Products of commuting nilpotent operators

Transcription

1 Electronic Journal of Linear Algebra Volume 16 Article Products of commuting nilpotent operators Damjana Kokol Bukovsek Tomaz Kosir Nika Novak Polona Oblak Follow this and additional works at: Recommended Citation Bukovsek, Damjana Kokol; Kosir, Tomaz; Novak, Nika; and Oblak, Polona (2007), "Products of commuting nilpotent operators", Electronic Journal of Linear Algebra, Volume 16 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact

2 PRODUCTS OF COMMUTING NILPOTENT OPERATORS DAMJANA KOKOL BUKOVŠEK, TOMAŽ KOŠIR, NIKA NOVAK, AND POLONA OBLAK Abstract Matrices that are products of two (or more) commuting square-zero matrices and matrices that are products of two commuting nilpotent matrices are characterized Also given are characterizations of operators on an infinite dimensional Hilbert space that are products of two (or more) commuting square-zero operators, as well as operators on an infinite-dimensional vector space that are products of two commuting nilpotent operators Key words Commuting matrices and operators, Nilpotent matrices and operators, Factorization, Products AMS subject classifications 15A23, 15A27, 47A68, 47B99 1 Introduction Is every complex singular square matrix a product of two nilpotent matrices? Laffey [5] and Sourour [8] proved that the answer is positive: any complex singular square matrix A (which is not 2 2 nilpotent with rank 1) is a product of two nilpotent matrices with ranks both equal to the rank of A Earlier Wu [9] studied the problem (Note that [9, Lem 3] holds but the decomposition given in its proof on [9, p 229] is not correct since the latter matrix given for the odd case is not always nilpotent) Novak [6] characterized all singular matrices in M n (F), where F is a field, which are a product of two square-zero matrices Related problem of existence of k-th root of a nilpotent matrix was studied by Psarrakos in [7] Similar results were proved for the set B(H) of all bounded (linear) operators on an infinite-dimensional separable Hilbert space H Fong and Sourour [3] proved that every compact operator is a product of two quasinilpotent operators and that a normal operator is a product of two quasinilpotent operators if and only if 0 is in its essential spectrum Drnovšek, Müller, and Novak [2] proved that an operator is a product of two quasinilpotent operators if and only if it is not semi-fredholm Novak [6] characterized operators that are products of two and of three square-zero operators Here we consider similar questions for products of commuting square-zero or commuting nilpotent operators on a finite dimensional vector space or on a infinitedimensional Hilbert or vector space The commutativity condition considerably restricts the set of operators that are such products Namely, if A = BC and B, C are commuting nilpotent operators then A is nilpotent as well and it commutes with both B and C If in addition B and C are square-zero then so is A Received by the editors 5 February 2007 Accepted for publication 10 August 2007 Handling Editor: Harm Bart This research was supported in part by the Research Agency of the Republic of Slovenia Department of Mathematics, Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia (damjanakokol@fmfuni-ljsi, tomazkosir@fmfuni-ljsi, nikanovak@fmfuni-ljsi) Department of Mathematics, Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubljana, Slovenia (polonaoblak@fmfuni-ljsi) 237

3 238 D Kokol Bukovšek et al In the paper we characterize the following sets of matrices and operators: Matrices that are products of k commuting square-zero matrices for each k 2 Matrices that are products of two commuting nilpotent matrices Operators on a Hilbert space that are products of k commuting square-zero operators for each k 2 Operators on an infinite-dimensional vector space that are products of two commuting nilpotent operators 2 When is a matrix a product of commuting square-zero matrices? First we consider the following question: Question 1 Which matrices A M n (F) can be written as a product A = BC, where B 2 = C 2 =0and BC = CB? Observe that if A, B and C are as above then B and C commute with A Example 21 It can be easily seen that E 13 = = , but E 13 cannot be written as a product of two commuting square-zero matrices Therefore the set of matrices that can be written as a product of two commuting square-zero matrices is not the same as the set of matrices that are products of two square-zero matrices Next,wehavethat E 14 = = = and = =0 Thus E 14 is a product of two commuting square-zero matrices We denote by J µ = J (µ1,µ 2,,µ t) = J µ1 J µ2 J µt the upper triangular nilpotent matrix in its Jordan canonical form with blocks of order µ 1 µ 2 µ t > 0 If A is similar to J µ then we call µ the partition corresponding to A Wealso

