On Numerical Radius of Some Matrices

International Journal of Mathematical Analysis Vol., 08, no., 9-8 HIKARI Ltd, www.m-hikari.com https://doi.org/0.988/ijma.08.75 On Numerical Radius of Some Matrices Shyamasree Ghosh Dastidar Department of Mathematics, Vidyasagar Evening College Kolkata 700006, India Gour Hari Bera Department of Mathematics, St. Paul s C. M. College Kolkata 700009, India Copyright c 08 Shyamasree Ghosh Dastidar and Gour Hari Bera. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we obtain numerical radius of a general complex matrix of order two without reducing it in triangular form. We also find numerical radius of nilpotent matrix of order three and different matrices of order three and four satisfying special property. Mathematics Subject Classification: Primary 7A, Secondary 5A60 Keywords: Numerical radius, Nilpotent matrix, Non-negative matrix. Introduction Let M n (C) denote the algebra of all n n complex matrices. For T M n (C), let W (T ), ω(t ) denote respectively the numerical range, numerical radius of T, where W (T ) = { T x, x : x H, x = }, sup{ T x, x : x H, x = }. The following properties of ω(t ) are immediate: (i) ω(αt ) = α ω(t ), for any α C, Corresponding author

0 Shyamasree Ghosh Dastidar and Gour Hari Bera (ii) ω(u T U) = ω(t ), for any unitary matrix U in M n (C). It is well known that ω(.) defines a norm on M n (C) and T ω(t ) T, The lower bound of this inequality is attained if T is -nilpotent and the upper bound is attained if T is a normal []. In 975, Goldberg and Tadmor [] proved that if T is a non-negative matrix, then ReT, probably this is the biggest class of operators for which we know the exact numerical radius. The numerical radius equalities have been studied by many mathematician over the years [], [], [5], [6]. In this paper we study numerical radius equalities of some matrices. In section, we obtain numerical radius of T M (C) without reducing it in upper triangular form. In section, we find numerical radius for a nilpotent matrix of order. We also obtain numerical radius for T M (C) satisfying the property ReT +, where T + = ( a ij ) for T = (a ij ). In section, we find necessary and sufficient conditions for a nilpotent matrix of order to satisfy the property ReT +.. On numerical radius of matrix of order ( For any angle θ, define the Hermitian matrix H(θ) = e iθ T + e iθ T ), then max λ maxh(θ) = max λ maxh(θ), 0 θ π 0 θ π where λmax denotes the algebraically largest eigenvalue of H(θ) []. Based on this result we find numerical radius of the following matrix: Theorem.. Let T M (C) with T = ( a b c), where a, b, c C, then [ ( + Re c) cos θ Im c sin θ + max 0 θ π p cos θ + q sin θ + r sin θ where p = ( Re c) + a + b +Re(ab), q = (Im c) + a + b Re (ab), r = Im c( Re c) Im(ab) and ω(t ) is attained at θ = θ 0, where θ 0 is a solution of {( + Re c) tan θ + Im c} {q tan θ + r tan θ + p} = {r + (q p) tan θ r tan θ}. ( ) cos θ Proof. Here H(θ) = (aeiθ + be iθ ) (ae iθ + be iθ ) (ceiθ + ce iθ. ) The characteristic equation of H(θ) is x {( + Re c) cos θ Im c sin θ}x + cos θ(re c cos θ Im c sin θ) ( a + b + abe iθ + a be iθ) = 0, x {( + Re c) cos θ Im c sin θ}x + cos θ(re c cos θ Im c sin θ) ( a + b + Re(ab) cos θ Im(ab) sin θ ) = 0. ],

The eigenvalues of H(θ) are On numerical radius of some matrices [( + Re c) cos θ Im c sin θ] ± (Re c cos θ Im c sin θ cos θ) + a + b + Re(ab) cos θ Im(ab) sin θ, = [( + Re c) cos θ Im c sin θ] ± p cos θ + q sin θ + r sin θ, where p = (Re c ) + a + b + Re(ab), q = (Im c) + a + b Re(ab), r = ( Re c)im c Im(ab). max 0 θ π Let { ( + Re c) cos θ Im c sin θ + f(θ) = ( + Re c) cos θ Im c sin θ + } p cos θ + q sin θ + r sin θ. p cos θ + q sin θ + r sin θ. Then ω(t ) is attained at θ = θ 0, where θ 0 is a solution of df dθ = 0 i.e., {(+Re c) tan θ 0 +Im c} (q tan θ 0 +r tan θ 0 +p) = {r+(q p) tan θ 0 r tan θ 0 }. This completes the proof. Remark.. For T = ( a b c d) ; a, b, c, d R, it is easy to check that a + d + (a d) + (b + c).. On numerical radius of different form of matrices of order Let T = (a ij ) M n (C) and T + = ( a ij ). Then we know that ω(t ) Re(T + ) []. If T is a non-negative matrix i.e., a matrix with non-negative real entries, then Goldberg and Tadmor [] proved that Re(T ). In this section with the help of these results we obtain numerical radius of some matrices T M (C) satisfying the property Re(T + ). Theorem.. Let T = 0 α β 0 0 γ, where α, β, γ C be a nilpotent 0 0 0 matrix. Then Re(T + ) and ( ) ( ) α + β + γ α + β + γ α β γ tan α β γ

