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On the M-Th Roots of a Complex Matrix The Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 9, pp. 32-41, April 2002 ELA http://math.technion.ac.il/iic/ela ON THE mth ROOTS OF A COMPLEX MATRIX∗ PANAYIOTIS J. PSARRAKOS† Abstract. If an n × n complex matrix A is nonsingular, then for everyinteger m>1,Ahas an mth root B, i.e., Bm = A. In this paper, we present a new simple proof for the Jordan canonical form of the root B. Moreover, a necessaryand sufficient condition for the existence of mth roots of a singular complex matrix A is obtained. This condition is in terms of the dimensions of the null spaces of the powers Ak (k =0, 1, 2,...). Key words. Ascent sequence, eigenvalue, eigenvector, Jordan matrix, matrix root. AMS subject classifications. 15A18, 15A21, 15A22, 47A56 1. Introduction and preliminaries. Let Mn be the algebra of all n × n complex matrices and let A ∈Mn. For an integer m>1, amatrix B ∈Mn is called an mth root of A if Bm = A.IfthematrixA is nonsingular, then it always has an mth root B. This root is not unique and its Jordan structure is related to k the Jordan structure of A [2, pp. 231-234]. In particular, (λ − µ0) is an elementary m k divisor of B if and only if (λ − µ0 ) is an elementary divisor of A.IfA is a singular complex matrix, then it may have no mth roots. For example, there is no matrix 01 B such that B2 = . As a consequence, the problem of characterizing the 00 singular matrices, which have mth roots, is of interest [1], [2]. m Consider the (associated) matrix polynomial P (λ)=Inλ − A, where In is the identity matrix of order n and λ is a complex variable. A matrix B ∈Mn is an mth root of A if and only if P (B)=Bm − A = 0. As a consequence, the problem of computation of mth roots of A is strongly connected with the spectral analysis of P (λ). The suggested references for matrix polynomials are [3] and [7]. A set of vectors {x0,x1,...,xk}, which satisfies the equations P (ω0)x0 =0 1 (1) P (ω0)x1 + P (ω0)x0 =0 1! . 1 (1) 1 (k) P (ω0)xk + P (ω0)xk−1 + ···+ P (ω0)x0 =0, 1! k! where the indices on P (λ) denote derivatives with respect to the variable λ, is called a Jordan chain of length k +1 of P (λ) corresponding to the eigenvalue ω0 ∈ C and n the eigenvector x0 ∈ C . The vectors in a Jordan chain are not uniquely defined and for m>1, they need not be linearly independent [3], [6]. If we set m =1, then the ∗Received bythe editors on 19 February2002. Accepted for publication on 9 April 2002. Handling Editor: Peter Lancaster. †Department of Mathematics, National Technical University, Zografou Campus, 15780 Athens, Greece ([email protected]). 32 The Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 9, pp. 32-41, April 2002 ELA http://math.technion.ac.il/iic/ela On the mth Roots of a Complex Matrix 33 Jordan structure of the linear pencil Inλ − A coincides with the Jordan structure of A, and the vectors of each Jordan chain are chosen to be linearly independent [2], [6]. Moreover, there exist a matrix J ⊕ξ I ω N k k ... k n , (1.1) A = j=1 kj j + kj ( 1 + 2 + + ξ = ) where Nk is the nilpotent matrix of order k having ones on the super diagonal and zeros elsewhere, and an n × n nonsingular matrix x ... x x ... x ... x ... x (1.2) XA = 1,1 1,k1 2,1 2,k2 ξ,1 ξ,kξ , where for every j =1, 2,...,ξ, {xj,1,xj,2,...,xj,kj } is a Jordan chain of A corre- sponding to ωj ∈ σ(A), such that (see [2], [4], [6]) −1 A = XA JA XA . The matrix JA is called the Jordan matrix of A, and it is unique up to permutations of the diagonal Jordan blocks Ikj ωj + Nkj (j =1, 2,...,ξ) [2], [4]. The set of all eigenvalues of P (λ), that is, σ(P )={µ ∈ C :detP (µ)=0}, is called the spectrum of P (λ). Denoting by σ(A)=σ(Inλ − A) the spectrum m of the matrix A, it is clear that σ(P )={µ ∈ C : µ ∈ σ(A)}.IfJA is the Jordan matrix of A in (1.1), then it will be convenient to define the J-spectrum of A, σJ (A)={ω1,ω2,...,ωξ}, where the eigenvalues of A follow exactly the order of their appearance in JA (obviously, repetitions are allowed). For example, the J-spectrum 01 11 of the matrix M = ⊕ 0 ⊕ is σJ (M)={ 0, 0, 1}. 