4 Products of Commuting Nilpotent Operators 239 say that µ is the Jordan canonical form of A For a finite sequence of natural numbers λ =(λ 1,λ 2,,λ t )wedenotebyord(λ) =µ the ordered sequence µ 1 µ 2 µ t Let ι(a) denote the index of nilpotency of matrix A For a nilpotent matrix A, define a sequence J (A) =(α 1,α 2,,α ι(a) )=(α 1,α 2,,α n ) where α i is the number of Jordan blocks of the size i and α j =0forj>ι(A) Note that n j=1 jα j = n If C commutes with J µ it is of the form C =[C ij ], where C ij M µi µ j and C ij are all upper triangular Toeplitz matrices (see e g [4, p 297]), ie for 1 i j t we have c 0 ji c 1 ji c µi 1 ji 0 0 c 0 ij c 1 ij c µi 1 0 c 0 ji ij C ij = 0 c 0 c 1 ij 0 c 1 and C ji ji = 0 c 0 ji ij 0 0 c 0 ij (21) If µ i = µ j then we omit the rows or columns of zeros in C ji or C ij above Proposition 22 AmatrixA is a product of two commuting square-zero matrices if and only if it has a Jordan canonical form (2 x, 1 n 2x ) for some x n 4, ie if and only if J (A) =(n 2x, x) for some x n 4 Proof SinceA 2 = B 2 C 2 = 0, it follows that also A is a square-zero matrix Since B 2 = 0, the Jordan canonical form of matrix B is equal to µ =(2 a, 1 n 2a )forsome 0 a n 2 Suppose that B = J µ is in its Jordan canonical form Since C commutes with B it is of the form C =[C ij ], where C ij are given in (21) Following Basili [1, p 60, Lemma 23], the matrix C is similar to U 1 X Y 0 U 1 0, 0 W U 2 where U 1,X M a a, Y M a (n 2a), W M (n 2a) a, U 2 M (n 2a) (n 2a) and U 1 and U 2 are strictly upper triangular matrices Note that B is transformed by the same similarity to B = 0 I

5 240 D Kokol Bukovšek et al Take an invertible matrix P 1 such that P 1 U 1 P1 1 = J λ and denote C = P P 1 0 U 1 X Y 0 U 1 0 P P1 1 0 = J λ X Y 0 J λ I 0 W U I 0 W U 2 Note that B does not change under the above similarity Since C 2 =0,also C 2 =0 and thus Jλ 2 = 0 Therefore, λ =(2x, 1 a 2x ), where 0 x a 2 n 4 Weseethat B C = 0 I J λ X Y 0 J λ 0 = 0 0 J λ W U Now, it easily follows that rk(a) =rk( B C) =x Since A 2 =0,weseethatA must have Jordan canonical form (2 x, 1 n 2x )forsomex n 4 Now, take a nilpotent matrix A with its Jordan canonical form (2 x, 1 n 2x ), where x n 4 Then there exists an invertible matrix Q such that QAQ 1 = } {{ } x n 4x In Example 21 we observed that the matrix E 14 is a product of two commuting square-zero matrices Then it follows that QAQ 1 and A are also products of two commuting square-zero matrices We have proved the proposition Theorem 23 AmatrixA is a product of k pairwise commuting square-zero matrices if and only if it has a Jordan canonical form (2 x, 1 n 2x ) for some x n, 2 k ie if and only if J (A) =(n 2x, x) for some x n 2 k Proof LetA be a matrix with Jordan canonical form (2 x, 1 n 2x )forsomex n 2 k Then it is similar to a matrix A = E 12 k E 12 k E 12 k 0 0 0, x n 2 k x where E 12 k M 2 k(c) is a matrix with only nonzero element (equal to 1) in the upper-right corner To prove that A is a product of k pairwise commuting squarezero matrices it is sufficient to show, that E 12 k is a product of k pairwise commuting square-zero matrices We define matrices [ ] 02 i 1 I C i = 2 i 1 M 0 2 i i(c) 2 i 1 for every i =1, 2,, k and let B i = C i C i C i 2 k i M 2k(C)

