1}𝑤(𝐴)‖𝑋‖²+𝑤(𝐷)‖𝑌‖²+(‖𝐵‖+‖𝐶‖)‖𝑋‖‖𝑌‖,=sup_{[]𝜃∈0,2𝜋}𝑤(𝐴)cos²𝜃+𝑤(𝐷)sin²𝐵‖+‖𝐶,≤1𝜃+(‖‖)cos𝜃sin𝜃2𝑤(𝐴)+𝑤(𝐷)+(𝑤(𝐴)−𝑤(𝐷))²+(‖𝐵‖+‖𝐶‖)². (2.5) This completes the first part of the proof.
(ii) Proceeding as in (i) we can prove the second part. This completes the proof of the theorem.

Remark 2.2. As an application of (i) in Theorem 2.1, ‖𝑇‖ has another estimation by ‖𝑇‖²=‖𝑇^∗𝑇‖=𝑤(𝑇^∗𝑇) as follows: ‖𝑇‖²≤12𝑤𝐴^∗𝐴+𝐶^∗𝐶+𝑤𝐵^∗𝐵+𝐷^∗𝐷+(𝑤(𝐴^∗𝐴+𝐶^∗𝐶)−𝑤(𝐵^∗𝐵+𝐷^∗𝐷))²+4𝐴^∗𝐵+𝐶^∗𝐷². (2.6) Furuta [2] obtained numerical radius for a bounded linear operator 𝑇 of the above form with 𝐴=𝑎𝐼_𝑟,  𝐵=𝑏𝐴,  𝐶=𝑐𝐴^∗,  𝐷=𝑑𝐼_𝑛−𝑟, and 𝑎,𝑏,𝑐,𝑑∈ℝ⁺. If we consider 𝐴=𝑎𝐼_𝑟,  𝐷=𝑑𝐼_𝑛−𝑟,  𝐶=0_{𝑛−𝑟,𝑟} where 𝑎,𝑑∈ℝ, then we can exactly calculate 𝑤(𝑇) and ‖𝑇‖ as proved in the next theorem.

Theorem 2.3. Suppose 𝐵∈𝑀_{𝑟,𝑛−𝑟}(ℂ) and 𝑇=𝑎𝐼_𝑟𝐵𝑂_{𝑛−𝑟,𝑟}𝑑𝐼_𝑛−𝑟. (2.7) Then(i)𝑤(𝑇)=(1/2)(|𝑎+𝑑|+(𝑎−𝑑)²+‖𝐵‖²)  and(ii)‖𝑇‖=(1/√2)(𝑎²+𝑑²+‖𝐵‖²)+(𝑎²−𝑑²−‖𝐵‖²)²+4𝑎²‖𝐵‖².

Proof. (i) Following the method employed in the previous theorem, we can show that 𝑤(𝑇)≤12|𝑎+𝑑|+(𝑎−𝑑)²+‖𝐵‖². (2.8) We only need to show that there exists 𝑧₀,  ‖𝑧₀‖=1 such that |⟨𝑇𝑧₀,𝑧₀⟩| equals the quantity in the RHS.
Suppose 𝐵 attains its norm at 𝑦 with ‖𝑦‖=1.
Let 𝑧=(𝐵𝑦𝑘𝑦)^𝑡 where 𝑘 is a scalar. Then, ‖𝑧‖²=‖𝐵‖²+|𝑘|². Now ⟨𝑇𝑧,𝑧⟩=𝑎𝐼_𝑟𝐵𝑂_{𝑛−𝑟,𝑟}𝑑𝐼_𝑛−𝑟𝐵_𝑦𝑘_𝑦,𝐵_𝑦𝑘_𝑦 (2.9) so that ⟩⋅1⟨𝑇𝑧,𝑧‖𝑧‖²=(𝑎+𝑘)‖𝐵‖²+𝑑𝑘²‖𝐵‖²+𝑘². (2.10) Thus for all scalar 𝑘, we get 𝑤(𝑇)≥(𝑎+𝑘)‖𝐵‖²+𝑑𝑘²‖𝐵‖²+𝑘². (2.11)

