Decompositions of Matrices into Potent and Square-Zero Matrices
aa r X i v : . [ m a t h . R A ] F e b DECOMPOSITIONS OF MATRICES INTOPOTENT AND SQUARE-ZERO MATRICES
PETER DANCHEV, ESTHER GARC´IA, AND MIGUEL G ´OMEZ LOZANO
Abstract.
In order to find a suitable expression of an arbitrary square matrixover an arbitrary finite commutative ring, we prove that every such a matrixis always representable as a sum of a potent matrix and a nilpotent matrixof order at most two when the Jacobson radical of the ring has zero-square.This somewhat extends results of ours in Lin. & Multilin. Algebra (2021)established for matrices considered on arbitrary fields. Our main theorem alsoimproves on recent results due to Abyzov et al. in Mat. Zametki (2017), ˇSterin Lin. Algebra & Appl. (2018) and Shitov in Indag. Math. (2019).
Key words : Finite commutative ring, Nilpotent matrix, Potent matrix, Lifting potent element : 15A24, 15B33, 16U99 Introduction and Fundamentals
We start the frontier of this paper by recalling that an element x of an arbitraryring R is said to be nilpotent if there is an integer i > x i = 0 whereasan element y from R is said to be potent , or more exactly m -potent , if there is anatural number m ≥ y m = y . In particular, all the idempotents are always2-potent elements.Our current work is devoted to the further study, firstly somewhat initiated in [5],of decomposing square matrices as a sum of a potent and a nilpotent. Concretely,a brief retrospection of the most important results in this direction is as follows:In [5] was proven that each matrix from the ring M n ( F ) of n × n matrices overthe field F of two elements is a sum of an idempotent matrix and a nilpotentmatrix – even something more, if the matrix ring M n ( F ) over an arbitrary field F possesses this property, then F ∼ = F . This result was substantially strengthened byˇSter in [13] who proved that M n ( F ) is actually a sum of an idempotent matrix anda nilpotent matrix of index at most 4. Lately, this result was significantly improvedby Shitov in [12] for certain matrix sizes n . Moreover, an important work was doneby de Seguins Pazzis in [7], where a valuable discussion on the decomposition of amatrix as a sum of an idempotent and a square-zero matrix is provided.On the other vein, Abyzov and Mukhametgaliev showed in [1] that, for all nat-urals n ≥
1, any element of the ring M n ( F ) is presented as a sum of a nilpotentand a q -potent element, provided that F is a field of cardinality q – specifically, in[1, Theorem 2] was showed that some square matrix over finite fields are express-ible as a sum of a potent and a nilpotent but the order of the existing nilpotent The first author was partially supported by the Bulgarian National Science Fund under GrantKP-06 No. 32/1 of December 07, 2019.The second two authors were partially supported by MTM2017-84194-P (AEI/FEDER, UE),and by the Junta de Andaluc´ıa FQM264. is, in general, greater than 2. Also, a recent paper [4] by Breaz deals with themore exact presentation of matrices over fields of odd cardinality q as a sum of a q -potent matrix and a nilpotent matrix of order 3. Besides, it was constructed in[4, Example 6] an ingenious example of a 3 × F of threeelements that cannot be presented as the sum of a 3-potent and a nilpotent matrixof order 2 (in other terms, the latter matrix is also called square-zero or, equiva-lently, zero-square ). Furthermore, improving the aforementioned results from [4],we establish in [8] that each square matrix over any infinite field as well as eachmatrix over some special finite fields can be expressed as a sum of a potent matrixand a square-zero matrix.So, a question which logically arises is of whether or not our results in [8] couldbe expanded for some kinds of (finite) commutative rings, that is, is every squarematrix over a finite commutative ring of square-prime characteristic decomposedas a sum of a potent matrix and a square-zero matrix (for example, for rings ofthe sort Z p for some fixed prime p )? To keep a record straight, we notice that asimilar representation of such a matrix ring over Z already exists in terms of anilpotent of order less than or equal to 8 and an idempotent (see, e.g., [13]). Evenmore generally, it was established in [1, Lemma 1] and [1, Theorem 4] that, forall n, m ∈ N , the matrices in M n ( Z p m ) are presentable as the sum of a nilpotentmatrix and a p -potent matrix, whenever p is a prime. However, the exact bound(of course, if it eventually exists) of the existing nilpotent matrix is not explicitlycalculated yet.And so, being seriously motivated by the present idea, in what follows, we shallcompletely resolve [8, Problem 2] even in a more general setting (see, e.g., Theo-rem 2.6) and, besides, we shall strengthen the previously mentioned achievementsfrom [1], [13] and [12], respectively.Likewise, for completeness of the introductory section, we refer to the bibliog-raphy of the cited by us articles and also concretize that some related results canbe found by the interested reader in [6] and [11] along with the given referencestherewith, respectively.2. Main Results and Conjecture
We begin here with the following simple but useful claim.