6 Products of Commuting Nilpotent Operators 241 It is easy to check that B 2 i =0andB ib j = B j B i for every i, j and that E 12 k = B 1 B 2 B k To prove the converse we have to show that every product of k pairwise commuting square-zero matrices has rank at most n We will show this by induction The 2 k assertion is true for k = 2 by the previous proposition Suppose that every product of k pairwise commuting square-zero matrices has rank at most n 2 and let k A = B 1 B 2 B k+1 where B 1,B 2,B k+1 are pairwise commuting square-zero matrices Denote by m the rank of B 1 SinceB1 2 =0wehavethatm n 2 NowthematrixB 1 is similar to amatrix B 1 = 0 m I m 0 0 m 0 m n m Again following Basili [1, p 60, Lemma 23], we transform the matrices B i simultaneously by similarity to the matrices B i = X i Y i Z i 0 X i 0 0 U i V i Here matrices X i are square-zero and they pairwise commute Now A = B 1 B 2 B k+1 = 0 I 0 X 2 Y 2 Z 2 X k+1 Y k+1 Z k+1 0 X 2 X k+1 0 = X X k+1 0 = U 2 V 2 0 U k+1 V k and the matrix X 2 X k+1 is a product of k pairwise commuting square-zero matrices, so it has the rank at most m 2 k and thus the rank of A is at most n 2 k+1 3 When is a matrix a product of two commuting nilpotent matrices? In this section we study the following question: Question 2 Which matrices A M n (F) canbewrittenasa = BC = CB, where B and C are nilpotent matrices? Clearly, A must be nilpotent Thus, not every singular matrix is a product of two commuting nilpotent matrices Moreover, suppose that rk(a) = n 1 anda = BC = CB with B and C nilpotent Then also rk(b) =rk(c) =n 1 and thus B = PJ n P 1 and P 1 CP = p(j n ), where p is a polynomial such that p(0) = 0 Then A = BC = PJ n p(j n )P 1 and thus rk(a) < n 1, which is a contradiction Hence not every nilpotent matrix is a product of two commuting nilpotent matrices (for example J n is not)

7 242 D Kokol Bukovšek et al [ ] Jm 0 Example 34 Suppose A =,wherem 3 Can A be written as 0 0 a product of two commuting nilpotent matrices? Assume that [ A = BC] is such a TB W product Since B and C commute with A it follows that B = B and C = V B U [ ] B TC W C,whereT V C U B,T C M m (F) are (strictly) upper triangular Toeplitz matrices, C U B,U C M k (F) are nilpotent matrices (see Basili [1]), W B,W C M m k (F) have the only nonzero entries in the first row and V B,V C M k m (F) have the only nonzero entries in the last column Since A = BC it follows that J m = T B T C + W B V C The product W B V C has the only nonzero entry in the first row and the last column, and T B and T C are strictly upper-triangular The assumption that m 3 is needed to conclude that T B T C + W B V C is upper triangular Toeplitz matrix with zero superdiagonal This contradicts the fact that J m has nonzero superdiagonal and implies that A is not a product of two commuting nilpotent matrices What is the Jordan canonical form of Jn t for t 2? It is an easy observation that the partition of n corresponding to Jn t is equal to (λ 1,λ 2,,λ t ), where λ 1 λ t 1 We denote this partition by r(n, t) If n = kt + r, where0 r<t,thenr(n, t) = ((k +1) r,k t r ) Note that k = n t It follows that J (J t n )=(0,,0,t r, r) n t 1 Proposition 35 If a nilpotent matrix A has a Jordan canonical form ord(r(n 1,t 1 ),r(n 2,t 2 ),,r(n m,t m ), 1 k ), where n = k + m i=1 n i and t i 2 for all i, thena can be written as a product of two commuting nilpotent matrices Proof Since the Jordan canonical form of Jn ti i n m J t1 n 1 J t2 n 2 J tm is r(n i,t i ), matrix A is similar to } {{ }, which is obviously equal to the product of two commuting nilpotent matrices J n1 J n2 J nm k k and J t1 1 n 1 J t2 1 n 2 Jn tm 1 m k Thus also A can be written as a product of two commuting nilpotent matrices In the following we show that the converse is true as well Theorem 36 For a nilpotent matrix A the following are equivalent: (a) A can be written as a product of two commuting nilpotent matrices,