Case 1 (𝑑+𝑎≥0). Define a function 𝜙∶ℝ⁺→ℝ by 𝜙(𝑘)=(𝑎+𝑘)‖𝐵‖²+𝑑𝑘²‖𝐵‖²+𝑘². (2.12) Then using elementary calculus, we can show that 𝜙(𝑘) attains its maximum at 𝑘₀=(𝑑−𝑎)+(𝑑−𝑎)²+‖𝐵‖² so that for 𝑧₀=(1/‖𝐵𝑦‖²+𝑘₀‖𝑦‖²)(𝐵𝑦𝑘₀𝑦)^𝑡 we get 𝑇𝑧₀,𝑧₀=12𝑎+𝑑+(𝑎−𝑑)²+‖𝐵‖². (2.13) Thus, we get 𝑤(𝑇)=12|𝑎+𝑑|+(𝑎−𝑑)²+‖𝐵‖². (2.14)

Case 2   (𝑑+𝑎≤0). As before we can show that there exists 𝑘₀=(𝑑−𝑎)−(𝑑−𝑎)²+‖𝐵‖² so that for 𝑧₀=(1/‖𝐵𝑦‖²+𝑘₀‖𝑦‖²)(𝐵𝑦𝑘₀𝑦)^𝑡 we get 𝑇𝑧₀,𝑧₀=12|𝑎+𝑑|+(𝑎−𝑑)²+‖𝐵‖². (2.15) Thus in all cases, we get 𝑤(𝑇)=12|𝑎+𝑑|+(𝑎−𝑑)²+‖𝐵‖². (2.16) This completes the proof of (i).
(ii) The proof is similar to the earlier one.
This completes the proof of the theorem.

Using Theorem 2.3, we can find numerical radius of an idempotent matrix 𝐴, that is, a matrix for which 𝐴²=𝐴 and also for a matrix for which 𝐴²=𝐼.

Corollary 2.4. Suppose 𝐴∈𝑀_𝑛(ℂ) with 𝐴²=𝐴. Then 𝑤(𝐴)=‖𝐴‖2+12. (2.17)

Proof. By Schur’s theorem,  𝐴  is unitarily equivalent to an upper triangular matrix. So without loss of generality, we can assume that 𝐼𝐴=^𝑟𝐵_{𝑟,𝑛−𝑟}O_{𝑛−𝑟,𝑟}0_𝑛−𝑟, (2.18) where 𝐼_𝑟 is the identity matrix, 𝐵_{𝑟,𝑛−𝑟} is any matrix. Using the last theorem, we get 𝑤(𝐴)=‖𝐴‖2+12. (2.19)

Corollary 2.5. Suppose 𝐴∈𝑀_𝑛(ℂ) and 𝐴²=𝐼. Then 𝑤(𝐴)=12‖𝐴‖+1‖𝐴‖. (2.20)

Proof. 𝐴  can be expressed as 𝐼𝐴=^𝑟𝐵_{𝑟,𝑛−𝑟}O_{𝑛−𝑟,𝑟}−𝐼_𝑛−𝑟, (2.21) where 𝐼_𝑟 is the identity matrix, 𝐵_{𝑟,𝑛−𝑟} is any matrix. By Theorem 2.3, we have ‖𝐴‖=11+2‖𝐵‖²+12‖𝐵‖4+‖𝐵‖²,𝑤(𝐴)=124+‖𝐵‖². (2.22) Therefore ‖𝐴‖²1=1+2‖𝐵‖²+12‖𝐵‖4+‖𝐵‖²,1‖𝐴‖²=11+(1/2)‖𝐵‖²+(1/2)‖𝐵‖4+‖𝐵‖²1=1+2‖𝐵‖²−12‖𝐵‖4+‖𝐵‖². (2.23) By adding, we get ‖𝐴‖²+1‖𝐴‖²=2+‖𝐵‖²⇒‖𝐴‖+1‖𝐴‖²=4+‖𝐵‖²(𝐴)=1⇒𝑤2‖𝐴‖+1‖𝐴‖. (2.24)

Corollary 2.6. Suppose 𝐴∈𝑀_𝑛(ℂ) with 𝐴^2𝑛=𝐼. Then 𝑤(𝐴)≥12𝐴^𝑛+1‖𝐴^𝑛‖^1/𝑛. (2.25)

Proof. It follows from the fact that (𝐴^𝑛)²=𝐼 and 𝑤(𝐴)^𝑛≥𝑤(𝐴^𝑛).