Lemma 2.1.
Let R be a finite unital commutative ring. For every invertible matrix A ∈ M n ( R ) there exists m ∈ N such that A m − = Id and A m = A .Proof. Let A be an invertible matrix in M n ( R ) and consider the set of matrices { A , A , . . . , A n , . . . } . Since this set is finite, there exists k < l such that A k = A l ,and since A is invertible Id = A l − k . The claim now follows by taking m − l − k . (cid:3) The following result generalizes [8, Corollary 3.2], where it was shown that everymatrix over a finite field is a sum of potent matrix and a zero-square matrix byusing a different approach. The result of this paper is entirely based on the primaryrational canonical form of a matrix ([10, VII.Corollary 4.7(ii)]), which states thatevery matrix A ∈ M n ( F ) where F is a field is similar to a direct sum of companionmatrices of prime power polynomials p m , . . . , p m sks s ∈ F [ x ] where each p i is prime(irreducible) in F [ x ]. The matrix A is uniquely determined except for the order ECOMPOSITIONS OF MATRICES INTO ... 3 of the companion matrices of the p m ij i along its main diagonal. The polynomials p m , . . . , p m sks s are called the elementary divisors of the matrix A . Proposition 2.2.
Let F be a finite field. For any matrix A ∈ M n ( F ) there exists k ∈ N such that A = P + N , where N = 0 , P k = P , E = P k − is an idempotentwith P E = EP = P and EN = N E = N .Proof. Let us consider the primary rational canonical form of the matrix A . Also,let us split our argument between elementary divisors q i ( x ) of A with q i (0) = 0 andthose with q i (0) = 0:(i) Any elementary divisor q i ( x ) with q i (0) = 0 gives rise to an invertiblecompanion matrix C i . By Lemma 2.1 there exists k i ∈ N such that C k i i = C i and C k i − i = Id. Let us denote P i := C i and define N i as the zero matrix.(ii) Let us suppose that q i ( x ) is an elementary divisor (power of an irreduciblepolynomial in F [ x ]) such that q i (0) = 0. This implies that q i ( x ) = x i s forcertain i s ∈ N and its associated companion matrix is of the form C i = . . .
01 0 .... . . . . .0 1 0 ∈ M i s ( F ) , i.e., it is a nilpotent Jordan block.(ii.1) If i s ≥
2, write P i = . . .