Shyamasree Ghosh Dastidar and Gour Hari Bera Proof. We have ω(t ) = max θ [0,π] Re(e iθ T ). 0 e iθ α e iθ β Here Re(e iθ T ) = e iθ α 0 e iθ γ. e iθ β e iθ γ 0 The characteristic equation of Re(e iθ T ) is x ( α + β + γ )x α β γ cos(θ α θ β + θ γ + θ) = 0, (..) where θ ξ = argξ. 0 α β Also T + = 0 0 γ and Re(T + ) = 0 0 0 The characteristic equation of Re(T + ) is 0 α β α 0 γ β γ 0. x ( α + β + γ )x α β γ = 0. (..) Now, for chosen θ = nπ (θ α θ β + θ γ ) for some n Z, (..) and (..) become identical. Since max θ [0,π] Re(e iθ T ) Re(T + ) and Re(e iθ 0 T ) = Re(T + ), for some θ 0 [0, π]. So, Re(T + ). Solving (..) by Cardan s method we get the largest eigenvalue of Re(T + ) as ( ) ( ) α + β + γ α + β + γ α β γ tan α β γ Hence Re(T + ) = ( α + β + γ ) ( ) α + β + γ α β γ tan α β γ = ω(t ). Remark.. A nilpotent matrix T M (C) is unitarily equivalent to T and ω( ) is unitarily invariant. So ω( T ) = ω(t ). Theorem.. Let T = 0 α 0 γ 0 β M (C), where α, β, γ, δ C. Then 0 δ 0 a necessary and sufficient condition for Re(T + ) is θ α θ β = θ δ θ γ, where θ ξ = arg ξ and [( α + γ ) + ( β + δ ) ]. Proof. We first prove the necessary part. Let Re(T + ). We have max θ [0,π] Re(e iθ T ).

On numerical radius of some matrices 0 αe iθ + γe iθ 0 Here Re(e iθ T ) = ᾱe iθ + γe iθ 0 βe iθ iθ + δe. 0 βe iθ + δe iθ 0 The characteristic equation of Re(e iθ T ) is x ( ) α + α γ cos(θ α + θ γ + θ) + β δ cos(θ β + θ δ + θ) x = 0. Also Re(T + ) = 0 α + γ 0 α + γ 0 β + δ 0 β + δ 0. (..) The characteristic equation of Re(T + ) is x ( ) α + α γ + β δ x = 0. (..) From the given condition, for some chosen θ, (..) and (..) must be identical. For this purpose, we must have θ α + θ γ = θ β + θ δ, i.e., θ α θ β = θ δ θ γ. Now we can choose θ = nπ θα+θγ for some n Z, then (..) and (..) both are identical. Therefore θ α θ β = θ δ θ γ is the necessary condition. Next we prove the sufficient part. Let θ α θ β = θ δ θ γ. Then the characteristic equation of Re(e iθ T ), i.e., (..) reduces to x ( ) α + α γ cos(θ α + θ γ + θ) + β δ cos(θ α + θ γ + θ) x = 0. (..) We now choose θ = nπ θα+θγ for some n Z, then (..) and (..) are identical and therefore Re(T + ). It is obvious from (..) that [( α + γ ) + ( β + δ ) ]. Remark.. The condition in Theorem. is also applicable for T = 0 α β γ 0 0, T = 0 0 α 0 0 β, where α, β, γ, δ C and ω(t ) = δ 0 0 γ δ 0 ω(t ) = ω(t ). Theorem.. Let T = 0 α β 0 0 γ M (C), where α, β, γ, δ C. Then δ 0 0 a necessary and sufficient condition for Re(T + ) is (θ α + θ γ ) = θ β + θ δ, where θ ξ = arg ξ and ( ) ( ) α α + β δ + β δ α γ ( β + δ ) tan α γ ( β + δ )

