The coefficients of the characteristic polynomial in terms of the eigenvalues and the elements of an n n matrix

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1 Alied Mathematics Letters 19 (2006) The coefficients of the characteristic olynomial in terms of the eigenvalues and the elements of an n n matrix Bernard P. Brooks Rochester Institute of Technology, Deartment of Mathematics and Statistics, College of Science, 85 Lomb Memorial Drive, Rochester, NY 14623, United States Received 7 March 2005; received in revised form 2 June 2005; acceted 11 July 2005 Abstract The coefficients of the characteristic olynomial of an n n matrix are derived in terms of the eigenvalues and in terms of the elements of the matrix. The connection between the two exressions allows the sum of the roducts of all sets of k eigenvalues to be calculated using cofactors of the matrix Elsevier Ltd. All rights reserved. Keywords: Derivative of a determinant; Eigenvalue; Jacobian 1. Motivations In order to derive general stability conditions for a first order three-dimensional discrete dynamic the coefficients of the characteristic olynomial of the Jacobian evaluated at equilibrium need to be exressed in terms of the three eigenvalues [1]. In order to extendthis stability analysis technique to a first order n-dimensional discrete dynamic the coefficients of the characteristic olynomial of an n n Jacobian evaluated at the equilibrium must first be exressed in terms of the eigenvalues and in terms of the elements of the Jacobian. Thus, given an n n matrix A [a ij ], a ij R, wewishtodetermine the relationshi between the eigenvalues of A and the coefficients of the characteristic olynomial C(x) address: bbsma@rit.edu. URL: htt:// bbsma /$ - see front matter 2005 Elsevier Ltd. All rights reserved. doi: /j.aml

2 512 B.P. Brooks / Alied Mathematics Letters 19 (2006) and the relationshi between the coefficients of the characteristic olynomial and the real elements of the matrix. 2. The coefficients of the characteristic olynomial in terms of the eigenvalues of A We can exress the characteristic olynomial as C(x) (x λ 1 )(x λ 2 ) (x λ k ) (x λ n ) where λ i are the eigenvalues of the matrix A.Exanding the above form clearly shows that the coefficient for x n 1 is the negative of the trace of A and the magnitude of the constant term is the determinant of A: C(x) x n tr(a)x n 1 ( 1) n det(a). Note that the signs of the coefficients alternate. We have, by convention, made the sign of the 1 coefficient of x n ositive: C(x) x n tr(a)x n 1 ( 1) k c k x n k ( 1) n det(a). Next we need to show by induction that c k is the sum of all roducts of k eigenvalues. Thus c k all sets of kλ s λ iλ j..λ l and c 1 trace, c n det(a). Consider the n 3 case: C(x) (x λ 1 )(x λ 2 )(x λ 3 ) x 3 (λ 1 λ 2 λ 3 )x 2 (λ 1 λ 2 λ 1 λ 3 λ 2 λ 3 )x λ 1 λ 2 λ 3 x 3 c 1 x 2 c 2 x c 3 x 3 tr(a)x 2 c 2 x det(a). Now assume that C(x) (x λ 1 )(x λ 2 ) (x λ k ) (x λ n ) C(x) x n c 1 x n 1 c 2 x n 2 ( 1) k c k x n k ( 1) n c n. Next we will consider a new exanded C(x) (x λ)c(x): C(x) (x λ)[x n c 1 x n 1 c 2 x n 2 ( 1) k c k x n k ( 1) n c n ] x n1 c 1 x n c 2 x n 1 ( 1) k c k x n k1 ( 1) n c n x λx n λc 1 x n 1 λc 2 x n 2 ( 1) k λc k x n k ( 1) n1 λc n. Now we collect on owers of x: C(x) x n1 [λ c 1 ]x n [λc 1 c 2 ]x n 1 ( 1) k [λc k 1 c k ]x n k1 ( 1) n1 λc k and therefore, by induction, the coefficient of x n k can be seen to be the sum of all roducts of k eigenvalues. The sign of that coefficient is ( 1) n k. Thus the characteristic olynomial can be exressed as [ ] [ ] [ ] C(x) x n x n 1 x n 2 ( 1) k λλ..λ x n k λ sets of 1 ( 1) n [ sets of n ] λλλ..λ. λλ sets of 2 sets of k