11 0 .... . . . . .0 1 0 and N i := . . . −
10 0 .... . . . . .0 0 0
Notice that P i is an invertible matrix and by 2.1 there exists k i ∈ N such that P k i i = P i and P k − i = Id with C i = P i + N i .(ii.2) If i s = 1, then C i = (cid:0) (cid:1) . Let Q be the invertible matrix in M n ( F ) such that Q − AQ is decomposed intoits primary rational canonical form (suppose without loss of generality that theblocks corresponding to (ii.2) are written together as the last zero block): Q − AQ = C · · · C · · · · · · C r
00 0 0 = P · · · P · · · · · · P r
00 0 0 | {z } P ′ + N · · · N · · · · · · N r
00 0 0 | {z } N ′ Since each P i satisfies P k i − i = Id, we have that ( P ′ ) k = P ′ for k = 1+ Q ri =1 ( k i −
1) and therefore E ′ := P ′ k − is an idempotent of M n ( F ) ( E ′ = ( P ′ k − ) = P ′ k P ′ k − = P ′ k − = E ′ ). By construction, N ′ = 0 and, since P ki = Id, ineach block it must be that E ′ N ′ = N ′ E ′ = N ′ .Finally, for P := QP ′ Q − , N := QN ′ Q − and E := QE ′ Q − we have that E isan idempotent of M n ( F ) satisfying the properties E = P k − , A = P + N , P k = P , P. DANCHEV, E. GARC´IA, AND M. G ´OMEZ LOZANO N = 0, EN = N E = N and EP = P E = P . In particular, P = EP E is invertiblein the subring E M n ( F ) E (note that E M n ( F ) E is a unital ring with unit E ). This,in turn, implies that P is strongly regular, as required. (cid:3) The next property of lifting idempotents is well-known, but we list the statementhere only for the sake of completeness and for the convenience of the readers.
Lemma 2.3. [2, 27.1]
Let R be a ring and let I be a nilpotent ideal of R . Thenany idempotent of R/I lifts to an idempotent of R . The following two technicalities on lifting special elements are the key for theestablishment of our further results.
Lemma 2.4.
Let R be a ring and let I be a nilpotent ideal of R . Let us supposethat a ∈ R/I has zero-square and that ¯ a is a von Neumann regular element in R/I .Then the element a lifts to an element of R with zero-square.Proof. Since a is a von Neumann regular element of zero square, there exists b ∈ R/I such that ¯ a ¯ b ¯ a = ¯ a , ¯ b ¯ a ¯ b = ¯ b and ¯ b = 0 (see [9, Lemma 2.4]). Let us consider theidempotent ¯ e = ¯ a ¯ b ∈ R/I . Notice that ¯ e ¯ a (1 − ¯ e ) = ¯ a , because ¯ a ¯ e = ¯0. Byconsulting with [2, 27.1], the element ¯ e lifts to an idempotent e ∈ R . If now wetake any representant a of ¯ a in R , we will have that ea (1 − e ) ∈ R has zero-squareand ea (1 − e ) = ¯ e ¯ a (1 − ¯ e ) = ¯ a , as claimed. (cid:3) For completeness of the exposition, let us recall now that a unital commutativering is said to be a local ring if it contains a unique maximal ideal, say M . In thatcase the factor-ring R/M is a field, called the residue field of R – cf. [3, Definition1.2.9]. Moreover, any finite commutative ring with identity R can be expressed asa direct sum of local rings and the decomposition is unique up to a permutation ofthe direct summands (see, e.g., [3, Theorem 3.1.4]). Lemma 2.5.
Let R be a unital commutative local ring such that its unique maximalideal M has M = 0 . Let S = M n ( R ) . Take P, E ∈ S such that E is an idempotentof S and P = EP E and suppose that there exists k ∈ N such that P k − = E ∈ S/rad ( S ) . Then there exists a prime p > such that P ( k − p = E . In particular, P ( k − p +1 = P and P is invertible in ESE .Proof.
Since P k − = E , there exists U ∈ rad ( S ) = M n ( M ) such that E = P k − + U . Multiplying on the left and on the right by E , we can suppose that U = EU E .We know that
R/M is a finite field of certain prime characteristic p . Thus p · ∈ M ,so we have that pM ⊂ M = 0. We, consequently, calculate that P ( k − p = ( E − U ) p = E + p X i =1 ( − i (cid:18) pi (cid:19) U i = E, as expected. (cid:3) So, we arrive at our central result on decomposing any matrix over special finitecommutative rings into a potent matrix and a zero-square matrix.
Theorem 2.6.
