Volume 2012, Article ID 129132,15pages doi:10.1155/2012/129132
Research Article
On Numerical Radius of a Matrix and Estimation of Bounds for Zeros of a Polynomial
Kallol Paul and Santanu Bag
Department of Mathematics, Jadavpur University, Kolkata 700032, India
Correspondence should be addressed to Kallol Paul,[email protected] Received 31 March 2012; Accepted 5 June 2012
Academic Editor: Teodor Bulboaca
Copyrightq2012 K. Paul and S. Bag. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We obtain inequalities involving numerical radius of a matrixA∈MnC. 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 matrixA∈MnCup to the desired degree of accuracy.
1. Introduction
Suppose A ∈ MnC. LetWA,σAdenote respectively the numerical range, spectrum ofAandwA,rσAdenote respectively the numerical radius, spectral radius ofA, that is,
WA {Ax, x:x1}, wA sup{|λ|:λ∈WA},
σA
λ:λis an eigenvalue ofA , rσA sup{|λ|:λ∈σA}.
1.1
It is well known that
iA/2≤wA≤ A.
Kittaneh1improved on the second inequality to prove that.
iiwA≤1/2A1/2A21/2.
Clearly,1/2A 1/2A21/2 ≤ A so that inequality ii is sharper than the second inequality ofi.
Letpz znan−1zn−1· · ·a1za0be a monic polynomial wherea0, a1, . . . , an−1are complex numbers and let
C p
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
−an−1 −an−2 · · −a1 −a0
1 0 · · 0 0
0 1 · · 0 0
0 0 · · 0 0
· · · ·
· · · · 1 0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
1.2
be the Frobenius companion matrix of the polynomialpz. Then, it is well known that zeros ofpare exactly the eigenvalues ofCp. ConsideringCpas an element ofMnC, we see that ifzis root of the polynomial equationpz 0, then
|z| ≤w C
p
, |z| ≤rσ
C p
. 1.3
Based on inequalityii, Kittaneh1 obtained an estimation for wCpwhich gives an upper bound for zeros of the polynomialpz.
InSection 1we 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 MnC and use it to obtain numerical radius for some special class of matrices.
Theorem 2.1. SupposeT ∈MnCand
T A B
C D
, 2.1
whereA∈MrC,B∈Mr,n−rC,C∈Mn−r,rCandD∈Mn−rC. Then, iwT≤1/2wA wD
wA−wD2 BC2and
iiT2 ≤ 1/2A2 B2 C2 D2 1/2
A2C2− B2− D224ABCD2. Proof. iLetZ∈Cnand
Z X
Y
, 2.2
whereX∈CrandY ∈Cn−rwithZ1.
Then,
TZ, Z
AXBY CXDY
,
X Y
AX, XBY, XCX, YDY, Y 2.3
and so
|TZ, Z| ≤ |AX, X||DY, Y|BXYCXY. 2.4
Therefore, we have
wT≤ sup
X2Y21
wAX2wDY2 BCXY ,
sup
θ∈0,2π
wAcos2θwDsin2θ BCcosθsinθ ,
≤ 1 2
wA wD
wA−wD2 BC2
.
2.5
This completes the first part of the proof.
iiProceeding as iniwe can prove the second part. This completes the proof of the theorem.
Remark 2.2. As an application ofiinTheorem 2.1,Thas another estimation by T2 T∗TwT∗Tas follows:
T2≤ 1 2
wA∗AC∗C wB∗BD∗D
wA∗AC∗C−wB∗BD∗D24A∗BC∗D2
.
2.6
Furuta2obtained numerical radius for a bounded linear operatorTof 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,C0n−r,r wherea, d∈R, then we can exactly calculatewTandTas proved in the next theorem.
Theorem 2.3. SupposeB∈Mr,n−rCand
T
aIr B On−r,r dIn−r
. 2.7
Then
iwT 1/2|ad|
a−d2B2and iiT 1/√
2
a2d2B2
a2−d2− B224a2B2.
Proof. iFollowing the method employed in the previous theorem, we can show that
wT≤ 1 2
|ad|
a−d2B2
. 2.8
We only need to show that there existsz0, z0 1 such that|Tz0, z0|equals the quantity in the RHS.