3. Bounds for Zeros of Polynomials

Let 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀ be a monic polynomial where 𝑎₀,𝑎₁,…,𝑎_𝑛−1 are complex numbers and let 𝐶𝑝=−𝑎_𝑛−1−𝑎_𝑛−2⋅⋅−𝑎₁−𝑎₀10⋅⋅0001⋅⋅0000⋅⋅00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅10 (3.1) be the Frobenius companion matrix of the polynomial 𝑝(𝑧). Then, it is well known that zeros of 𝑝 are exactly the eigenvalues of 𝐶(𝑝). Considering 𝐶(𝑝) as a linear operator on 𝐶^𝑛, we see that if  𝑧  is root of the polynomial equation 𝑝(𝑧)=0 then |𝑧|≤𝑤𝐶𝑝as𝜎𝐶𝑝⊂𝑊𝐶𝑝, (3.2) where 𝜎(𝐶(𝑝)) is the spectrum of operator 𝐶(𝑝). Estimation of the roots of zeros of the polynomial 𝑝(𝑧) has been done by many mathematicians over the years, some of them are mentioned below. Let 𝜆 be a root of the polynomial equation 𝑝(𝑧)=0.(i)Carmichael and Mason [3] proved that |𝜆|≤1+𝑎₀²+𝑎₁²+⋯+𝑎_𝑛−1^21/2. (3.3)(ii)Montel [4, 5] proved that |𝜆|≤𝑎₀+𝑎₀−𝑎₁+⋯+𝑎_𝑛−2−𝑎_𝑛−1+𝑎_𝑛−1,|𝜆|≤()+𝑎+1𝑛−1₀+𝑎₁+⋯+𝑎_𝑛−1. (3.4)(iii)Cauchy [3] proved that |𝜆|≤1+max𝑎₀,𝑎₁,…,𝑎_𝑛−1. (3.5)(iv)Fujii and Kubo [6, 7] proved that |𝜆|𝜋≤cos+1𝑛+12^𝑛−2_𝑖=0𝑎_𝑖^21/2+𝑎_𝑛−1. (3.6)(v)Alpin et al. [8] proved that |𝜆|≤max_{1≤𝑘≤𝑛}1+𝑎_𝑛−11+𝑎_𝑛−2⋯1+𝑎_𝑛−𝑘^1/𝑘. (3.7)(vi)Kittaneh [1] proved that |𝜆|≤12𝐶𝑝+𝐶𝑝^21/2. (3.8)We develop an inequality involving numerical radius with the help of which we estimate the zeros of the polynomial 𝑝. We show with examples that our estimation is better than the estimations mentioned above.

Theorem 3.1. If 𝜆 is a zero of the polynomial  𝑝(𝑧), then |𝜆|≤𝑎_𝑛−1𝑛+12𝜋cos𝑛+cos²𝜋𝑛+1+^𝑛−2_𝑟=0𝛼_𝑟²², (3.9) where 𝛼_𝑟=∑_𝑛^𝑛−𝑟_𝑘=0𝐶_𝑘(−𝑎_𝑛−1/𝑛)^𝑘𝑎_𝑛−𝑘,𝑟=0,1,2,…,𝑛−2.

