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1、On Taylors formula for the resolvent of a complex matrixMatthew X. Hea, Paolo E. Ricci b,_Article history:Received 25 June 2007Received in revised form 14 March 2008Accepted 25 March 2008Keywords: Powers of a matrixMatrix invariantsResolvent1. IntroductionAs a consequence of the Hilbert identity in
2、1, the resolvent = of a nonsingular square matrix ( denoting the identity matrix) is shown to be an analytic function of the parameter in any domain D with empty intersection with the spectrum of . Therefore, by using Taylor expansion in a neighborhood of any fixed , we can find in 1 a representatio
3、n formula for using all powers of .In this article, by using some preceding results recalled, e.g., in 2, we write down a representation formula using only a finite number of powers of . This seems to be natural since only the first powers of are linearly independent.The main tool in this framework
4、is given by the multivariable polynomials (;) (see 26), depending on the invariants of ); here m denotes the degree of the minimal polynomial.2. Powers of matrices and functionsWe recall in this section some results on representation formulas for powers of matrices (see e.g. 26 and the references th
5、erein). For simplicity we refer to the case when the matrix is nonderogatory so that .Proposition 2.1. Let be an complex matrix, and denote by the invariants of , and by.its characteristic polynomial (by convention ); then for the powers of with nonnegative integral exponents the following represent
6、ation formula holds true:. (2.1)The functions that appear as coefficients in (2.1) are defined by the recurrence relation, (2.2)and initial conditions: . (2.3)Furthermore, if is nonsingular , then formula (2.1) still holds for negative values of n, provided that we define the function for negative v
7、alues of n as follows:，.3. Taylor expansion of the resolventWe consider the resolvent matrix defined as follows:. (3.1)Note that sometimes there is a change of sign in Eq. (3.1), but this of course is not essential.It is well known that the resolvent is an analytic (rational) function of in every do
8、main D of the complex plane excluding the spectrum of , and furthermore it is vanishing at infinity so the only singular points (poles) of are the eigenvalues of .In 6 it is proved that the invariants of are linked with those of by the equations ,. （3.2）As a consequence of Proposition 2.1, and Eq. (
9、3.2), the integral powers of can be represented as follows.Theorem 3.1 For every and , , (3.3)where the are given by Eq.(3.2). Denoting by the spectral radius of , for every , such that the Hilbert identity holds true(see 1): . (3.4)Therefore for every , we have, (3.5)and in general, (3.6)so, for ev
10、ery can be expanded in the Taylor series, (3.7)which is absolutely and uniformly convergent in D. Defining , (3.8), (3.9)where the are defined by Eq. (3.2), we can prove the following theorem.Theorem 3.2 The Taylor expansion (3.7) of the resolvent in a neighborhood of any regular point can be writte
11、n in the form. (3.10)Therefore we can derive as a consequence:Corollary 3.1 For every and the series expansions (3.11)are convergent.Proof. Recalling (3.3), we can write，Therefore, taking into account the initial conditions (2.3) we can write,so (3.10) holds true. The convergence of series expansion
12、s (3.11) is a trivial consequence of the convergence of the initial expansion (3.7).4. Concluding remarksIt is worth noting that the resolvent is a keynote element for representing analytic functions of a matrix . In fact,denoting by a function of the complex variable , analytic in a domain containi
13、ng the spectrum of , and denoting by the distinct eigenvalues of with multiplicities , the LagrangeSylvester formula (see 4) is given by,where is the projector associated with the eigenvalue , and.Denoting by a Jordan curve, the boundary of the domain , separating a fixed from all other eigenvalues,
14、 recalling the Riesz formula, it follows that.When is only known approximately, this projector cannot be derived by using the residue theorem.In this case it is necessary to integrate along (being possibly a Gershgorin circle), by using the known representation of the resolvent (see 3), (4.1)or by substituting with its Taylor expansion, and assuming as initial point any inside .Which is the best formula depends on the relevant stability and computational cost. From the theoretical point of view,formulas (3.7), (3.10) and (4.1) seem to be equivalent from the stability point of view, sin
