aa r X i v : . [ m a t h . R A ] J un Preserving zeros of a polynomial ∗ A. E. Guterman aFaculty of Algebra, Department of Mathematics and Mechanics, Moscow StateUniversity, GSP-2, 119992 Moscow, Russia.
B. Kuzma b Institute of Mathematics, Physics, and Mechanics, Jadranska 19, 1000 Ljubljana,Slovenia. University of Primorska, Faculty of Education, Cankarjeva 5, 6000 Koper, Slovenia.
Abstract
We study non-linear surjective mappings on subsets of M n ( F ),which preserve the zeros of some fixed polynomials in noncommutingvariables. Mathematics subject classification (2000):
Keywords:
Matrix algebra, Multilinear polynomials, Preservers.
The theory of transformations preserving different properties and invariantsdates back to the works by Frobenius, [21], Schur, [34], and Dieudonn´e, [19],and is an intensively developing part of algebra nowadays. A characteri-zation of maps preserving zeros of polynomials plays a central role in thisarea. The most known results of this type were devoted to the characteriza-tion of linear maps on matrix algebras, preserving the following polynomials: p ( x ) = x k , which correspond to nilpotent matrices, see [8], p ( x ) = x − x , ∗ The research of the first author was supported by the grant from Russian Ministryof Science and Technology, MK-1417.2005.1. The research of the second author was sup-ported by grants from the Ministry of Science of Slovenia. a E-mail: [email protected] b E-mail: [email protected] p ( x ) = x k − x , see [10], and matrices of finite order, i.e., the zeros of p ( x ) = x k −
1, see [31, 23]. In 1980 Howard [27] proved the general clas-sification theorem for bijective linear transformations on matrix algebras,preserving zeros of a polynomial in one variable with at least two distinctzeros. This result, together with the main theorem from [8], provides thecomplete characterization of bijective linear maps on matrix algebras overalgebraically closed fields, preserving zeros of polynomials in one variable.Later in [28] Li and Pierce investigated the possibility to remove the invert-ibility assumption from Howard’s theorem and proved some related results.In parallel since 1976, a question of characterizing linear transformationspreserving zeros of multilinear polynomials in several noncommuting vari-ables was considered. In particular, Watkins, [35], characterized bijectivelinear transformations preserving commutativity, i.e., zeros of the polyno-mial p ( x, y ) = xy − yx , in [36, 37] Wong classified operators preserving zeroproducts, i.e., zeros of p ( x, y ) = xy . During the last years there was a biginterest to this question, see [3, 6, 9, 13, 14, 20, 30, 33, 38] and referencestherein. Also several related topics were intensively investigated. Namely,additive transformations Φ on prime rings satisfying the stronger condition p (Φ( x ) , . . . , Φ( x k )) = Φ( p ( x , . . . , x k )) for all x , . . . , x m , were studied in [5].Linear maps that preserve operators on infinitely dimensional algebras anni-hilated by a polynomial in one variable, were treated in [25, 32]. Bijectivelinear operators on the matrix algebra preserving zeros of the involutorypolynomial p ( x, y ) = xy − yx ∗ are classified in [14]. Additive surjections oncertain classes of algebras, preserving zeros of p ( x ) = x or preserving ze-ros of the Jordan polynomial p ( x, y ) = xy + yx , were characterized in [15,Theorem 4.1], [26, Lemma 2.3], and in [38].In spite of the constant interest to the characterization problems for lineartransformations, preserving zeros of matrix polynomials, there were no gen-eral answers in the multivariable case, like the Howard’s theorem concerninglinear preservers of zeros of polynomials in one variable. In particular, allresults were proved for some specific polynomials.In [14] Chebotar, Fong and Lee posed the question about the generalform of linear preservers for zeros of multivariable polynomials explicitly asan open problem: Problem . [14, Problem 1.1] Let p ( x , . . . , x k ) be a polynomial over afield F in noncommuting indeterminates x , . . . , x k of the degree deg p > M n ( F ) → M n ( F ) be a linear map on the matrix algebra M n ( F ).Suppose that p ( A , . . . , A k ) = 0 implies p (Φ( A ) , . . . , Φ( A k )) = 0. Is itpossible to describe such Φ? 2he authors of [14] have conjectured that if the size of matrices, n , isbig enough comparing with the number of variables, k , and Φ is linear andbijective, then Φ is a sum of a scalar multiple of Jordan homomorphism anda transformation that maps into the center of algebra.The present paper is devoted to the solution of the above mentionedproblem for certain sufficiently large classes of polynomials of a general type.Several remarks are in order. Firstly, our results are non-linear in nature,we even do not assume additivity of a transformation under considerationin advance. Secondly, the transformation is not necessary required to beinvertible and we