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

Authors: Kallol Paul; Santanu Bag
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We obtain inequalities involving numerical radius of a matrixA∈Mn(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∈Mn(ℂ) up to the desired degree of accuracy.


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## Body

## 1. Introduction

SupposeA∈Mn(ℂ). Let  W(A), σ(A) denote respectively the numerical range, spectrum of  A  and w(A), rσ(A) denote respectively the numerical radius, spectral radius of A, that is,
(1.1)W(A)={〈Ax,x〉:∥x∥=1},w(A)=sup{|λ|:λ∈W(A)},σ(A)={λ:λisaneigenvalueofA},rσ(A)=sup{|λ|:λ∈σ(A)}.
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).Letp(z)=zn+an-1zn-1+⋯a1z+a0 be a monic polynomial where a0,a1,…,an-1 are complex numbers and let
(1.2)C(p)=(-an-1-an-2··-a1-a010··0001··0000··00··········10)
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 Mn(ℂ), we see that if z is root of the polynomial equation p(z)=0, then
(1.3)|z|≤w(C(p)),|z|≤rσ(C(p)).
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 Section1 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 inMn(ℂ) and use it to obtain numerical radius for some special class of matrices.Theorem 2.1.
SupposeT∈Mn(ℂ) and
(2.1)T=(ABCD),
where A∈Mr(ℂ),  B∈Mr,n-r(ℂ),  C∈Mn-r,r(ℂ)  and  D∈Mn-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) LetZ∈ℂn and
(2.2)Z=(XY),
where X∈ℂr and  Y∈ℂn-r with ∥Z∥=1.
Then,(2.3)〈TZ,Z〉=〈(AX+BYCX+DY),(XY)〉=〈AX,X〉+〈BY,X〉+〈CX,Y〉+〈DY,Y〉
and so
(2.4)|〈TZ,Z〉|≤|〈AX,X〉|+|〈DY,Y〉|+∥B∥∥X∥∥Y∥+∥C∥∥X∥∥Y∥.
Therefore, we have
(2.5)w(T)≤sup∥X∥2+∥Y∥2=1[w(A)∥X∥2+w(D)∥Y∥2+(∥B∥+∥C∥)∥X∥∥Y∥],=supθ∈[0,2π][w(A)cos2θ+w(D)sin2θ+(∥B∥+∥C∥)cosθsinθ],≤12[w(A)+w(D)+(w(A)-w(D))2+(∥B∥+∥C∥)2].
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:
(2.6)∥T∥2≤12[{(w(A*A+C*C)-w(B*B+D*D))2+4∥A*B+C*D∥2}w(A*A+C*C)+w(B*B+D*D)+(w(A*A+C*C)-w(B*B+D*D))2+4∥A*B+C*D∥2].
Furuta [2] obtained numerical radius for a bounded linear operator T of the above form with A=aIr,  B=bA,  C=cA*,  D=dIn-r, and a,b,c,d∈ℝ+. If we consider A=aIr,  D=dIn-r,  C=0n-r,r where a,d∈ℝ, then we can exactly calculate w(T) and ∥T∥ as proved in the next theorem.Theorem 2.3.
SupposeB∈Mr,n-r(ℂ) and
(2.7)T=(aIrBOn-r,rdIn-r).
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(2.8)w(T)≤12(|a+d|+(a-d)2+∥B∥2).
We only need to show that there exists z0,  ∥z0∥=1 such that |〈Tz0,z0〉| equals the quantity in the RHS.
SupposeB attains its norm at y with ∥y∥=1.
Letz=(Byky)t where k is a scalar. Then, ∥z∥2=∥B∥2+|k|2. Now
(2.9)〈Tz,z〉=〈(aIrBOn-r,rdIn-r)(Byky),(Byky)〉
so that
(2.10)|〈Tz,z〉·1∥z∥2|=|(a+k)∥B∥2+dk2|∥B∥2+k2.
Thus for all scalar k, we get
(2.11)w(T)≥|(a+k)∥B∥2+dk2|∥B∥2+k2.Case 1 (d+a≥0).
Define a functionϕ:ℝ+→ℝ by
(2.12)ϕ(k)=(a+k)∥B∥2+dk2∥B∥2+k2.
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
(2.13)|〈Tz0,z0〉|=12(a+d+(a-d)2+∥B∥2).
Thus, we get
(2.14)w(T)=12(|a+d|+(a-d)2+∥B∥2).Case 2   (d+a≤0).
As before we can show that there existsk0=(d-a)-(d-a)2+∥B∥2 so that for z0=(1/∥By∥2+k0∥y∥2)(Byk0y)t we get
(2.15)|〈Tz0,z0〉|=12(|a+d|+(a-d)2+∥B∥2).
Thus in all cases, we get
(2.16)w(T)=12(|a+d|+(a-d)2+∥B∥2).
This completes the proof of (i).
(ii) The proof is similar to the earlier one.
This completes the proof of the theorem.Using Theorem2.3, 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.
SupposeA∈Mn(ℂ) with A2=A. Then
(2.17)w(A)=∥A∥2+12.Proof.
By Schur’s theorem,A  is unitarily equivalent to an upper triangular matrix. So without loss of generality, we can assume that
(2.18)A=(IrBr,n-rOn-r,r0n-r),
where Ir is the identity matrix, Br,n-r is any matrix. Using the last theorem, we get
(2.19)w(A)=∥A∥2+12.Corollary 2.5.
SupposeA∈Mn(ℂ) and A2=I. Then
(2.20)w(A)=12(∥A∥+1∥A∥).Proof.
A  can be expressed as
(2.21)A=(IrBr,n-rOn-r,r-In-r),
where Ir is the identity matrix, Br,n-r is any matrix. By Theorem 2.3, we have
(2.22)∥A∥=1+12∥B∥2+12∥B∥4+∥B∥2,w(A)=12(4+∥B∥2).
Therefore
(2.23)∥A∥2=1+12∥B∥2+12∥B∥4+∥B∥2,1∥A∥2=11+(1/2)∥B∥2+(1/2)∥B∥4+∥B∥2=1+12∥B∥2-12∥B∥4+∥B∥2.
By adding, we get
(2.24)∥A∥2+1∥A∥2=2+∥B∥2⇒(∥A∥+1∥A∥)2=4+∥B∥2⇒w(A)=12(∥A∥+1∥A∥).Corollary 2.6.
SupposeA∈Mn(ℂ) with A2n=I. Then
(2.25)w(A)≥(12(∥An∥+1∥An∥))1/n.Proof.
It follows from the fact that(An)2=I and w(A)n≥w(An).