8 (b) A has a Jordan canonical form Products of Commuting Nilpotent Operators 243 ord(r(n 1,t 1 ),r(n 2,t 2 ),,r(n m,t m ), 1 k ), where n = k + m i=1 n i and t i 2 for all i, (c) J (A) does not include a subsequence (0, 1, 1,,1, 0) for any l 1 2l 1 We first prove the following lemma and propositions Lemma 37 If J (A) =(0,,0, 1,,1), wherel 1 and m 2l, thena is m 2l+1 2l 1 not a product of two commuting nilpotent matrices Proof Suppose that A = BC = CB is nilpotent matrix with J (A) asinthe statement of the lemma Let us denote s =2l 1 and let us assume that A = J (m,m 1,,m s+1) Then B 11 B 12 B 1s C 11 C 12 C 1s B 21 B 22 B 2s B = and C = C 21 C 22 C 2s, B s1 B s2 B ss C s1 C s2 C ss where all B ij and C ij are upper triangular Toeplitz and we use the notation introduced in (21) Since J m = B 11 C 11 + B 12 C B 1s C s1 = C 11 B 11 + C 12 B C 1s B s1 and the only possible summands with nonzero superdiagonal are B 12 C 21 and C 12 B 21, it follows that b 0 12 c0 21 = c0 12 b0 21 =1 SinceJ m 1 = B 21 C 12 + B 22 C B 2s C s2 = C 21 B 12 + C 22 B C 2s B s2 and the only possible summands with nonzero superdiagonals are B 21 C 12 + B 23 C 31 and C 21 B 12 + C 23 B 31, it follows that b 0 21c b 0 23c 0 32 = c 0 21 b c0 23 b0 32 = 1 and therefore b0 23 c0 32 = c0 23 b0 32 =0 Similarly, we show by induction, that b 0 i,i+1 c0 i+1,i = c0 i,i+1 b0 i+1,i = 0 for all even i and b 0 i,i+1 c0 i+1,i = c 0 i,i+1 b0 i+1,i = 1 for all odd i In particular, it follows that b 0 s 1,s c0 s,s 1 = c0 s 1,s b0 s,s 1 =0 Furthermore, J m s+1 = B s1 C 1s +B s2 C 2s ++B ss C ss = C s1 B 1s +C s2 B 2s ++ C ss B ss and the only possible summands with nonzero superdiagonals are B s,s 1 C s 1,s and C s,s 1 B s 1,s It follows that the superdiagonal of J m s+1 is equal to 1 = b 0 s,s 1 c0 s 1,s = c0 s,s 1 b0 s 1,s = 0, which is a contradiction Proposition 38 If J (A) =(α 1,,α m 2l, 0, 1,,1, 0,α m+2,,α n ),where 2l 1 l 1, thena is not a product of two commuting nilpotent matrices Proof Denoteµ =(n αn, (n 1) αn 1,,(m +2) αm+2 ), λ =(m, m 1,,m 2l +2)andµ =((m 2l) α m 2l, (m 2l 1) α m 2l 1,,1 α1 ) Suppose that A = BC = CB with J (A) as in the statement Then we can assume that A = J µ J λ J µ, B = B 11 B 12 B 13 B 21 B 22 B 23 and C = C 11 C 12 C 13 C 21 C 22 C 23 B 31 B 32 B 33 C 31 C 32 C 33 in the same block partition, where all B ij,c ij are block upper triangular Toeplitz

