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The Computation of the Inverse of a Square Polynomial Matrix ∗ The Computation of the Inverse of a Square Polynomial Matrix ∗ Ky M. Vu, PhD. AuLac Technologies Inc. c 2008 Email: [email protected] Abstract 2 The Inverse Polynomial An approach to calculate the inverse of a square polyno- The inverse of a square nonsingular matrix is a square ma- mial matrix is suggested. The approach consists of two trix, which by premultiplying or postmultiplying with the similar algorithms: One calculates the determinant polyno- matrix gives an identity matrix. An inverse matrix can be mial and the other calculates the adjoint polynomial. The expressed as a ratio of the adjoint and determinant of the algorithm to calculate the determinant polynomial gives the matrix. A singular matrix has no inverse because its deter- coefficients of the polynomial in a recursive manner from minant is zero; we cannot calculate its inverse. We, how- a recurrence formula. Similarly, the algorithm to calculate ever, can always calculate its adjoint and determinant. It is, the adjoint polynomial also gives the coefficient matrices therefore, always better to calculate the inverse of a square recursively from a similar recurrence formula. Together, polynomial matrix by calculating its adjoint and determi- they give the inverse as a ratio of the adjoint polynomial nant polynomials separately and obtain the inverse as a ra- and the determinant polynomial. tio of these polynomials. In the following discussion, we will present algorithms to calculate the adjoint and deter- minant polynomials. Keywords: Adjoint, Cayley-Hamilton theorem, determi- nant, Faddeev’s algorithm, inverse matrix. 2.1 The Determinant Polynomial To begin our discussion, we consider the following equa- 1 Introduction tion that is the determinant of a square matrix of dimension n, viz Modern control theory consists of a large part of matrix the- n n−1 n−2 n jλI − Cj = λ − p1(C)λ + p2(C)λ ··· (−1) pn(C); ory. A control theorist is usually an expert in matrix theory, n X i n−i a field of applied algebra. There are books written entirely = (−1) pi(C)λ ; p0(C) = 1; on matrix theory for control engineers. We can cite two i=0 books: one by S. Barnett (1960) and the other by T. Kac- = (λ − λ1)(λ − λ2) ··· (λ − λn); zorek (2007). The reason for this fact is that matrices are = f(λ): used to represent models of a control system. Since matri- ces are used in control theory, their manipulation and calcu- The coefficients pi(C)’s are called the characteristic coef- lation are of paramount importance for control engineers. ficients of the matrix C. They are related to the zeros λi’s The most significant task is the calculation of the inverse of the polynomial f(λ) as of a polynomial matrix. There are algorithms, discussed in textbooks, for this task. We can cite the algorithms in p1(C) = λ1 + λ2 + ··· + λn; V. Kuceraˇ (1979) and T. Kaczorek (2007). These algo- = trC; rithms are poor and inefficient in the sense that they cannot p2(C) = λ1λ2 ··· + λ1λn + λ2λ3 ··· + λ2λn ··· + λn−1λn; be used for similar types of polynomial matrix. In this pa- . per, we will present algorithms to calculate the determinant . and adjoint polynomials of an inverse polynomial matrix. pn(C) = λ1λ2 ··· λn; The algorithms are an extension of a known algorithm by = jCj: K. Vu (1999) to calculate the characteristic coefficients of a pencil. The paper is organized as follows. Section one The characteristic coefficients are given recursively as is the introduction section. The algorithms are discussed m in section two. In section three, some examples are given. 