only assume that it is surjective. In addition we foundsome conditions which replace the surjectivity assumption and also providethe examples showing that the assumptions on the transformations, we haveposed, are indispensable. Moreover, the developed technique is characteristicfree and allows us to work without restrictions on the number of variablesof a polynomial. This is done by the exclusion of polynomials which donot provide sufficient restrictions on the transformation to be classified. Forexample, this is the case with polynomial identities of M n ( F ). Say, thepolynomial p ( x , . . . , x n ) := P σ ∈S n sign ( σ ) x σ (1) · · · x σ (2 n ) is an identity on M n ( F ) by the famous Amitsur-Levitzki’s theorem, see [2]. This polynomialvanishes on the whole matrix algebra and therefore it gives no condition on Φ.Therefore we divide our considerations in two parts. Firstly we consider the generic case , where the sum of coefficients of a multilinear polynomial p isnon-zero, and thus p can not be an identity in M n ( F ). Then we investigatethe derogatory case , where the sum of coefficients of p is zero and polynomialidentities may appear.Throughout, n ≥ M n ( F ) will be the algebra of n × n -matrices over an arbitrary field F . Let E ij be its standard basis. LetGL n ( F ) ⊂ M n ( F ) denote the group of invertible matrices, with identity Id.Let I ⊆ M n ( F ) be the set of all rank-one idempotents. Given a fieldhomomorphism ϕ : F → F ( i.e., an additive and multiplicative function on F ), we let X ϕ be a matrix, obtained from X by applying ϕ entry-wise. Inaddition, let X tr be the transposed matrix of X . Matrices P, Q ∈ M n ( F )are called orthogonal to each other if P Q = QP = 0.Lastly, let S k be the set of all permutations of the set { , . . . , k } . Definition 1.1.
Let k ≥
2. We say that a matrix k -tuple ( A , . . . , A k ) is azero of a homogeneous multilinear polynomial p ( x , . . . , x k ) := X σ ∈S k α σ x σ (1) · · · x σ ( k ) p ( A , . . . , A k ) := X σ ∈S k α σ A σ (1) · · · A σ ( k ) = 0 . The set of all such k -tuples will be denoted by S p ⊆ M n ( F ) × · · · × M n ( F ). Definition 1.2.
Suppose p , p are two homogenous multilinear polynomials.A transformation Φ : M n ( F ) → M n ( F ) maps the zeros of p to the zeros of p whenever the implication p ( A , . . . , A k ) = 0 = ⇒ p (Φ( A ) , . . . , Φ( A k )) = 0holds ( equivalently, whenever Φ( S p ) ⊆ S p ). If p = p =: p then Φ pre-serves the zeros of p . In addition, if p ( x, y ) = xy − yx then Φ preservescommutativity . Definition 1.3.
A transformation Φ : M n ( F ) → M n ( F ) strongly maps thezeros of p to the zeros of p whenever the equivalence p ( A , . . . , A k ) =0 ⇐⇒ p (Φ( A ) , . . . , Φ( A k )) = 0 holds. If p = p =: p then Φ stronglypreserves the zeros of p .Our paper is organized as follows.In Section 2 we study the mappings between certain matrix subspaces,including the map from the whole matrix algebra to itself, which stronglypreserve zeros of homogeneous multilinear polynomials with nonzero sum ofcoefficients.In Section 3 we study the transformations that map zeros of a homoge-neous multilinear polynomials of arbitrary many variables with zero sum ofcoefficients to zeros of another such polynomial. To avoid the obstructionswhich come from the polynomial identities of matrix algebra, it is necessaryto restrict the set of permutations. The general problem is then reduced tothe already well-studied commutativity preservers, see [33, 20, 9, 30, 6] fortheir characterization. We remark that some ideas that we are using in thissection came from our recent paper [1].Section 4 contains a number of examples showing that our assumptionare indispensable without posing some additional conditions on Φ or on p ( x , . . . , x k ). In the present section we investigate surjective mappings on certain matrixsubspaces, in particular on the whole matrix space, that strongly preservezeros of a polynomial with non-vanishing sum of coefficients.4e also refer to Chan, Li, and Sze [13], where the zeros of a polynomial p ( x, y ) := xy were considered, and, similarly to our results below, the nicestructure was obtained solely on matrices of rank-one. We will show in thelast section that in a way our results cannot be further improved, withoutimposing additional hypothesis, say additivity of Φ.However, if Φ strongly preserves the zeros of a polynomial with at leastthree variables, then we were able to deduce a structural result holding forall matrices from the defining set of Φ.We now list the main results of the present section. Let D ⊆ M n ( F )and D ⊆ M n ( F ) be subsets that contain all matrices of rank-one and allidempotents of rank n −
1. Also in this section we assume that a homogeneousmultilinear polynomial p ( x , . . . , x k ) := X σ ∈S k α σ x σ (1) · · · x σ ( k ) ; ( α σ ∈ F ) (1)satisfies P α σ = 0.The most general form of the main result of this section is the following: Theorem 2.1.