Let R be a finite commutative ring such that its Jacobson radicalhas zero-square. Then every matrix A in M n ( R ) can be expressed as P + N , where P is a potent matrix and N is a nilpotent matrix with N = 0 . ECOMPOSITIONS OF MATRICES INTO ... 5
Proof.
We know with the aid of the comments alluded to above that R is a directsum R = L R i , where each R i is a local ring. Then one finds that the decomposition M n ( R ) = L i M n ( R i ) holds, so that we can express A as a direct sum of matricesover local rings.Suppose without loss of generality that R is a local ring. Let us denote S := M n ( R ) and let us decompose A ∈ M n ( R ) into a potent and a zero-square ma-trix. Let I be the unique maximal ideal of R . By hypothesis, one calculatesthat I = 0 because I coincides with the Jacobson radical rad ( R ) of R . Clearly, J := rad ( M n ( R )) = M n ( rad ( R )) = M n ( I ) and hence J = 0.Let us consider the residue class of A modulo J : In fact, ¯ A ∈ S/J ∼ = M n ( R/I ).Since
R/I is a finite field, by virtue of Proposition 2.2 there exists k ∈ N suchthat ¯ A = ˆ P + ˆ N with ˆ P k = ˆ P , ˆ N = 0, and ˆ E := ˆ P k − is an idempotent withˆ E ˆ N = ˆ N ˆ E = ˆ N and ˆ E ˆ P = ˆ P ˆ E = ˆ P .Now, with Lemma 2.3 at hand, there exists an idempotent E of S such that E = ˆ E . Let us consider P ∈ S such that P = ˆ P . Since ˆ E ˆ P = ˆ P ˆ E = ˆ P , we have that EP = P E = P and we can suppose (by replacing P by EP E ) that EP = P E = P .Applying Lemma 2.4 there exists N ∈ S such that N = 0 with N = ˆ N and wecan suppose, again replacing N by EN E , that N = EN = N E . Furthermore,employing Lemma 2.5, there exists a prime p > P ( k − p = E .Let us take V ∈ J such that A = P + N + V , and write A = P + EV E + (1 − E ) V E + EV (1 − E ) | {z } + N + (1 − E ) V (1 − E ) | {z } . ( ∗ )(1) Let us show that P + EV E + (1 − E ) V E + EV (1 − E ) is a potent element of S : In fact, notice that for any n ∈ N (( P + EV E ) + (1 − E ) V E + EV (1 − E )) n = ( P + EV E ) n + (1 − E ) V E ( P + EV E ) n − + ( P + EV E ) n − EV (1 − E )because EV E , (1 − E ) V E , EV (1 − E ) and (1 − E ) V (1 − E ) belong to an ideal of zero-square. Moreover, since P is invertible in ESE , the matrix P + EV E is invertibleand, therefore, it is k -potent. Furthermore, P + EV E satisfies the conditions ofLemma 2.5, so we detect that (( P + EV E ) ( k − p ) = E and ( P + EV E ) ( k − p +1 = P + EV E . Then(( P + EV E ) + (1 − E ) V E + EV (1 − E )) ( k − p +1 = ( P + EV E ) ( k − p +1 + (1 − E ) V E ( P + EV E ) ( k − p + ( P + EV E ) ( k − p EV (1 − E )= P + EV E + (1 − E ) V E + EV (1 − E ) . (2) Let us show that N + (1 − E ) V (1 − E ) has zero-square: Indeed, since EN = N E = N , one has that N (1 − E ) V (1 − E ) = 0 = (1 − E ) V (1 − E ) N , and since(1 − E ) V (1 − E ) belongs to an ideal of zero-square, one has that( N + (1 − E ) V (1 − E )) = N + ((1 − E ) V (1 − E )) = 0 . Therefore, equality ( ∗ ) provides the desired decomposition of A into a potentmatrix and a zero-square matrix, as wanted. (cid:3) Since Z p is a unital commutative local ring whose Jacobson radical has zero-square, we immediately obtain the following consequence, which completely resolves[8, Problem 2] when p = 2, i.e., for the ring Z . P. DANCHEV, E. GARC´IA, AND M. G ´OMEZ LOZANO
Corollary 2.7.