1 X p (C) = (−1)i−1p (C)p (Ci): Section four concludes the discussion of the paper. m m m−i 1 i=1 i ∗A new-and-improved version of this paper can be found in the The coefficients p1(C )’s are the Newton sums of the func- Proceedings of the IEEE Multi-conference on Systems and Control on tion f(λ). Now, consider the case September 3-5th, 2008 in San Antonio, TX under the title An Extension 2 r of the Faddeev’s Algorithms. C = A + µB1 + µ B2 + ··· + µ Br: (1) i In this case, the function pn(C) is a polynomial of degree The derivatives of the matrix C can be calculated re- n × r in the parameter µ. Similarly, the characteristic co- cursively. In the following discussion, we will present an efficient pm(C) is a polynomial of degree m × r in the approach to calculate them numerically. We write parameter µ. Therefore, we can write Ci = Ci−1C; m×r X k then differentiate both sides with respect to µ to obtain pm(C) = qm;k(C)µ ; m = 1; ··· n: j k=0 djCi X j dj−lCi−1 dlC = ; From the last equation, we can obtain dµj l dµj−l dµl l=0 k j 1 d pm(C) djCi−1 X j dj−lCi−1 dlC qm;k(C) = k ; k = 0; ··· m × r: = C + ; k! dµ µ=0 dµj l dµj−l dµl l=1 Some special values for these coefficients are given below j i−2 2 j j−l i−k l d C X X j d C d C = C2 + Ck−1: dµj l dµj−l dµl qm;0(C) = pm(A); k=1 l=1 q0;0(C) = 1: To be able to calculate the derivatives numerically in a re- cursive manner, we have to eliminate the jth derivative of The coefficient q (C) can be calculated recur- m;k the matrix Ci−2 on the right hand side of the above equa- sively, and the last equation serves as the initial condition tion. We, therefore, will make repeated substitutions to re- for the calculation. In the following discussion, we will duce the power of the matrix Ci−2 with matrices of lower prove this fact. powers until we know the jth derivative of this matrix is From the relation of the characteristic coefficients zero. This matrix can be an identity matrix, so we have m 1 X i−1 i eventually pm(C) = (−1) pm−i(C)p1(C ) m i j i=1 djCi X X j dj−lCi−k dlC = Ck−1: dµj l dµj−l dµl and the derivative of a product of two scalars k=1 l=1 k i k k−j j i The lth derivative of the matrix C can be evaluated as d pm−i(C)p1(C ) X k d pm−i(C) d p1(C ) r k = k−j j ; l l dµ j dµ dµ d C d X s j=0 = [A + B µ ]; dµl dµl s s=1 we can write r X s−l = s(s − 1) ··· (s − l + 1)Bsµ : (3) qm;k(C) = s=l m k k−j j i 1 X i−1 X k d pm−i(C) d p1(C ) (−1) : By putting this derivative into the previous equation, we get k!m j dµk−j dµj i=1 j=0 µ=0 i j r djCi X X j dj−lCi−k X = s(s − 1) ··· (s − l + 1) By canceling and redistributing the factorials, we get dµj l dµj−l k=1 l=1 s=l m k k−j j i B µs−lCk−1; 1 X X d pm−i(C) d p1(C ) s q (C) = (−1)i−1 ; m;k k−j j i j j−l i−k r m (k − j)!dµ j!dµ µ=0 X X j! d C X i=1 j=0 = s(s−1) ··· (s−l+1) l!(j − l)! dµj−l m k j i k=1 l=1 s=l 1 X X i−1 d p1(C ) = (−1) qm−i;k−j(C) : s−l k−1 m j!dµj Bsµ C : (4) i=1 j=0 µ=0 Now, by setting µ = 0, Eq. (3) becomes i The derivatives of the characteristic coefficient p1(C ) are l r not difficult to obtain, but it is complicated to obtain a gen- d C X s−l = s(s − 1) ··· (s − l + 1)Bsµ ; eral formula for them. We, therefore, will solve the prob- dµl µ=0 s=l lem as follows. By noting that we can interchange the order µ=0 of the operators: trace and derivative, we write = s!Bl; = l!Bl: djp (Ci) djCi 1 = p ; j!dµj 1 j!dµj Therefore, by setting µ = 0 and dividing both sides of Eq. (4) by j!, we can calculate the derivatives of the matrix Ci then obtain recursively as m k j i j i i j j−l i−k 1 X X i−1 d C d C X X d C k−1 qm;k(C) = (−1) qm−i;k−j(C) p1 (2) = BlA m j!dµj j!dµj (j − l)!dµj−l i=1 j=0 µ=0 µ=0 k=1 l=1 µ=0 with the initial condition By differentiating both sides k times, dividing both sides by k! and setting µ = 0, we get d0Ci = Ai: 0 n−1 k dµ µ=0 X i 1 d i Dk = (−1) pn−1−i(C)C ; k! dµk With the derivatives obtained, we can take their traces and i=0 µ=0 calculate, using Eq. (2), all the coefficients q (C)’s of n−1 k k−j j i m;k X X i 1 k d pn−1−i(C) d C = (−1) ; the characteristic coefficient pm(C) recursively. The deter- k! j dµk−j dµj i=0 j=0 µ=0 minant corresponds to the coefficient pn(C).
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