Let F be an arbitrary algebraically closed field. Assume n ≥ and k ≥ . If a surjection Φ : D → D strongly preserves the zerosof p ( x , . . . , x k ) then there exists a field isomorphism ϕ : F → F , a function γ : D \{ } → F ∗ := F \{ } , and an invertible matrix T such that (i) Φ( A ) = γ ( A ) T A ϕ T − for all rank-one matrices A , or (ii) Φ( A ) = γ ( A ) T ( A ϕ ) tr T − for all rank-one matrices A . Remark 2.2.
The converse to Theorem 2.1 does not hold without imposingadditional constraints on isomorphism ϕ . Namely, there are transformationsof types (i) and (ii) which do not preserve the zeros of p . We refer the readerto the last section for examples. When char F = 2, and the dimension of matrices is n ≥
4, and thepolynomial has at least three variables we have a nice structural result holding for all matrices from the defining set D of Φ. In particular, for all matricesif D = M n ( F ). We merely add scalar matrices to conclusions (i) and (ii) ofTheorem 2.1: Corollary 2.3.
Under the assumptions of Theorem 2.1, assume further char F = 2 , and n ≥ , and k ≥ . Then, (i) Φ( A ) = γ ( A ) T A ϕ T − + µ ( A ) Id for all A ∈ D , or A ) = γ ( A ) T ( A ϕ ) tr T − + µ ( A ) Id for all A ∈ D . The situation is completely different when k = 2, see Example 4.3 below.However, if p ( x, y ) := xy + yx is a polynomial of Jordan multiplication wecan still get the characterization for some special matrices A ∈ D . Corollary 2.4.
Under the assumptions of Theorem 2.1, assume further p ( x, y ) := xy + yx and char F = 2 . Then, the conclusions (i) and (ii) alsohold for diagonalizable A ∈ D , with the spectrum, Sp( A ) = { λ, − λ } . Moreover, we may remove the surjectively assumption from the Theo-rem 2.1, at least for some polynomials:
Corollary 2.5.
Under the assumptions of Theorem 2.1, suppose that a pos-sibly nonsurjective
Φ : D → D strongly preserves the zeros of polynomial p ,defined in Eq. (1), but such that the matrix Cof ( p ) := P σ (1)=1 α σ P σ (1)=2 α σ . . . P σ (1)= k α σ P σ (2)=1 α σ P σ (2)=2 α σ . . . P σ (2)= k α σ ... ... . . . ... P σ ( k )=1 α σ P σ ( k )=2 α σ . . . P σ ( k )= k α σ ∈ M k ( F ) is invertible. Then, the conclusions (i) and (ii) remain valid, with the excep-tion that a field homomorphism ϕ : F → F might be nonsurjective. We first rewrite p . Namely, its coefficients satisfy X α σ = X σ (1)=1 α σ + X σ (2)=1 α σ + · · · + X σ ( k )=1 α σ , and since the left-hand side is nonzero, so must be at least one of the sum-mands on the right. Say, 0 = P σ ( ı )=1 α σ . By dividing p we may assume P σ ( ı )=1 α σ = 1. Moreover, we may also assume ı = 1. Otherwise wewould regard the polynomial p ′ ( x , . . . , x k ) := p ( x τ (1) , . . . , x τ ( k ) ) for permu-tation τ := (1 , ı ). Obviously, Φ would still strongly preserve the zeros of p ′ .We can now rewrite p into the form p ( x , . . . , x k ) := X σ (1)=1 α σ x σ (1) . . . x σ ( k ) + X σ (1) =1 α σ x σ (1) . . . x σ ( k ) ; (2)6here P σ (1)=1 α σ = 1, and where ξ := P σ (1) =1 α σ = −