9 244 D Kokol Bukovšek et al We compute that J λ = B 21 C 12 + B 22 C 22 + B 23 C 32 = C 21 B 12 + C 22 B 22 + C 23 B 32 Since m 2l +2>m 2l + 1, it follows that superdiagonals of all blocks of J λ must be equal to superdiagonals of B 22 C 22 and by symmetry to superdiagonals of C 22 B 22 We have already seen in the proof of Lemma 37 that this is not possible Proposition 39 If J (A) does not include a subsequence (0, 1, 1,,1, 0) for 2l 1 any l 1, then the Jordan canonical form of a matrix A is equal to ord(r(n 1,t 1 ),r(n 2,t 2 ),,r(n m,t m ), 1 k ), where n = k + m i=1 n i and t i 2 for all i Proof IfJ (A) does not include a subsequence of the form (0, 1, 1,,1, 0), then 2l 1 for any subsequence of J (A) oftheform (0,α t,α t+1,,α s, 0), (32) for some 2 t s n, whereα i 0fori = t, t +1,,s holds either (a) s t + 1 is even or (b) s t +1isoddandthereexistsj, t j s, such that α j 2 So, the matrix A can be written as a direct sum A 1 A 2 A r,whereeacha i has one of the following forms: (i) J (A i )=(α 1 ) and the Jordan canonical form of A i is equal to (1 α1 ) (ii) J (A i )=(0,,0,α qi ), where α qi 2 and the Jordan canonical form of A i q i 1 is equal to r(q i α qi,α qi ) (iii) J (A i )=(0,,0,α qi,α qi+1) and the Jordan canonical form of A i is equal q i 1 to r(q i α qi +(q i +1)α qi+1,α qi + α qi+1) (iv) J (A i )=(0,,0,α qi 1,α qi,α qi+1), where α qi 2 and the Jordan canonical q i 2 form of A i is equal to ord(r((q i 1)α qi 1+q i (α qi 1),α qi 1+α qi 1),r(q i +(q i +1)α qi+1, 1+α qi+1)) In case (a) we can write each block corresponding to subsequences of the form (32) as a direct sum of blocks of type (iii) In the case (b) we use types (ii) and (iii) if there is an odd i 1 such that α t 1+i 2 and types (iii) and (iv) otherwise If α 1 1 then the block corresponding to the subsequence (α 1,α 2,,α s, 0), α i 1, is decomposed as a direct sum of blocks of type (iii) if s is even and a combination of types (i) and (iii) if s is odd This proves the proposition Proof (of Theorem 36) Since Proposition 38 holds also for m = n, theimplication (a) (c) follows from Proposition 38 The implication (c) (b) is the statement of Proposition 39 and the implication (b) (a) is the statement of Proposition 35

10 Products of Commuting Nilpotent Operators When is an operator a product of commuting square-zero operators? In this section we assume that H is an infinite-dimensional, separable, real or complex Hilbert space We denote by B(H) the algebra of all operators (ie, bounded linear transformations) on H Question 3 Which operators A B(H) canbewrittenasaproductoftwo commuting square-zero operators? Similarly as in the finite-dimensional case we notice that also A is square-zero Therefore im A ker A Sothespaceim A +kera is closed Theorem 410 Let A B(H) Then A = BC = CB, whereb 2 = C 2 =0,if and only if dim(ker A ker A )= and A 2 =0 Proof If A is a product of two square-zero operators it follows by [6] that dim(ker A im A) = Since kera im A =kera ( im A ) =kera ker A we have that dim(ker A ker A )= and A 2 =0 It remains to prove the converse We can choose a decomposition of H as a direct sum of infinite-dimensional subspaces H 1 and H 2 such that [ H 2 ] ker A ker A The D 0 matrix of A relative to this decomposition is of the form SinceA =0also D 2 = 0 Therefore we can find a decomposition of space H 1 = H 11 H 12,where both subspaces are infinite-dimensional [ ] and im D H 11 ThematrixofDrelative to 0 D1 this decomposition is Since H is infinite-dimensional space, we can write it as a direct sum of two infinite-dimensional subspaces H 21 and H 22 TheformofA relative to the decomposition H 11 H 12 H 21 H 22 is 0 D Define operators B and C on H by 0 0 D I B = and C = I D It is evident that A = BC = CB and B 2 = C 2 =0 The factorization in the proof above is based on the factorization in the finitedimensional case Since H 2 is an infinite-dimensional space, we can write it as a direct sum of k infinite-dimensional subspaces Using the factorization in the proof of Theorem 23 we get the following result Corollary 411 An operator A is a product of two commuting square-zero operators if and only if A is a product of k square-zero operators