Proof. Putting 𝑧=𝜉+ℎ in the polynomial equation 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛𝑧^𝑛−1+⋯𝑎₂𝑧+𝑎₁=0, we get (𝜉+ℎ)^𝑛+𝑎_𝑛−1(𝜉+ℎ)^𝑛−1+⋯+𝑎₁(𝜉+ℎ)+𝑎₀=0. (3.10) Substituting ℎ=−𝑎_𝑛−1/𝑛, we get 𝜉^𝑛+𝛼_𝑛−2𝜉^𝑛−2+𝛼_𝑛−3𝜉^𝑛−3+⋯+𝛼₁𝜉+𝛼₀=0, (3.11) where 𝛼_𝑟=∑_𝑛^𝑛−𝑟_𝑘=0𝐶_𝐾(−𝑎_𝑛−1/𝑛)^𝑘𝑎_𝑛−𝑘,𝑟=0,1,2,…,𝑛−2.
Let 𝐶(𝜉) be the Frobenius companion matrix of the polynomial 𝑞(𝜉)=𝜉^𝑛+𝛼_𝑛−2𝜉^𝑛−2+𝛼_𝑛−3𝜉^𝑛−3+⋯+𝛼₁𝜉+𝛼₀.
Then 𝐶(𝜉)=^𝐴𝐵_𝐶𝐷,where  𝐴=(0)₁,𝐵=(−𝛼_𝑛−2,−𝛼_𝑛−3,…,−𝛼₀)_1,𝑛−110⋅⋅0𝐶=_𝑛−1,1,𝐷=00⋅⋅010⋅⋅0⋅⋅⋅⋅0⋅⋅⋅⋅000⋅10_{𝑛−1,𝑛−1}. (3.12) Using Theorem 2.1, we get 𝑤(𝐶(𝜉≤1))2𝑤(𝐴)+𝑤(𝐷)+(𝑤(𝐴)−𝑤(𝐷))²+(‖𝐵‖+‖𝐶‖)²(𝐶(𝜉≤1⇒𝑤))2𝜋cos𝑛+cos²𝜋𝑛+1+^𝑛−2_𝑟=0𝛼_𝑟²². (3.13) This shows that if 𝜉₀ is a zero of the polynomial 𝑞(𝜉), then 𝜉₀≤12𝜋cos𝑛+cos²𝜋𝑛+1+^𝑛−2_𝑟=0𝛼_𝑟²². (3.14) Thus if 𝜆 is a zero of the polynomial 𝑝(𝑧), then |𝜆|≤𝑎_𝑛−1𝑛+12𝜋cos𝑛+cos²𝜋𝑛+1+^𝑛−2_𝑟=0𝛼_𝑟²². (3.15) This completes the proof of the theorem.

Example 3.2. Consider the polynomial equation 𝑝(𝑧)=𝑧³−3𝑧²+2𝑧=0. Then the bounds estimated by different mathematicians are as shown in Table 1.

Table 1 

But our estimation shows that if 𝜆 is a zero of the polynomial then |𝜆|≤2.280776406 which is much better than all the estimations mentioned above.

The companion matrix of the polynomial after removing the second term can be written as =𝐴010100010^2,2𝐵_2,1𝐶_1,1𝐷_1,1. (3.16) Then using the above theorems, it is easy to show that |𝜆|≤2.207 which is even better estimation.

Example 3.3. Consider the polynomial equation 𝑝(𝑧)=𝑧⁵−8𝑧⁴+25𝑧³−38𝑧²+28𝑧−8=0. Then, the bounds estimated by different mathematicians are as shown in Table 2.

Table 2 

But our estimation shows that if 𝜆 is a zero of the polynomial then |𝜆|≤2.703669110 which is much better than all the estimations mentioned above.

Theorem 3.4. Let 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀ having 𝛼_𝑖(𝑖=1,2,…,𝑛) as zeros and for each 𝑚∈𝑁,  𝑝_𝑚(𝑧)=𝑧^𝑛+𝑎^(𝑚)_𝑛−1𝑧^𝑛−1+⋯𝑎₂^(𝑚)𝑧+𝑎₁^(𝑚)𝑧+𝑎₀^(𝑚) is a polynomial having 𝛼^2𝑚_𝑖(𝑖=1,2,…,𝑛) as zeros. If 𝜆 is a zero of the polynomial  𝑝(𝑧), then for all  𝑚|𝜆|≤12𝑎^(𝑚)_𝑛−1𝜋+cos𝑛+𝑎^(𝑚)_𝑛−1𝜋−cos𝑛²+1+^𝑛_𝑘=2𝑎^(𝑚)_𝑛−𝑘^221/2𝑚. (3.17)

Proof. We first prove the lemma which shows that the coefficients of 𝑝_𝑚(𝑧) can be expressed in terms of coefficients of 𝑝(𝑧).