00 01 In this article, we study the Jordan structure of the mth roots (m>1) of a complex matrix. In Section 2, we consider a nonsingular matrix and present a new constructive proof for the Jordan canonical form of its mth roots. This proof is simple and based on spectral analysis of matrix polynomials [2], [3], [7]. Furthermore, it yields directly the Jordan chains of the mth roots. We also generalize a known uniqueness statement [5]. In Section 3, using a methodology of Cross and Lancaster [1], we obtain a necessary and sufficient condition for the existence of mth roots of a singular matrix. 2. The nonsingular case. Consider a nonsingular matrix A ∈Mn and an integer m>1. If A is diagonalizable and S ∈Mn is a nonsingular matrix such that iφ1 iφ2 iφn −1 A = S diag{r1e ,r2e ,...,rne } S , where rj > 0, φj ∈ [0, 2π)(j =1, 2,...,n), then for every n-tuple (s1,s2,...,sn), sj ∈{1, 2,...,m} (j =1, 2,...,n), the matrix 1 φ1+2(s1−1)π 1 φ2+2(s2−1)π 1 φn+2(sn−1)π m i m i m i −1 B = S diag{r1 e m ,r2 e m ,...,rn e m } S is an mth root of A. Hence, the investigation of the mth roots of a nonsingular (and not diagonalizable) matrix A via the Jordan canonical form of A arises in a natural way [2]. The following lemma is necessary and of independent interest. The Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 9, pp. 32-41, April 2002 ELA http://math.technion.ac.il/iic/ela 34 P.J. Psarrakos Lemma 2.1. Let {x0,x1,...,xk} be a Jordan chain of A ∈Mn (with linearly φ independent terms) corresponding to a nonzero eigenvalue ω0 = r0ei 0 ∈ σ(A)(r0 > m 0,φ0 ∈ [0, 2π)), and let P (λ)=Inλ − A. Then for every eigenvalue 1 φ0+2(t−1)π m i r0 e m ∈ σ(P ); t =1, 2,...,m, the matrix polynomial P (λ) has a Jordan chain of the form y0 = x0 y1 = a1,1x1 (2.1) y2 = a2,1x1 + a2,2x2 . yk = ak,1x1 + ak,2x2 + ···+ ak,kxk, where the coefficients ai,j (1 ≤ j ≤ i ≤ k) depend on the integer t and for ev- m−1 φ0+2(t−1)π m i(m−1) i ery i =1, 2,...,k, ai,i =(mr0 e m ) =0 . Moreover, the vectors y0,y1,...,yk are linearly independent. Proof.Since{x0,x1,...,xk} is a Jordan chain of the matrix A corresponding to the eigenvalue ω0 =0 , we have (A − Inω0)x0 =0 and (A − Inω0)xi = xi−1 ; i =1, 2,...,k. m Let µ0 be an eigenvalue of P (λ) such that µ0 = ω0. By the equation m (Inω0 − A)x0 =(Inµ0 − A)x0 =0, it is obvious that y0 = x0 is an eigenvector of P (λ) corresponding to µ0 ∈ σ(P ). n Assume now that there exists a vector y1 ∈ C such that (1) P (µ0) P (µ0)y1 + y0 =0. 1! Then m m−1 (Inµ0 − A)y1 = − mµ0 y0, or equivalently, m−1 (Inω0 − A)y1 = mµ0 (Inω0 − A)x1. m−1 Hence, we can choose y1 = a1,1x1, where a1,1 = mµ = 0. Similarly, if we consider the equation (1) (2) P (µ0) P (µ0) P (µ0)y2 + y1 + y0 =0, 1! 2! The Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 9, pp. 32-41, April 2002 ELA http://math.technion.ac.il/iic/ela On the mth Roots of a Complex Matrix 35 then it follows m m−1 m(m − 1) m−2 (Inµ0 − A)y2 = − mµ y1 − µ y0, 0 2 0 or equivalently, m−1 2 m(m − 1) m−2 (Inω0 − A)y2 =(Inω0 − A) (mµ ) x2 + µ x1 . 0 2 0 m(m−1) m−2 Thus, we can choose y2 = a2,1x1 + a2,2x2, where a2,1 = 2 µ0 and a2,2 = 2 mµm−1 = 0. Repeating the same steps implies that the matrix polynomial P (λ) has a Jordan chain {y0,y1,...,yk} as in (2.1). Define the n × (k + 1) matrices X0 = x0 x1 ··· xk and Y0 = y0 y1 ··· yk . n Since the vectors x0,x1,...,xk ∈ C are linearly independent, rank(X0)=k +1. Moreover, Y0 = X0 T0, where the (k +1)× (k + 1) upper triangular matrix 10 0··· 0 0 a1,1 a2,1 ··· ak,1 a ··· a T0 = 00 2,2 k,2 . . .. 00 0··· ak,k is nonsingular. As a consequence, rank(Y0)=k +1, and the proof is complete.
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