SupposeBattains its norm atywithy1.
Letz By kytwherekis a scalar. Then,z2B2|k|2. Now
Tz, z
aIr B On−r,r dIn−r
By
ky
,
By
ky
2.9
so that
Tz, z · 1 z2
akB2dk2
B2k2 . 2.10
Thus for all scalark, we get
wT≥
akB2dk2
B2k2 . 2.11
Case 1da≥0. Define a functionφ:R → Rby
φk akB2dk2
B2k2 . 2.12
Then using elementary calculus, we can show thatφkattains its maximum atk0 d−a d−a2B2so that forz0 1/
By2k0y2By k0ytwe get
|Tz0, z0| 1 2
ad
a−d2B2
. 2.13
Thus, we get
wT 1
2
|ad|
a−d2B2
. 2.14
Case 2da≤0. As before we can show that there existsk0 d−a−
d−a2B2so that forz0 1/
By2k0y2By k0ytwe get
|Tz0, z0| 1 2
|ad|
a−d2B2
. 2.15
Thus in all cases, we get
wT 1
2
|ad|
a−d2B2
. 2.16
This completes the proof ofi.
iiThe proof is similar to the earlier one.
This completes the proof of the theorem.
UsingTheorem 2.3, we can find numerical radius of an idempotent matrixA, that is, a matrix for whichA2Aand also for a matrix for whichA2I.
Corollary 2.4. SupposeA∈MnCwithA2A. Then
wA A
2 1
2. 2.17
Proof. By Schur’s theorem,Ais unitarily equivalent to an upper triangular matrix. So without loss of generality, we can assume that
A
Ir Br,n−r
On−r,r 0n−r
, 2.18
whereIris the identity matrix,Br,n−r is any matrix. Using the last theorem, we get
wA A
2 1
2. 2.19
Corollary 2.5. SupposeA∈MnCandA2I. Then
wA 1
2
A 1
A
. 2.20
Proof. Acan be expressed as
A
Ir Br,n−r
On−r,r −In−r
, 2.21
whereIris the identity matrix,Br,n−r is any matrix. ByTheorem 2.3, we have
A
11
2B2 1 2B
4B2,
wA 1
2
4B2 .
2.22
Therefore
A211
2B21
2B
4B2, 1
A2 1
1 1/2B2 1/2B
4B2 11
2B2−1
2B
4B2.
2.23
By adding, we get
A2 1
A2 2B2 ⇒
A 1
A 2
4B2
⇒wA 1 2
A 1
A
.
2.24
Corollary 2.6. SupposeA∈MnCwithA2nI. Then
wA≥ 1
2
An 1 An
1/n
. 2.25
Proof. It follows from the fact thatAn2IandwAn≥wAn.
3. Bounds for Zeros of Polynomials
Letpz znan−1zn−1· · ·a1za0be a monic polynomial wherea0, a1, . . . , an−1are complex numbers and let
C p
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
−an−1 −an−2 · · −a1 −a0
1 0 · · 0 0
0 1 · · 0 0
0 0 · · 0 0
· · · ·
· · · · 1 0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
3.1
be the Frobenius companion matrix of the polynomialpz. Then, it is well known that zeros ofpare exactly the eigenvalues ofCp. ConsideringCpas a linear operator onCn, we see that ifzis root of the polynomial equationpz 0 then
|z| ≤w C
p asσ
C p
⊂W C
p
, 3.2
where σCp is the spectrum of operator Cp. Estimation of the roots of zeros of the polynomialpzhas been done by many mathematicians over the years, some of them are mentioned below. Letλbe a root of the polynomial equationpz 0.