Shyamasree Ghosh Dastidar and Gour Hari Bera Proof. We first prove the necessary part. Let Re(T + ). We have max θ [0,π] Re(e iθ T ). The characteristic equation of Re(e iθ T ) is x ( ) α + β δ cos(θ β + θ δ + θ) x ( α β γ cos(θ α θ β +θ γ +θ)+ α γ δ cos(θ α +θ γ +θ δ +θ)) = 0. (..) The characteristic equation of Re(T + ) is x ( ) α + β δ x ( α β γ + α γ δ ) = 0. (..) According to given condition, for some chosen θ, (..) and (..) must be identical. For this purpose, we must have θ α + θ γ + θ δ = (θ α θ β + θ γ ), i.e., (θ α + θ γ ) = θ β + θ δ. (..) So, θ β + θ δ = θ β + θ α + θ γ θ β = (θ α θ β + θ γ ) by (..). If we choose θ = nπ (θ α θ β + θ γ ) for some n Z, then (..) and (..) both are identical under the condition (..). Therefore (..) is the necessary condition. Next we prove the sufficient part. Let (θ α + θ γ ) = θ β + θ δ. Then the characteristic equation of Re(e iθ T ), i.e., (..) reduces to x ( ) α + β δ cos((θ α θ β + θ γ + θ)) x ( α β γ cos(θ α θ β + θ γ + θ) + α γ δ cos((θ α θ β + θ γ + θ))) = 0. (..) Now we choose θ = nπ (θ α θ β + θ γ ) for some n Z, then (..) and (..) become identical and therefore Re(T + ). Solving (..) by Cardan s method we get the largest eigenvalue of Re(T + ) and so ( ) ( ) α α + β δ + β δ α γ ( β + δ ) tan α γ ( β + δ ) Remark.. The result in Theorem. is also applicable for T = 0 0 α γ 0 β. 0 δ 0

On numerical radius of some matrices 5 Theorem.. Let T = 0 α β γ 0 0 M (C), where α, β, γ, δ C.Then 0 δ 0 a necessary and sufficient condition for Re(T + ) is (θ β + θ δ ) = θ α + θ γ, where θ ξ = arg ξ and ( ) ( ) α α + α γ + α γ β δ ( α + γ ) tan β δ ( α + γ ) Proof. The proof is similar to that of Theorem.. Theorem.5. Let T = 0 α β δ 0 γ M (C), where α, β, γ, δ C. Then 0 0 0 a necessary and sufficient condition for Re(T + ) is (θ β θ γ ) = θ α θ δ, where θ ξ = arg ξ and ( ) ( ) α α + α δ + α δ β γ ( α + δ ) tan β γ ( α + δ ) Proof. We first prove the necessary part. Let Re(T + ). We have max θ [0,π] Re(e iθ T ). The characteristic equation of Re(e iθ T ) is x ( ) α + α δ cos(θ α + θ δ + θ) x ( α β γ cos(θ α θ β +θ γ +θ) + β γ δ cos(θ β θ γ + θ δ + θ)) = 0. (.5.) The characteristic equation of Re(T + ) is x ( ) α + α δ x β γ ( α + δ ) = 0. (.5.) According to given condition, for some chosen θ, (.5.) and (.5.) must be identical. For this purpose, we must have θ α θ β + θ γ = θ β θ γ + θ δ, i.e., (θ β θ γ ) = θ α θ δ. (.5.) So by (.5.), we have θ α + θ δ = θ α (θ β θ γ ) = (θ α θ β + θ γ ). Now we choose suitable θ = nπ (θ α θ β + θ γ ) for some n Z, then (.5.) and (.5.) become identical under the condition (.5.). Therefore (.5.) is the necessary condition.

6 Shyamasree Ghosh Dastidar and Gour Hari Bera We next prove the sufficient part. Let (θ β θ γ ) = θ α θ δ. Then the characteristic equation of Re(e iθ T ), i.e., (.5.) reduces to x ( ) α + α δ cos((θ α θ β + θ γ + θ)) x ( α β γ cos(θ α θ β + θ γ + θ) + β γ δ cos(θ α θ β + θ γ + θ)) = 0. (.5.) Now we choose θ = nπ (θ α θ β + θ γ ) for some n Z, then (.5.) and (.5.) both are identical and therefore Re(T + ). Solving (.5.) by Cardan s method we can find the largest eigenvalue of Re(T + ) and so ( ) ( ) α α + α δ + α δ β γ ( α + δ ) tan β γ ( α + δ ) Remark.. The result in Theorem.5 is also applicable for T = 0 0 α γ 0 β, δ 0 0 where α, β, γ, δ C. Theorem.6. Let T = 0 α β 0 0 γ M (C), where α, β, γ, δ C. Then 0 δ 0 a necessary and sufficient condition for Re(T + ) is (θ α θ β ) = θ δ θ γ, where θ ξ = arg ξ and ( ) ( ) α α + γ δ + γ δ α β ( γ + δ ) tan α β ( γ + δ ) Proof. The proof is similar to that of Theorem.5. On numerical radius of a nilpotent matrix of order 0 α β γ Theorem.. Let T = 0 0 δ ξ 0 0 0 η M (C), be a nilpotent matrix, 0 0 0 0 where α, β, γ, δ, ξ, η C and atleast five of them are non-zero. Then a necessary and sufficient condition for Re(T + ) is θ α +θ ξ θ β θ η = 0, θ β + θ ξ θ γ θ δ = 0, where θ φ = arg φ.