Lemma 3.5. Suppose 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀ is a monic polynomial, where 𝑎₀,𝑎₁,…,𝑎_𝑛−1 are complex numbers and 𝛼_𝑖,(𝑖=1,2,…,𝑛)  are the zeros of this polynomial. If 𝑝₁(𝑧)=𝑧^𝑛+𝑎⁽¹⁾_𝑛−1𝑧^𝑛−1+⋯𝑎₁⁽¹⁾𝑧+𝑎₀⁽¹⁾ is the polynomial having 𝛼²_𝑖(𝑖=1,2,…,𝑛) as zeros, then for 𝑟=1,2,…,𝑛: 𝑎_𝑟⁽¹⁾=(−1)^2𝑛−𝑟𝑎²_𝑟+2^𝑛−𝑟_𝑘=1(−1)^𝑘𝑎_𝑟+𝑘𝑎_𝑟−𝑘,𝑤ℎ𝑒𝑟𝑒𝑎_𝑛=1,𝑎_𝑛+𝑘=𝑎_𝑛−𝑘=0. (3.18)

Proof. We have det𝑧²𝐼−𝐶𝑝²=det𝑧𝐼−𝐶𝑝det𝑧𝐼+𝐶𝑝⇒𝑝₁𝑧²=𝑝(𝑧)𝑝(−𝑧)⇒𝑧^2𝑛+𝑎⁽¹⁾_𝑛−1𝑧^2(𝑛−1)+⋯𝑎₁⁽¹⁾𝑧²+𝑎₀⁽¹⁾=(−1)^𝑛𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀×𝑧^𝑛−𝑎_𝑛−1𝑧^𝑛−1+⋯+(−1)^𝑛−1𝑎₁𝑧+(−1)^𝑛𝑎₀. (3.19) Comparing the coefficient of 𝑧^2𝑟, we get for 𝑟=1,2,…𝑛: 𝑎_𝑟⁽¹⁾=(−1)^2𝑛−𝑟𝑎²_𝑟+2^𝑛−𝑟_𝑘=1(−1)^𝑘𝑎_𝑟+𝑘𝑎_𝑟−𝑘,where𝑎_𝑛=1,𝑎_𝑛+𝑘=𝑎_𝑛−𝑘=0. (3.20) This completes the proof of lemma.

The companion matrix of the monic polynomial 𝑝_𝑚(𝑧) is 𝐶𝑝_𝑚=−𝑎^(𝑚)_𝑛−1−𝑎^(𝑚)_𝑛−2⋅⋅−𝑎₁^(𝑚)−𝑎₀^(𝑚)10⋅⋅0001⋅⋅0000⋅⋅00⋅⋅⋅⋅⋅⋅00⋅⋅10. (3.21) We have 𝑤𝐶𝑝_𝑚≥𝑟_𝜎𝐶𝑝_𝑚=𝑟_𝜎𝐶𝑝^2𝑚. (3.22) So 𝑟_𝜎𝐶𝑝≤𝑤𝐶𝑝_𝑚^1/2𝑚. (3.23) Using Theorem 2.1, we get 𝑤𝐶𝑝_𝑚≤12𝑎^(𝑚)_𝑛−1𝜋+cos𝑛+𝑎^(𝑚)_𝑛−1𝜋−cos𝑛²+1+^𝑛_𝑘=2𝑎^(𝑚)_𝑛−𝑘²². (3.24) Thus if 𝜆 is a zero of the polynomial 𝑝(𝑧), then |𝜆|≤12𝑎^(𝑚)_𝑛−1𝜋+cos𝑛+𝑎^(𝑚)_𝑛−1𝜋−cos𝑛²+1+^𝑛_𝑘=2𝑎^(𝑚)_𝑛−𝑘^221/2𝑚. (3.25) This completes the proof of the theorem.

We next prove the theorem.

Theorem 3.6. Suppose 𝑝(𝑧)=𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+⋯𝑎₁𝑧+𝑎₀ is a monic polynomial and 𝛼_𝑖 are the roots of this equation 𝑖=1,2,…,𝑛,   where 𝑎₀,𝑎₁,…,𝑎_𝑛−1 are complex numbers with |𝛼₁|>1>|𝛼₂|>⋯>|𝛼_𝑛|. If the equation having roots 𝛼^2𝑚_𝑖 for 𝑖=1,2,…,𝑛 is 𝑝_𝑚(𝑧)=𝑧^𝑛+𝑎^(𝑚)_𝑛−1𝑧^𝑛−1+⋯𝑎₁^(𝑚)𝑧+𝑎₀^(𝑚), then there exists 𝑚₀∈ℕ such that(1)|𝑎^(𝑚)_𝑛−𝑘|≤|𝑎^(𝑚)_𝑛−1|  whenever   𝑚≥𝑚₀  and  for   𝑘=2,3,…,𝑛;(2)[𝑤(𝐶_𝑚(𝑝))]^1/2𝑚  converges  to  𝑟_𝜎[𝐶(𝑝)].