iCarmichael and Mason3proved that
|λ| ≤
1|a0|2|a1|2· · ·|an−1|21/2
. 3.3
iiMontel4,5proved that
|λ| ≤ |a0||a0−a1|· · ·|an−2−an−1||an−11|,
|λ| ≤n−1 |a0||a1|· · ·|an−1|. 3.4
iiiCauchy3proved that
|λ| ≤1max{|a0|,|a1|, . . . ,|an−1|}. 3.5
ivFujii and Kubo6,7proved that
|λ| ≤cos π n1 1
2
⎡
⎣ n−2
i0
|ai|2 1/2
|an−1|
⎤
⎦. 3.6
vAlpin et al.8proved that
|λ| ≤max
1≤k≤n1|an−1|1|an−2|· · ·1|an−k|1/k. 3.7
viKittaneh1proved that
|λ| ≤ 1 2
!!C
p!!!!!C p2!!!1/2
. 3.8
We develop an inequality involving numerical radius with the help of which we estimate the zeros of the polynomialp. We show with examples that our estimation is better than the estimations mentioned above.
Theorem 3.1. Ifλis a zero of the polynomialpz, then
|λ| ≤ an−1
n 1
2
⎡
⎢⎢
⎢⎣cosπ n
#$
$$
$%cos2π n
⎛
⎝1
#$
$%n−2
r0|αr|2
⎞
⎠
2
⎤
⎥⎥
⎥⎦, 3.9
whereαr 'n−r
k0nCk−an−1/nkan−k, r 0,1,2, . . . , n−2.
Proof. Puttingzξhin the polynomial equationpz znanzn−1· · ·a2za10, we get ξhnan−1ξhn−1· · ·a1ξh a0 0. 3.10
Substitutingh−an−1/n, we get
ξnαn−2ξn−2αn−3ξn−3· · ·α1ξα00, 3.11
whereαr'n−r
k0nCK−an−1/nkan−k, r0,1,2, . . . , n−2.
LetCξbe the Frobenius companion matrix of the polynomialqξ ξnαn−2ξn−2 αn−3ξn−3· · ·α1ξα0.
ThenCξ A B
C D
,whereA 01, B −αn−2,−αn−3, . . . ,−α01,n−1
C
⎛
⎜⎜
⎜⎜
⎜⎝ 1 0
·
· 0
⎞
⎟⎟
⎟⎟
⎟⎠
n−1,1
, D
⎛
⎜⎜
⎜⎜
⎜⎝
0 0 · · 0 1 0 · · 0
· · · · 0
· · · · 0 0 0 · 1 0
⎞
⎟⎟
⎟⎟
⎟⎠
n−1,n−1
. 3.12
UsingTheorem 2.1, we get wCξ≤ 1
2
wA wD
wA−wD2 BC2
⇒wCξ≤ 1 2
⎡
⎢⎢
⎢⎣cosπ n
#$
$$
$%cos2π n
⎛
⎝1
#$
$%n−2
r0
|αr|2
⎞
⎠
2
⎤
⎥⎥
⎥⎦.
3.13
This shows that ifξ0is a zero of the polynomialqξ, then
|ξ0| ≤ 1 2
⎡
⎢⎢
⎢⎣cosπ n
#$
$$
$%cos2π n
⎛
⎝1
#$
$%n−2
r0
|αr|2
⎞
⎠
2
⎤
⎥⎥
⎥⎦. 3.14
Thus ifλis a zero of the polynomialpz, then
|λ| ≤ an−1
n 1
2
⎡
⎢⎢
⎢⎣cosπ n
#$
$$
$%cos2π n
⎛
⎝1
#$
$%n−2
r0
|αr|2
⎞
⎠
2
⎤
⎥⎥
⎥⎦. 3.15
This completes the proof of the theorem.
Example 3.2. Consider the polynomial equationpz z3−3z22z 0. Then the bounds estimated by different mathematicians are as shown inTable 1.
But our estimation shows that ifλis a zero of the polynomial then|λ| ≤ 2.280776406 which is much better than all the estimations mentioned above.
The companion matrix of the polynomial after removing the second term can be written as
⎛
⎝0 1 0 1 0 0 0 1 0
⎞
⎠
A2,2 B2,1
C1,1 D1,1
. 3.16
Then using the above theorems, it is easy to show that |λ| ≤ 2.207 which is even better estimation.
Example 3.3. Consider the polynomial equationpz z5−8z425z3−38z228z−8 0.
Then, the bounds estimated by different mathematicians are as shown inTable 2.