3 B.P. Brooks / Alied Mathematics Letters 19 (2006) An alternative roof can be found in [2]. Next the characteristic olynomial will be exressed using the elements of the matrix A, C(x) ( 1) n det [A xi], withthesign factor, ( 1) n,usedsothatthe coefficient of x n is 1. The coefficients will now be generated by differentiating C(x) as a determinant. The formula for the kth derivative of a generaldeterminant will now be shown. 3. The kth derivative of adeterminant Theorem 1. The kth derivative of a determinant of the n nmatrixm[m ij ] is given by the formula d k dt k m 11 (t) m 12 (t) m 1n (t) m 21 (t) m 22 (t) m n1 (t) m n2 (t) m nn (t) k k i 0 i, i 1, 2,..., k 1 k 2 k k where m ij (t) are real functions of t R. d k 1m 11 (t) d k 1m 12 (t) d k 1m 1n (t) dt k 1 dt k 1 dt k 1 d k 2m 21 (t) d k 2m 22 (t) dt k 2 dt k 2 d k m n1 (t) d k m n2 (t) d k m nn (t) dt k dt k dt k Proof. Exress the determinant as a sum of roducts: m 11 (t) m 12 (t) m 1n (t) det(m) m 21 (t) m 22 (t) sgn(π) m iπi. πs m n1 (t) m n2 (t) m nn (t) n i1 Take the kth derivative with resect to t using the generalizedleibniz rule to differentiate the roducts: d k dt k det(m) πs n sgn(π) k k i 0 i, i 1, 2,..., k 1 k 2 k k k k i 0 i, i 1, 2,..., k 1 k 2 k k πs n sgn(π) i1 i1 d k i m iπi dt k i d k i m iπi dt k i

4 514 B.P. Brooks / Alied Mathematics Letters 19 (2006) d k 1m 11 (t) d k 1m 12 (t) d k 1m 1n (t) dt k 1 dt k 1 dt k 1 k d k 2m 21 (t) d k 2m 22 (t) k k i 0 1!k 2! k! dt k 2 dt k 2. i, i 1, 2,..., k 1 k 2 k k d k m n1 (t) d k m n2 (t) d k m nn (t) dt k dt k dt k 4. The coefficients of the characteristic olynomial in terms of the elements of A Next let us exress the characteristic olynomial as C(x) ( 1) n det[a xi]. It is known that the coefficient of the x n 1 term in C(x) is the negative of the trace of the matrix A and the constant term is the determinant of A. Tocalculate the coefficient of the x k 1 term, dk C dx k x0 will be constructed from elements of the matrix, A, usingtheabove-derived derivative of a determinant formula. The elements of A are all real constants and all the determinants in the sum below when k i 0 or 1 are zero: 1 d k dx k ( 1)n det[a xi] ( 1)n d k a 11 x a 12 a 1n a 21 a 22 x dx k a n1 a n2 a nn x 1 d k n C dx k x0 ( 1) n 21. n1 nn i 1...k ii 1, ij ji 0 else mj a mj ( Thus the coefficient of the x k term in C(x) can be generated by summing the n k) determinants of the matrices created by relacing k of the diagonal elements of matrix A with 1andtheremaining elements in those corresonding rows and columns with 0. Examle 1. Consider the matrix A

5 B.P. Brooks / Alied Mathematics Letters 19 (2006) The ( coefficient of the x 2 term in the characteristic olynomial of A can be calculated by summing the 5 2) 10 determinants: and multilying by ( 1) 5.Thus the coefficient of x 2 is 273. We also know that the sum of all ten roducts of three eigenvalues is 273. This is of course valid regardless of the eigenvalues being real or comlex or distinct. 5. Conclusions Of course in the above examle the characteristic olynomial can be calculated directly. The real strength of this concet occurs when each real element in the matrix in question is a comlicated exression involving arameters, such as occurs in many mathematical models. This method can be used in the derivation of the stability conditions for a first order two-dimensional discrete dynamic [3] and for the derivation of the stability conditions for a first order three-dimensional discrete dynamic [1]. In an ucoming work the stability conditions for a first order four-dimensional discrete dynamic and stability conditions for a first order n-dimensional discrete dynamic will be derived using the results of this work. References [1] B.P. Brooks, Linear stability conditions for a first order 3-dimensional discrete dynamic, Alied Mathematics Letters 17 (4)(2004) [2] R.R.Silva, The trace formulas yield the inverse metric formula, Journal of Mathematical Physics 39 (1998) [3] J.M.T.Thomson, Non-Linear Dynamics and Chaos, John Wiley & Sons, Toronto, 1986.