## 3. Bounds for Zeros of Polynomials

Letp(z)=zn+an-1zn-1+⋯a1z+a0 be a monic polynomial where a0,a1,…,an-1 are complex numbers and let
(3.1)C(p)=(-an-1-an-2··-a1-a010··0001··0000··00··········10)
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 Cn, we see that if  z  is root of the polynomial equation p(z)=0 then
(3.2)|z|≤w(C(p))asσ(C(p))⊂W(C(p)),
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
(3.3)|λ|≤[1+|a0|2+|a1|2+⋯+|an-1|2]1/2.(ii)
Montel [4, 5] proved that
(3.4)|λ|≤|a0|+|a0-a1|+⋯+|an-2-an-1|+|an-1+1|,|λ|≤(n-1)+|a0|+|a1|+⋯+|an-1|.(iii)
Cauchy [3] proved that
(3.5)|λ|≤1+max{|a0|,|a1|,…,|an-1|}.(iv)
Fujii and Kubo [6, 7] proved that
(3.6)|λ|≤cosπn+1+12[(∑i=0n-2‍|ai|2)1/2+|an-1|].(v)
Alpin et al. [8] proved that
(3.7)|λ|≤max1≤k≤n[(1+|an-1|)(1+|an-2|)⋯(1+|an-k|)]1/k.(vi)
Kittaneh [1] proved that
(3.8)|λ|≤12[∥C(p)∥+∥C(p)2∥1/2].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.
Ifλ is a zero of the polynomial  p(z), then
(3.9)|λ|≤|an-1n|+12[cosπn+cos2πn+(1+∑r=0n-2‍|αr|2)2],
where αr=∑k=0n-r‍nCk(-an-1/n)kan-k,r=0,1,2,…,n-2.Proof.
Puttingz=ξ+h in the polynomial equation p(z)=zn+anzn-1+⋯a2z+a1=0, we get
(3.10)(ξ+h)n+an-1(ξ+h)n-1+⋯+a1(ξ+h)+a0=0.
Substituting h=-an-1/n, we get
(3.11)ξn+αn-2ξn-2+αn-3ξn-3+⋯+α1ξ+α0=0,
where αr=∑k=0n-rnCK(-an-1/n)kan-k,r=0,1,2,…,n-2.
LetC(ξ) be the Frobenius companion matrix of the polynomial q(ξ)=ξn+αn-2ξn-2+αn-3ξn-3+⋯+α1ξ+α0.
ThenC(ξ)=(ABCD),where  A=(0)1,B=(-αn-2,-αn-3,…,-α0)1,n-1(3.12)C=(10··0)n-1,1,D=(00··010··0····0····000·10)n-1,n-1.
Using Theorem 2.1, we get
(3.13)w(C(ξ))≤12[w(A)+w(D)+(w(A)-w(D))2+(∥B∥+∥C∥)2]⇒w(C(ξ))≤12[cosπn+cos2πn+(1+∑r=0n-2‍|αr|2)2].
This shows that if ξ0 is a zero of the polynomial q(ξ), then
(3.14)|ξ0|≤12[cosπn+cos2πn+(1+∑r=0n-2‍|αr|2)2].
Thus if λ is a zero of the polynomial p(z), then
(3.15)|λ|≤|an-1n|+12[cosπn+cos2πn+(1+∑r=0n-2‍|αr|2)2].
This completes the proof of the theorem.Example 3.2.
Consider the polynomial equationp(z)=z3-3z2+2z=0. Then the bounds estimated by different mathematicians are as shown in Table 1.Table 1
Carmichael and Mason
3.741657387
Montel
7
Cauchy
4
Fujii  and  Kubo
4.0098824
Yuri, Chien and Yeh
3.464101615
Kittaneh
3.44572894But our estimation shows that ifλ 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(3.16)(010100010)=(A2,2B2,1C1,1D1,1).
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 equationp(z)=z5-8z4+25z3-38z2+28z-8=0. Then, the bounds estimated by different mathematicians are as shown in Table 2.Table 2