But our estimation shows that ifλis a zero of the polynomial then|λ| ≤ 2.703669110 which is much better than all the estimations mentioned above.
Theorem 3.4. Letpz znan−1zn−1· · ·a1za0havingαi i1,2, . . . , nas zeros and for each m∈N,pmz znamn−1zn−1· · ·am2 zam1 zam0 is a polynomial havingα2im i1,2, . . . , n as zeros. Ifλis a zero of the polynomialpz, then for allm
|λ| ≤
⎛
⎜⎜
⎝1 2
⎡
⎢⎢
⎣amn−1cos π
n
#$
$$
%
amn−1−cos π
n 2
⎛
⎝1
#$
$%n
k2
amn−k2
⎞
⎠
2⎤
⎥⎥
⎦
⎞
⎟⎟
⎠
1/2m
.
3.17
Table 1
Carmichael and Mason 3.741657387
Montel 7
Cauchy 4
Fujii and Kubo 4.0098824
Yuri, Chien and Yeh 3.464101615
Kittaneh 3.44572894
Table 2
Carmichael and Mason 54.60769176
Montel 215
Cauchy 39
Fujii and Kubo 32.16529279
Yuri, Chien and Yeh 9
Proof. We first prove the lemma which shows that the coefficients ofpmzcan be expressed in terms of coefficients ofpz.
Lemma 3.5. Supposepz znan−1zn−1· · ·a1za0is a monic polynomial, wherea0, a1, . . . , an−1
are complex numbers and αi, i 1,2, . . . , nare the zeros of this polynomial. If p1z zn a1n−1zn−1· · ·a11 za10 is the polynomial havingα2i i1,2, . . . , nas zeros, then forr 1,2, . . . , n:
a1r −12n−r
a2r2
n−r
k1
−1karkar−k
, wherean1, ankan−k0. 3.18
Proof. We have
det
z2I−C p2
det zI−C
p det
zIC p ⇒p1
z2
pzp−z
⇒z2na1n−1z2n−1· · ·a11 z2a10 −1n
znan−1zn−1· · ·a1za0
×
zn−an−1zn−1· · · −1n−1a1z −1na0
.
3.19
Comparing the coefficient ofz2r, we get forr1,2, . . . n:
a1r −12n−r
a2r2 n−r k1
−1karkar−k
, wherean1, ankan−k0. 3.20
This completes the proof of lemma.
The companion matrix of the monic polynomialpmzis
C pm
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
−amn−1 −amn−2 · · −am1 −am0
1 0 · · 0 0
0 1 · · 0 0
0 0 · · 0 0
· · · ·
0 0 · · 1 0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
. 3.21
We have
w C
pm
≥rσ
C pm
rσ
C p2m
. 3.22
So
rσ
C p
≤ w
C pm
1/2m
. 3.23
UsingTheorem 2.1, we get
w C
pm
≤ 1 2
⎡
⎢⎢
⎣amn−1cos π
n
#$
$$
%
amn−1−cos π
n 2
⎛
⎝1
#$
$%n
k2
amn−k2
⎞
⎠
2⎤
⎥⎥
⎦. 3.24
Thus ifλis a zero of the polynomialpz, then
|λ| ≤
⎛
⎜⎜
⎝1 2
⎡
⎢⎢
⎣amn−1cos π
n
#$
$$
%
amn−1−cos π
n 2
⎛
⎝1
#$
$%n
k2
amn−k2
⎞
⎠
2⎤
⎥⎥
⎦
⎞
⎟⎟
⎠
1/2m
.
3.25
This completes the proof of the theorem.
We next prove the theorem.
Theorem 3.6. Supposepz znan−1zn−1· · ·a1za0is a monic polynomial andαiare the roots of this equationi1,2, . . . , n, wherea0, a1, . . . , an−1are complex numbers with|α1|>1>|α2|>· · ·>
|αn|. If the equation having rootsα2im fori1,2, . . . , nispmz znamn−1zn−1· · ·am1 zam0 , then there existsm0∈Nsuch that
1|amn−k| ≤ |amn−1|whenever m≥m0 and for k2,3, . . . , n;
2 wCmp1/2m converges torσCp.