Proof. (1) We prove this for 𝑘=2 and the rest are similar.
First observe that 𝑎^(𝑚)_𝑛−1=^𝑛_𝑘=1𝛼^2𝑚_𝑘≥𝛼₁^2𝑚−^𝑛_𝑘=2𝛼_𝑘^2𝑚,𝑎^(𝑚)_𝑛−2=^𝑛_{𝑗≠𝑘=1}𝛼^2𝑚_𝑗𝛼^2𝑚_𝑘≤𝛼₁^2𝑚𝑛_𝑘=2𝛼_𝑘^2𝑚+^𝑛_{𝑗≠𝑘=2}𝛼_𝑗^2𝑚𝛼_𝑘^2𝑚. (3.26) Now in order to have 𝑎^(𝑚)_𝑛−2≤𝑎^(𝑚)_𝑛−1. (3.27) We get 𝛼₁^2𝑚−^𝑛_𝑘=2𝛼_𝑘^2𝑚≥𝛼₁^2𝑚𝑛_𝑘=2𝛼_𝑘^2𝑚+^𝑛_{𝑗≠𝑘=2}𝛼_𝑗^2𝑚𝛼_𝑘^2𝑚, (3.28) that is, 𝛼₁^2𝑚≥1+𝛼₁^2𝑚𝑛_𝑘=2𝛼_𝑘^2𝑚+^𝑛_{𝑗≠𝑘=2}𝛼_𝑗^2𝑚𝛼_𝑘^2𝑚, (3.29) that is, 𝛼₁^2𝑚1+𝛼₁^2𝑚≥∑^𝑛_𝑘=2𝛼_𝑘^2𝑚1+𝛼₁^2𝑚+∑^𝑛_{𝑗≠𝑘=2}𝛼_𝑗^2𝑚𝛼_𝑘^2𝑚1+𝛼₁^2𝑚. (3.30) Clearly, this inequality holds good as the left-hand side converges to 1, but the right-hand side converges to 0.
(2) We have 𝑤𝐶_𝑚𝑝≤12𝑎^(𝑚)_𝑛−1𝜋+cos𝑛+𝑎^(𝑚)_𝑛−1𝜋−cos𝑛²+1+^𝑛_𝑘=2𝑎^(𝑚)_𝑛−𝑘²², (3.31) that is, 𝑤𝐶_𝑚𝑝≤12𝑎^(𝑚)_𝑛−1+1+𝑎^(𝑚)_𝑛−1²+(1+√𝑛−2𝑎^(𝑚)_𝑛−1², (3.32) that is, 𝑤𝐶_𝑚𝑝≤12𝑎^(𝑚)_𝑛−1+𝑎^(𝑚)_𝑛−1+𝑎^(𝑚)_𝑛−1²+(𝑎^(𝑚)_𝑛−1+√𝑛−2𝑎^(𝑚)_𝑛−1². (3.33) So we get 𝑤𝐶_𝑚𝑝≤𝐾𝑎^(𝑚)_𝑛−1≤𝐾^𝑛_𝑖=1𝛼^2𝑚_𝑖≤𝐾𝛼₁^2𝑚1+^𝑛_𝑖=2𝛼_𝑖𝛼₁^2𝑚. (3.34) Now 𝑟_𝜎𝐶𝑝^2𝑚=𝑟_𝜎𝐶_𝑚𝑝≤𝑤𝐶_𝑚𝑝≤𝐾𝛼₁^2𝑚1+^𝑛_𝑖=2𝛼_𝑖𝛼₁^2𝑚. (3.35) Therefore, 𝑟_𝜎𝐶𝑝≤𝑤𝐶_𝑚𝑝^1/2𝑚≤𝛼₁𝐾1+^𝑛_𝑖=2𝛼_𝑖𝛼₁^2𝑚1/2𝑚. (3.36) As the terms inside the bracket on the RHS converges to 1, we get the desired result.
This completes the proof of the theorem.