For all natural numbers n and primes p , every matrix in M n ( Z p ) can be expressed as P + N , where P is a potent matrix and N is a matrix with N = 0 . The next construction sheds a further light on the more concrete decompositionof such a type.
Example 2.8.
Let us check in an example how this decomposition successfullyworks. Suppose S = M ( Z ) and consider the following matrix A = | {z } ¯ A +2 B, where B is any matrix in S , B = a b c de f g hi j k lm n p q , where a, b, . . . , q denote matrices of the appropriate sizes.If we regard A as a matrix over Z , we will obtain ¯ A , whose elementary divisorsare x + x + 1, x , x , x . As in the proof of Theorem 2.6, we add and subtract theelement 1 in the adequate position of the second diagonal box in order to transformthe companion matrix of x into an invertible matrix plus a zero-square matrix: A = | {z } ˆ P = ¯ A + e , + | {z } ˆ N =3 e , +2 B ECOMPOSITIONS OF MATRICES INTO ... 7
The matrix ˆ P = ¯ A + e , satisfiesˆ P = (it is worthwhile noticing that the first diagonal box to the 7 th -power is the identitywhen regarded as a matrix over Z , the second diagonal box to the 3 rd -power isthe identity when regarded as a matrix over Z , so globally we need 7 × × Z ).Following the proof of the theorem, let us denote E = ˆ P , which is clearly anidempotent of S .Now A = ˆ P + E (2 B ) E + (1 − E )(2 B ) E + E (2 B )(1 − E ) | {z } P + ˆ N + (1 − E )(2 B )(1 − E ) | {z } N where • P = ˆ P + E (2 B ) E + (1 − E )(2 B ) E + E (2 B )(1 − E ) == + 2 a b c de f g hi j m n • N = ˆ N + (1 − E )(2 B )(1 − E ) == + 2 k l p q and P = P and N = 0.In this example, ¯ A was already in its rational canonical form over Z . For other-wise, one should consider the appropriate invertible matrix Q such that Q − ¯ AQ is adirect sum of blocks corresponding to elementary divisors and adapt the idempotentmatrix ˆ E and the nilpotent matrix ˆ N .The following construction unambiguously illustrates that the square-prime char-acteristic of the ring is an essential condition and cannot be dropped off. P. DANCHEV, E. GARC´IA, AND M. G ´OMEZ LOZANO
Remark . There are matrices over Z that do not admit a decomposition intopotent + zero-square. For example, the matrix A = 2 Id ∈ M n ( Z )does not admit such a decomposition. Otherwise, since A = 0 there would exista non-zero potent matrix P and a zero-square matrix N such that A = P + N .Then P = (( A − N ) ) = (4 Id − N ) = 0, which is not possible if P is potentand non-zero, thus establishing our claim.On the other side, Theorem 2.6 remains no longer true for finite commutativerings of characteristic p for some arbitrary but fixed prime p . In fact, it sufficesto find a finite commutative ring R of characteristic p having an element a with a = 0 and a = 0. For example, consider the ring R = Z [ x ] /I where I is theideal generated by the polynomial ( x + x + 1) . The characteristic of R is thenexactly 4. Choose a = ( x + x + 1) + I ∈ R , and let us consider similarly to abovethe matrix A = a Id ∈ M n ( R ) for some n ∈ N . This matrix A has the properties A = 0 and A = 0, whence with the help of the same argument as above it surelycannot be decomposed into the sum of a potent and a zero-square nilpotent. Thisconcludes our arguments.In order to generalize Theorem 2.6 to commutative rings of the form Z p r for somenatural number r ≥
2, we first are going to show that potent elements lift moduloa nilpotent ideal. Our proof follows the ideas of the classical lifting of idempotents(see, for instance, [2, Proposition 27.1]).
Proposition 2.10.