Proof. 1We prove this fork2 and the rest are similar.
First observe that amn−1
n k1
α2km
≥ |α1|2m−n
k2
|αk|2m, amn−2
n j /k1
α2jmα2km ≤ |α1|2m
n
k2
|αk|2m
n
j /k2
αj2m|αk|2m.
3.26
Now in order to have
amn−2≤amn−1. 3.27
We get
|α1|2m −n
k2
|αk|2m ≥ |α1|2m n
k2
|αk|2m
n
j /k2
αj2m|αk|2m, 3.28
that is,
|α1|2m ≥
1|α1|2m(n
k2
|αk|2m )
n
j /k2
αj2m|αk|2m, 3.29
that is,
|α1|2m 1|α1|2m ≥
'n
k2|αk|2m 1|α1|2m
'n
j /k2αj2m|αk|2m
1|α1|2m . 3.30
Clearly, this inequality holds good as the left-hand side converges to 1, but the right-hand side converges to 0.
2We have
w Cm
p
≤ 1 2
⎡
⎢⎢
⎣amn−1cos π
n
#$
$$
%
amn−1−cos π
n 2
⎛
⎝1
#$
$%n
k2
amn−k2
⎞
⎠
2⎤
⎥⎥
⎦, 3.31
that is,
w Cm
p
≤ 1 2
⎡
⎣amn−11
amn−12 1√
n−2amn−1 2⎤
⎦, 3.32
that is,
w Cm
p
≤ 1 2
⎡
⎣amn−1amn−1
amn−12 amn−1√
n−2amn−1 2⎤
⎦. 3.33
So we get
w Cm
p
≤Kamn−1≤K
n i1
α2im
≤K|α1|2m
1n
i2
αi
α1
2
m
. 3.34
Now
rσ
C p2m
rσ
Cm
p
≤w Cm
p
≤K|α1|2m
1n
i2
αi
α1
2m
. 3.35
Therefore,
rσ
C p
≤* w
Cm
p+1/2m
≤ |α1|
K
1n
i2
αi
α1
2m
1/2m
. 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 matrixAof ordern.
Step 1. We first find the characteristic polynomialqz znbn−1zn−1· · ·b1zb0. Suppose αi, i1,2, . . . , rare the distinct roots ofqz 0 with|α1|>|α2|>· · ·>|αr|.
Step 2. Findpz qz/gcdqz, qz zrar−1zr−1· · ·a1za0. Then, roots ofqz 0 areαi, i 1,2, . . . , r without multiplicity. Letpmz znamn−1zn−1· · ·am1 zam0 be the polynomial havingα2im fori1,2, . . . , ras its zeros.
Step 3. Since|α2| < |α1|, taking < 1/4|α1| − |α2|, we can see that|α2| < |α1 −2 . Again using a result of9, we get|amn−1|1/2m converging to|α1|. So for this there exists anm0 ∈N such that|α1| − <|amn−1|1/2m <|α1| for allm≥m0. Therefore,
|α2| <|α1| − <amn−11/2m <|α1| ∀m≥m0. 3.37
Table 3
No. of iterations am4 am3 am2 am1 am0 rσCqz
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
Step 4. Lett|amn−10|1/2m0 − . Findsz zr'r
k1ar−k/tkzr−kzrcr−1zr−1· · ·c1zc0. If the roots ofsz 0 areβi, thenβiαi/t, i1,2, . . . , rand
β1α1
t
>1>α2
t
β2>β3>· · ·>βr. 3.38
Then,szsatisfies all the criterion ofTheorem 3.6.
Step 5. The required sequence is xm twCms1/2m, which converges to the spectral radius of matrixA.
Example 3.7. Consider the 5th-degree polynomialqz z52z41.
By Rouche’s theorem, it is easy to see that all the roots except one are enclosed by the simple closed curve|z|2.
Considersz z5z4 1/25and then iterating the coefficints ofQzwe 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 Alpin8. The iterations are shown inTable 3.
Acknowledgments
The authors would like to thank the referees for their suggestions, one of the referees suggested the inclusion ofRemark 2.2followingTheorem 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.
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