Application. As an application we can exactly find the spectral radius of a given matrix. Consider a given matrix 𝐴 of order 𝑛.

Step 1. We first find the characteristic polynomial 𝑞(𝑧)=𝑧^𝑛+𝑏_𝑛−1𝑧^𝑛−1+⋯𝑏₁𝑧+𝑏₀. Suppose 𝛼_𝑖,𝑖=1,2,…,𝑟 are the distinct roots of 𝑞(𝑧)=0 with |𝛼₁|>|𝛼₂|>⋯>|𝛼_𝑟|.

Step 2. Find 𝑝(𝑧)=𝑞(𝑧)/gcd[𝑞(𝑧),𝑞^′(𝑧)]=𝑧^𝑟+𝑎_𝑟−1𝑧^𝑟−1+⋯𝑎₁𝑧+𝑎₀. Then, roots of 𝑞(𝑧)=0 are 𝛼_𝑖,𝑖=1,2,…,𝑟 without multiplicity. Let 𝑝_𝑚(𝑧)=𝑧^𝑛+𝑎^(𝑚)_𝑛−1𝑧^𝑛−1+⋯𝑎₁^(𝑚)𝑧+𝑎₀^(𝑚) be the polynomial having 𝛼^2𝑚_𝑖 for 𝑖=1,2,…,𝑟 as its zeros.

Step 3. Since |𝛼₂|<|𝛼₁|, taking 𝜖<(1/4)(|𝛼₁|−|𝛼₂|), we can see that |𝛼₂|<|𝛼₁−2𝜖. Again using a result of [9], we get |𝑎^(𝑚)_𝑛−1|^1/2𝑚 converging to |𝛼₁|. So for this 𝜖 there exists an 𝑚₀∈ℕ such that |𝛼₁|−𝜖<|𝑎^(𝑚)_𝑛−1|^1/2𝑚<|𝛼₁|+𝜖 for all 𝑚≥𝑚₀. Therefore, 𝛼₂+𝜖<𝛼₁−𝜖<𝑎^(𝑚)_𝑛−1^1/2𝑚<𝛼₁+𝜖∀𝑚≥𝑚₀. (3.37)

Step 4. Let 𝑡=|𝑎^(𝑚0)_𝑛−1|^1/2𝑚0−𝜖.
Find 𝑠(𝑧)=𝑧^𝑟+∑^𝑟_𝑘=1(𝑎_𝑟−𝑘/𝑡^𝑘)𝑧^𝑟−𝑘=𝑧^𝑟+𝑐_𝑟−1𝑧^𝑟−1+⋯𝑐₁𝑧+𝑐₀. If the roots of 𝑠(𝑧)=0 are 𝛽_𝑖, then 𝛽_𝑖=𝛼_𝑖/𝑡,𝑖=1,2,…,𝑟 and 𝛽₁=𝛼₁𝑡𝛼>1>²𝑡=𝛽₂>𝛽₃>⋯>𝛽_𝑟. (3.38) Then, 𝑠(𝑧) satisfies all the criterion of Theorem 3.6.

Step 5. The required sequence is 𝑥_𝑚=𝑡[𝑤(𝐶_𝑚(𝑠))]^1/2𝑚, which converges to the spectral radius of matrix  𝐴.

Example 3.7. Consider the 5th-degree polynomial 𝑞(𝑧)=𝑧⁵+2𝑧⁴+1.
By Rouche’s theorem, it is easy to see that all the roots except one are enclosed by the simple closed curve |𝑧|=2.
Consider 𝑠(𝑧)=𝑧⁵+𝑧⁴+(1/2⁵) and then iterating the coefficints of 𝑄(𝑧) we get the following.
The highest absolute value of the zeros of the polynomial is 2.055 and by 5th iteration we get 2.05. Continuing the above process, we can find the highest absolute value of the zeros of the polynomial up to the desired degree of accuracy. The previous best result for this is known to be 2.414 given by Alpin [8]. The iterations are shown in Table 3.

Table 3 

Acknowledgments

The authors would like to thank the referees for their suggestions, one of the referees suggested the inclusion of Remark 2.2 following Theorem 2.1. The research of the first author is partially supported by PURSE-DST, Govt. of India and the research of the second author is supported by CSIR, India.