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

We obtain inequalities involving numerical radius of a matrix . 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 up to the desired degree of accuracy. 1. Introduction Suppose . Let？？ , denote respectively the numerical range, spectrum of？？ ？？and , denote respectively the numerical radius, spectral radius of , that is, It is well known that(i) . Kittaneh [1] improved on the second inequality to prove that.(ii) . Clearly, so that inequality (ii) is sharper than the second inequality of (i). Let be a monic polynomial where are complex numbers and let be the Frobenius companion matrix of the polynomial . Then, it is well known that zeros of？？ ？？are exactly the eigenvalues of . Considering as an element of , we see that if is root of the polynomial equation , then Based on inequality (ii), Kittaneh [1] obtained an estimation for which gives an upper bound for zeros of the polynomial . 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 and use it to obtain numerical radius for some special class of matrices. Theorem 2.1. Suppose and where ,？？ ,？？ ？？and？？ . Then,(i) ？？and (ii) ？？+？？ . Proof. (i) Let and where and with . Then, and so Therefore, we have This completes the first part of the proof. (ii) Proceeding as in we can prove the second part. This completes the proof of the theorem. Remark 2.2. As an application of in Theorem 2.1, has another estimation by as follows: Furuta [2] obtained numerical radius for a bounded linear operator of the above form with ,？？ ,？？ ,？？ , and . If we consider ,？？ ,？？ where , then we can exactly calculate and as proved in the next theorem. Theorem 2.3. Suppose and Then(i) ？？and(ii) . Proof. (i) Following the method employed in the previous theorem, we can show that We only need to show that there exists ,？？ such that equals the quantity in the RHS. Suppose attains its norm at with . Let where is a scalar. Then, . Now so that Thus for all scalar , we get Case 1 ( ). Define a function by Then using elementary calculus, we can show that attains its maximum at so that for we get Thus, we get Case 2 ？？( ). As before we can show that there exists so that for we get Thus in all cases, we get 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