1. Similarly, thereexists = such that ξ k := X σ ( )= k α σ = 0 , for otherwise 0 = P σ ( k )= k α σ + P σ ( k − k α σ + · · · + P σ (1)= k α σ = P α σ , acontradiction! In the sequel, we will always use the equivalent form (2) ofpolynomial p .It will be beneficial for our further considerations to associate with eachmatrix A the two sets: Ω • A and Ω A • A , defined via the polynomial p byΩ • A := { X ∈ M n ( F ); p ( X, A, . . . , A ) = 0 } , (3)Ω A • A := { X ∈ M n ( F ); p ( A, . . . , A, X , A, . . . , A ) = 0 } , (4)in the last equation, X is at the position. Note that Ω A • A = Ω • A if = 1( say for polynomial p ( x , x ) := x x − x x ). Clearly, Ω • A and Ω A • A are bothvector subspaces of M n ( F ). Moreover, we can rewrite the condition X ∈ Ω • A ,respectively, X ∈ Ω A • A , as XA k − + ( β AXA k − + · · · + β k A k − X ) = 0 , (cid:16) β i := X σ (1)= i α σ (cid:17) (5)respectively, ξ k A k − X + ( ˜ β k − A k − XA + · · · + ˜ β XA k − ) = 0 , (cid:16) ˜ β i := X σ ( )= i α σ (cid:17) . (6)In particular, Ω • A equals the null space of the elementary operator T • A : M n ( F ) → M n ( F ), defined by X XA k − + ( β AXA k − + · · · + β k A k − X ).Similarly, Ω A • A equals the null space of of the elementary operator T A • A ,defined by X A k − X + ( ˆ β k − A k − XA + · · · + ˆ β XA k − ), where ˆ β i :=˜ β i /ξ k .We proceed with a series of lemmas. The first lemma allows us to computethe spectrum of elementary operators, in particular, of the operators T • A and T A • A . We present the easy proof for the sake of convenience. Lemma 2.6.
Let F be an arbitrary field. Let L ∈ M m ( F ) and M ∈M n ( F ) be matrices with Sp( L ) = { λ } and Sp( M ) = { µ } . Then, the spec-trum of elementary operator T L,M : M m × n ( F ) → M m × n ( F ) , defined by X β t L t X + β t − L t − XM + · · · + β XM t is a singleton: Sp( T L,M ) = { β t λ t + β t − λ t − µ + · · · + β µ t } . roof. Let us consider the decomposition L = S L ˆ LS − L , where ˆ L is the upper-triangular Jordan form of L , and M = S M ˘ M S − M , where ˘ M is lower-triangularJordan form of M . It is well-known that the matrix representation of X L t − k XM k , relative to basis E , . . . , E n , E , . . . , E n , . . . , E n , . . . , E nn equals the tensor (Kronecker) product ( M k ) tr ⊗ L t − k . Moreover, the matrices( M k ) tr ⊗ L t − k ; ( k = 0 , , . . . ) are simultaneously similar to ( ˘ M k ) tr ⊗ ˆ L t − k viasimilarity ( S − M ) tr ⊗ S L .Hence, the spectrum of ( M k ) tr ⊗ L t − k equals the spectrum of ( ˘ M k ) tr ⊗ ˆ L t − k . Now, ( ˘ M k ) tr , as well as ˆ L t − k are both upper-triangular matrices,with µ k , respectively, λ t − k on the diagonal. Hence, their tensor product re-mains upper-triangular, with λ t − k µ k on the diagonal. Consequently, T L,M = P β k ( M k ) tr ⊗ L t − k is similar to the upper-triangular matrix P β k ( ˘ M k ) tr ⊗ ˆ L t − k , with the number P β k λ t − k µ k on main diagonal. This number is, hence,the only eigenvalue of T L,M .We remark that over the field of complex numbers this fact follows fromthe results of Lumer and Rosenblum [29] and Curto [18], where it was provenin a different way for linear operators on Hilbert spaces, possibly infinitedimensional. We also remark that in the case T • A = A k − X + XA k − a shortproof of the corresponding result for Hilbert spaces is presented by Bhatiaand Rosenthal in [7, p. 2] and some further properties of the spectrum canbe found in [18], some additional properties of T • A are investigated by Chuaiand Tian in [17]. Lemma 2.7.
Let µ, ν, µ ′ , ν ′ ∈ F be scalars, and let µ + ν = 0 . If P X + µ P XP + νXP = 0 = XP + µ ′ P XP + ν ′ P X holds for some idempotent P then P X = 0 = XP .Proof. Postmultiply the equation
P X + µ P XP + νXP = 0 with P andsubtract. We derive P X = P XP . Likewise
P XP = XP , from the secondequation. Using P X = P XP = XP again in the first equation, we get(1 + µ + ν ) P XP = 0, so XP = P XP = P X = 0.