Let R be a finite ring and let I be a nilpotent ideal of R ofindex n . Let us suppose A ∈ R is such that A ∈ R/I is a potent element of
R/I .Then there exists B ∈ R such that A = B and B is potent in R .Proof. Suppose that A t = A ∈ R/I . Then A t − A ∈ I and, therefore,0 = ( A t − A ) n = n X k =0 ( − n − k (cid:18) nk (cid:19) A kt A ( n − k ) = ( − n A n − n X k =1 ( − n − k +1 (cid:18) nk (cid:19) A n +( t − k = ( − n A n − A n +( t − n X k =1 ( − n − k +1 (cid:18) nk (cid:19) A ( k − t − ! Let T := P nk =1 ( − n − k +1 (cid:0) nk (cid:1) A ( k − t − . Therefore, A n = A n +( t − T . Define E := A n ( t − T n . Notice that EA = AE . Let us show that E is an idempotent of R . In fact, one sees that E = A n ( t − T n = A n ( t − t − T n +1 = A n ( t − t − T n +2 = · · · == A n ( t − n ( t − T n + n = E From the above calculations, we may write that E = ( EA )( EA n ( t − − T n ) , and so we get that EA is an invertible element in ERE (note that
ERE is a unitalring with identity element E ). Decompose A = EA + (1 − E ) A . Since EA n = A n , ECOMPOSITIONS OF MATRICES INTO ... 9 it must be that ((1 − E ) A ) n = 0. On the other hand, if we choose k such that t k ≥ n and taking into account that ¯ A is t -potent, it follows that¯ A = ¯ A t = · · · = ¯ A t k = ( EA + (1 − E ) A ) t k = EA t k + ((1 − E ) A t k = EA t k ∈ ER, so ¯ A = EA + (1 − E ) A ∈ ER implies ¯ A = EA .To conclude that EA is potent, it suffices to consider the finite set { EA, ( EA ) , . . . , ( EA ) r , . . . } to get that ( EA ) l = ( EA ) m for some l < m ∈ N , and from the invertibility of EA we get that ( EA ) m − l = E , so ( EA ) m − l +1 = EA , as asserted. (cid:3) So, we are ready to proceed by proving with the promised generalization.
Corollary 2.11.
Let n, r be two natural numbers. Then every matrix A in M n ( Z p r ) can be expressed as P + N , where P is a potent matrix and N is a matrix such that N ∈ M n ( p Z p r ) .Proof. Let R = M n ( Z p r ) and let us consider the nilpotent ideal I of R generated by p Id, i.e., I = M n ( p Z p r ). Then R/I ∼ = M n ( Z p ) and thus it satisfies the hypothesisof Theorem 2.6. Consequently, there exists P , N ∈ R/I such that ¯ A = ¯ P + ¯ N , where P is potent and N = 0. By making use of Proposition 2.10, we can lift ¯ P to apotent matrix P of R . Take any matrix N ∈ R such that N modulo I coincideswith ¯ N (in particular we have that N ∈ I ). Finally, there exists V ∈ I such that A = P + N + V , where • P is potent • ( N + V ) = N + N V + V N + V ∈ I = M n ( p Z p r ) . (cid:3) It is worth noticing that this result extends Corollary 2.7 in a more general vision(notice that when r = 2, the ideal M n ( p Z p ) is zero and so N has zero-square).We finish off our work with the following conjecture which addresses Remark 2.9quoted above. Conjecture.
Suppose m, n ≥ p is a prime. Then everymatrix in M n ( Z p m ) is a sum of a potent and of a nilpotent of order at most m .Note that our results stated above (especially Corollary 2.7) completely settledthe problem for m = 2. In this aspect, can we refine our machinery and results forfinite commutative rings of characteristic p m ? References [1] A. N. Abyzov and I. I. Mukhametgaliev. On some matrix analogues of the little Fermat theo-rem.
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Email address : [email protected] Departamento de Matem´atica Aplicada, Ciencia e Ingenier´ıa de los Materiales yTecnolog´ıa Electr´onica, Universidad Rey Juan Carlos, 28933 M´ostoles (Madrid), Spain
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