On numerical radius of some matrices 7 Proof. We first prove the necessary part. Let Re(T + ). We have max θ [0,π] Re(e iθ T ). The characteristic equation of Re(e iθ T ) is x α x ( α β δ cos(θ α θ β +θ δ +θ)+ α γ ξ cos(θ α θ γ +θ ξ +θ) + δ ξ η cos(θ δ θ ξ + θ η + θ) + β γ η cos(θ β θ γ + θ η + θ)) x + 6 [ α η + β ξ + δ γ α β ξ η cos(θ α + θ ξ θ β θ η ) β γ ξ δ cos(θ β + θ ξ θ γ θ δ ) α δ γ η cos(θ α + θ δ θ γ + θ η + θ)] = 0. (..) The characteristic equation of Re(T + ) is x α x ( α β δ + α γ ξ + β γ η + ξ δ η )x + 6 ( α η + β ξ + δ γ α γ δ η α β ξ η β γ δ ξ ) = 0. (..) According to given condition, for some chosen θ, (..) and (..) must be identical. For this purpose, we must have θ α + θ ξ θ β θ η = 0 (..) θ β + θ ξ θ γ θ δ = 0. (..) From (..) and (..), we get θ ξ + θ α θ γ θ δ θ η = 0, θ ξ = θ γ + θ δ + θ η θ α. (..5) and θ α θ β θ η + θ γ + θ δ = 0, θ β = θ α + θ γ + θ δ θ η. (..6) So, θ α θ β + θ δ = θα θγ+θ δ+θ η by (..6), θ α θ γ + θ ξ = θα θγ+θ δ+θ η by (..5), θ δ θ ξ + θ η = θ δ+θ η+θ α θ γ by (..5), θ β θ γ + θ η = θα+θη+θ δ θ γ by (..6). Now we choose θ = nπ θα+θ δ+θ η θ γ, for some n Z, then (..) and (..) become identical under the conditions (..) and (..). Therefore (..) and (..) are the necessary conditions. We next prove the sufficient part. Let θ α +θ ξ θ β θ η = 0 and θ β +θ ξ θ γ θ δ = 0. Then the characteristic equation of Re(e iθ T ), i.e., (..) reduces to x α x ( α β δ + α γ ξ + δ ξ η + β γ η ) ( ) θα + θ δ θ γ + θ η cos + θ x + 6 [ α η + β ξ + δ γ α β ξ η

8 Shyamasree Ghosh Dastidar and Gour Hari Bera β γ ξ δ α δ γ η cos(θ α + θ δ θ γ + θ η + θ)] = 0. (..7) Now we choose θ = nπ θα+θ δ+θ η θ γ for some n Z, then (..) and (..7) both are identical and therefore Re(T + ). Remark.. If the number of non-zero variables is less than five, then there is no condition need to satisfy. For some chosen θ we always have identical characteristic equations, i.e., Re(T + ). References [] T. Furuta, Applications of polar decompositions of idempotent and -nilpotent operators, Linear and Multilinear Algebra, 56 (008), 69-79. https://doi.org/0.080/00808070009 [] M. Goldberg and E. Tadmor, On the numerical radius and its application, Linear Algebra and its Applications, (98), 6-8. https://doi.org/0.06/00-795(8)9055-0 [] K.E. Gustafson and D.K.M. Rao, Numerical Range, Springer-Verlag, Newyork, Inc., 997. https://doi.org/0.007/978--6-898- [] R.A. Horn and C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, Cambridge, 99. https://doi.org/0.07/cbo97805807 [5] K. Paul and S. Bag, On numerical Radius of a Matrix and Estimation of Bounds for Zeros of a Polynomial, International Journal of Mathematics and Mathematical Sciences, 0 (0), Article ID 9. https://doi.org/0.55/0/9 [6] K. Paul and S. Bag, Estimation of Bounds for the Zeros of a Polynomial Using Numerical Radius, Applied Mathematics and Computation, (0), -. https://doi.org/0.06/j.amc.0.07.06 Received: November 9, 07; Published: January 7, 08