Carmichael and Mason
54.60769176
Montel
215
Cauchy
39
Fujii and Kubo
32.16529279
Yuri, Chien and Yeh
9But our estimation shows that ifλ is a zero of the polynomial then |λ|≤2.703669110 which is much better than all the estimations mentioned above.Theorem 3.4.
Letp(z)=zn+an-1zn-1+⋯a1z+a0 having αi(i=1,2,…,n) as zeros and for each m∈N,  pm(z)=zn+an-1(m)zn-1+⋯a2(m)z+a1(m)z+a0(m) is a polynomial having αi2m(i=1,2,…,n) as zeros. If λ is a zero of the polynomial  p(z), then for all  m(3.17)|λ|≤(12[|an-1(m)|+cos(πn)+(|an-1(m)|-cos(πn))2+(1+∑k=2n‍|an-k(m)|2)2])1/2m.Proof.
We first prove the lemma which shows that the coefficients ofpm(z) can be expressed in terms of coefficients of p(z).Lemma 3.5.
Supposep(z)=zn+an-1zn-1+⋯a1z+a0 is a monic polynomial, where a0,a1,…,an-1 are complex numbers and αi,(i=1,2,…,n)  are the zeros of this polynomial. If p1(z)=zn+an-1(1)zn-1+⋯a1(1)z+a0(1) is the polynomial having αi2(i=1,2,…,n) as zeros, then for r=1,2,…,n:
(3.18)ar(1)=(-1)2n-r(ar2+2∑k=1n-r‍(-1)kar+kar-k),wherean=1,an+k=an-k=0.Proof.
We have(3.19)det(z2I-C(p)2)=det(zI-C(p))det(zI+C(p))⇒p1(z2)=p(z)p(-z)⇒z2n+an-1(1)z2(n-1)+⋯a1(1)z2+a0(1)=(-1)n(zn+an-1zn-1+⋯a1z+a0)×(zn-an-1zn-1+⋯+(-1)n-1a1z+(-1)na0).
Comparing the coefficient of z2r, we get for r=1,2,…n:
(3.20)ar(1)=(-1)2n-r(ar2+2∑k=1n-r‍(-1)kar+kar-k),wherean=1,an+k=an-k=0.
This completes the proof of lemma.The companion matrix of the monic polynomialpm(z) is
(3.21)C(pm)=(-an-1(m)-an-2(m)··-a1(m)-a0(m)10··0001··0000··00······00··10).
We have
(3.22)w(C(pm))≥rσ(C(pm))=rσ(C(p))2m.
So
(3.23)rσ(C(p))≤(w(C(pm)))1/2m.
Using Theorem 2.1, we get
(3.24)w(C(pm))≤12[|an-1(m)|+cos(πn)+(|an-1(m)|-cos(πn))2+(1+∑k=2n‍|an-k(m)|2)2].
Thus if λ is a zero of the polynomial p(z), then
(3.25)|λ|≤(12[|an-1(m)|+cos(πn)+(|an-1(m)|-cos(πn))2+(1+∑k=2n‍|an-k(m)|2)2])1/2m.
This completes the proof of the theorem.We next prove the theorem.Theorem 3.6.
Supposep(z)=zn+an-1zn-1+⋯a1z+a0 is a monic polynomial and αi are the roots of this equation i=1,2,…,n,   where a0,a1,…,an-1 are complex numbers with |α1|>1>|α2|>⋯>|αn|. If the equation having roots αi2m for i=1,2,…,n is pm(z)=zn+an-1(m)zn-1+⋯a1(m)z+a0(m), then there exists m0∈ℕ such that(1)
|an-k(m)|≤|an-1(m)|  whenever   m≥m0  and  for   k=2,3,…,n;(2)
[w(Cm(p))]1/2m  converges  to  rσ[C(p)].Proof.
(1) We prove this for k=2 and the rest are similar.