11 246 D Kokol Bukovšek et al 5 When is an operator a product of two commuting nilpotent operators? Let V be an infinite-dimensional vector space and A : V V a nilpotent operator with index of nilpotency n We proceed to define the sequence J (A) =(α 1,α 2,,α n ), where now α i N {0, } For k =0, 1,,n 1 we choose subspaces V n k such that: 1 For k =0wehaveV =kera n =kera n 1 V n 2 For k > 0 we have kera n k = kera n k 1 AW n k+1 V n k, where W n k+1 = AW n k+2 V n k+1 and W n+1 =0 Then we define α i =dimv i, i =1, 2,,n Observe that if dim V < then this definition of J (A) coincides with the one given in 2 Observe that if an operator A is a product of two commuting nilpotent operators, then A is also a nilpotent operator Theorem 512 A nilpotent operator A with J (A) =(α 1,α 2,,α n ) is a product of two commuting nilpotent operators if and only if a matrix B with J (B) = (β 1,β 2,,β n ),whereβ i =min{α i, 2}, is a product of two commuting nilpotent matrices Before we prove the theorem let us show the following two lemmas Lemma 513 If J (A) =(0,,0, ), thena is a product of two commuting nilpotent operators Proof Sinceα i =0fori n, it follows that all the indecomposable blocks are of size n Then A is similar to k=1 J k,n = k=1 (J 2k 1,n J 2k,n ), where J k,n, k N, are indecomposable blocks of A each of them similar to J n SinceJ 2k 1,n J 2k,n is a product of two commuting nilpotent matrices by Theorem 36, the assertion follows Lemma 514 Let A be a nilpotent matrix with J (A) =(α 1,α 2,,α n ),where there exists an index j such that α j 2 Suppose that B is a matrix with J (B) = (β 1,β 2,,β n ),whereβ i = α i for i j and β j = α j +1 Then A is a product of two commuting nilpotent matrices if and only if B is a product of two commuting nilpotent matrices Proof By Theorem 36 a matrix A is a product of two commuting nilpotent matrices if and only if J (A) is not of the form (α 1,,α i, 0, 1,,1, 0,α k,,α n ) (53) 2l 1 It follows easily that matrices A and B are simultaneously the products of two commuting nilpotent matrices Proof (of Theorem 512) Suppose that A is a product of two commuting nilpotent operators Similarly as in the finite-dimensional case we can show that J (A) can not be of the form (53) Hence also J (B) is not of that form and therefore B is a product of two commuting nilpotent matrices To prove the converse write A as a direct sum of A 1 and A 2 with J (A j )=(α 1j,α 2j,,α nj )

12 Products of Commuting Nilpotent Operators 247 so that α i1 = α i and α i2 =0ifα i is finite, and α i1 =2andα i2 = otherwise It suffices to show that A 1 and A 2 are the products of commuting nilpotent operators, which follows from the lemmas above REFERENCES [1] R Basili On the irreducibility of commuting varieties of nilpotent matrices J Algebra, 268:58 80, 2003 [2] R Drnovšek, V Müller, and N Novak An operator is a product of two quasi-nilpotent operators if and only if it is not semi-fredholm ProcRoySocEdinburghSectionA, 136: , 2006 [3] C K Fong and A R Sourour Sums and products of quasi-nilpotent operators Proc Roy Soc Edinburgh Section A, 99: , 1984 [4] I Gohberg, P Lancaster, and L Rodman Invariant Subspaces of Matrices with Applications Wiley-Interscience, New York, 1986 [5] T J Laffey Products of matrices InGenerators and Relations in Groups and Geometries NATO ASI Series, Kluwer, 1991 [6] N Novak Products of square-zero operators J Math Anal Appl To appear [7] PJPsarrakosOnthemth Roots of a Complex Matrix Electron J Linear Algebra, 9:32 41, 2002 [8] A R Sourour Nilpotent factorization of matrices Linear Multilinear Algebra, 31: , 1992 [9] P Y Wu Products of nilpotent matrices Linear Algebra Appl, 96: , 1987