First observe that(3.26)|an-1(m)|=|∑k=1n‍αk2m|≥|α1|2m-∑k=2n‍|αk|2m,|an-2(m)|=|∑j≠k=1n‍αj2mαk2m|≤|α1|2m(∑k=2n‍|αk|2m)+∑j≠k=2n‍|αj|2m|αk|2m.
Now in order to have
(3.27)|an-2(m)|≤|an-1(m)|.
We get
(3.28)|α1|2m-∑k=2n‍|αk|2m≥|α1|2m(∑k=2n‍|αk|2m)+∑j≠k=2n‍|αj|2m|αk|2m,
that is,
(3.29)|α1|2m≥(1+|α1|2m)[∑k=2n‍|αk|2m]+∑j≠k=2n‍|αj|2m|αk|2m,
that is,
(3.30)|α1|2m1+|α1|2m≥∑k=2n|αk|2m1+|α1|2m+∑j≠k=2n|αj|2m|αk|2m1+|α1|2m.
Clearly, this inequality holds good as the left-hand side converges to 1, but the right-hand side converges to 0.
(2) We have
(3.31)w(Cm(p))≤12[|an-1(m)|+cos(πn)+(|an-1(m)|-cos(πn))2+(1+∑k=2n‍|an-k(m)|2)2],
that is,
(3.32)w(Cm(p))≤12[|an-1(m)|+1+(|an-1(m)|2+(1+n-2|an-1(m)|)2],
that is,
(3.33)w(Cm(p))≤12[|an-1(m)|+|an-1(m)|+(|an-1(m)|2+(|an-1(m)|+n-2|an-1(m)|)2].
So we get
(3.34)w(Cm(p))≤K|an-1(m)|≤K|∑i=1n‍αi2m|≤K|α1|2m(1+∑i=2n|αiα1|2m‍).
Now
(3.35)(rσ(C(p)))2m=rσ(Cm(p))≤w(Cm(p))≤K|α1|2m(1+∑i=2n‍|αiα1|2m).
Therefore,
(3.36)rσ(C(p))≤[w(Cm(p))]1/2m≤|α1|(K(1+∑i=2n|αiα1|2m‍))1/2m.
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 polynomialq(z)=zn+bn-1zn-1+⋯b1z+b0. Suppose αi,i=1,2,…,r are the distinct roots of q(z)=0 with |α1|>|α2|>⋯>|αr|.Step 2.
Findp(z)=q(z)/gcd[q(z),q′(z)]=zr+ar-1zr-1+⋯a1z+a0. Then, roots of q(z)=0 are αi,i=1,2,…,r without multiplicity. Let pm(z)=zn+an-1(m)zn-1+⋯a1(m)z+a0(m) be the polynomial having αi2m 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 |an-1(m)|1/2m converging to |α1|. So for this ϵ there exists an m0∈ℕ such that |α1|-ϵ<|an-1(m)|1/2m<|α1|+ϵ for all m≥m0. Therefore,
(3.37)|α2|+ϵ<|α1|-ϵ<|an-1(m)|1/2m<|α1|+ϵ∀m≥m0.Step 4.
Lett=|an-1(m0)|1/2m0-ϵ.
Finds(z)=zr+∑k=1r(ar-k/tk)zr-k=zr+cr-1zr-1+⋯c1z+c0. If the roots of s(z)=0 are βi, then βi=αi/t,i=1,2,…,r and
(3.38)|β1|=|α1t|>1>|α2t|=|β2|>|β3|>⋯>|βr|.
Then, s(z) satisfies all the criterion of Theorem 3.6.Step 5.
The required sequence isxm=t[w(Cm(s))]1/2m, which converges to the spectral radius of matrix  A.Example 3.7.
Consider the 5th-degree polynomialq(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.
Considers(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
No. of iterations
a
4
(
m
)
a
3
(
m
)
a
2
(
m
)
a
1
(
m
)
a
0
(
m
)
r
σ
C
(
(
q
(
z
)
)
)
0
−1
0
0
0
-
2
-
5
2.85
1
−1
0
-
2
-
4
0
-
2
-
10
2.40
2
−1
-
2
-
3
-
2
-
9
-
2
-
13
-
2
-
20
2.20
3
−1.25
31
×
2
-
12
13
×
2
-
19
3
×
2
-
28
-
2
-
40
2.11
4
1.547363281
0.000119
0
0
0
2.07
5
2.394094544
0
0
0
0
2.05



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*Source: 129132-2012-10-02.xml*