Lemma 2.8.
Suppose D ⊆ M n ( F ) contains all matrices of rank-one. Then, A ∈ M n ( F ) is a zero matrix if and only if p ( A, X, . . . , X ) = 0 for each X ∈ D . roof. We prove only the nontrivial implication. Indeed, substituting X := E ii we have0 = p ( A, E ii , . . . , E ii ) = X σ (1)=1 α σ AE ii + X σ (1) ,k } α σ E ii AE ii + X σ (1)= k α σ E ii A = AE ii + αE ii AE ii + β k E ii A for i = 1 , . . . , n . Premultiply with idempotent E ii , and compare the twoequations. We get AE ii = E ii AE ii . Hence, 0 = AE ii + ( αAE ii + β k E ii A ).We may sum-up these n equations to get 0 = A Id +( αA Id + β k Id A ) =(1 + α + β k ) A , so A = 0. Corollary 2.9.
Let Φ satisfy conditions of Theorem 2.1. Then, ∈ D ifand only if ∈ D . Moreover, Φ( A ) = 0 if and only if A = 0 .Proof. If A = 0 then, by Lemma 2.8, there exists some X ∈ D with p ( A, X, . . . , X ) = 0. Consequently, p (Φ( A ) , Φ( X ) , . . . , Φ( X )) = 0, and soΦ( A ) = 0. Hence, if 0 D then 0 / ∈ Im Φ = D . By surjectivity we likewisesee that Φ( A ) = 0 implies A = 0.We now characterize when two rank-one nilpotents are scalar multiple ofeach other in terms of Ω • A ∩ Ω A • A , i.e., in terms of the zeros of polynomial p . Lemma 2.10.
Let n ≥ , let ϕ : F → F be a nonzero field homomorphism,and let N , N ∈ M n ( F ) be rank-one nilpotents. Assume that the followingcondition (i) is satisfied:(i) N ∈ Ω • P ∩ Ω P • P ⇐⇒ N ∈ Ω • P ϕ ∩ Ω P ϕ • P ϕ holds for every rank-oneidempotent P .Then N = λN ϕ for some nonzero scalar λ ∈ F .Proof. Pick a similarity S such that N = SE S − . Then, the rank-oneidempotents P := SE S − , . . . , P n := SE nn S − , and P n +1 := S ( E n + E nn ) S − are orthogonal to N . Hence, N ∈ Ω • P i ∩ Ω P i • P i , which by (i)implies N ∈ Ω • P ϕi ∩ Ω P ϕi • P ϕi for i = 3 , . . . , ( n + 1). Using the equivalentexpressions (5)–(6), we can easily rewrite this into0 = (cid:0) ( S − ) ϕ N S ϕ (cid:1) ◦ • E (cid:0) ( S − ) ϕ N S ϕ (cid:1) • ◦ E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .0 = (cid:0) ( S − ) ϕ N S ϕ ) ◦ • E nn (cid:0) ( S − ) ϕ N S ϕ ) • ◦ E nn (cid:0) ( S − ) ϕ N S ϕ (cid:1) ◦ • ( E n + E nn ) 0 = (cid:0) ( S − ) ϕ N S ϕ (cid:1) • ◦ ( E n + E nn ) , X ◦ • P := XP + βP XP + β k P X , and where X • ◦ P := P X + ˆ β P XP +ˆ β XP for scalars β := ( β + · · · + β k − ), ˆ β := ( ˜ β k − + · · · + ˜ β ) /ξ k , andˆ β := ˜ β /ξ k .By Lemma 2.7, the above equations imply orthogonality between thenilpotent ( S − ) ϕ N S ϕ and idempotents E , . . . , E nn , ( E n + E nn ). More pre-cisely, the first ( n −
2) equalities give that ( S − ) ϕ N S ϕ can be nonzero onlyin the upper-left 2 × S − ) ϕ N S ϕ = ςE + λE . Since N is nilpotent, ς = 0. Thus N = λS ϕ E ( S − ) ϕ = λS ϕ E ϕ ( S − ) ϕ = λN ϕ , as desired.Similarly we can prove the following: Lemma 2.11.
Under the assumptions of Lemma 2.10, suppose the followingcondition (i’) is satisfied:(i’) N ∈ Ω • P ∩ Ω P • P ⇐⇒ N ∈ Ω • ( P ϕ ) tr ∩ Ω ( P ϕ ) tr • ( P ϕ ) tr holds for everyrank-one idempotent P .Then N = λ ( N ϕ ) tr for some nonzero scalar λ ∈ F .Proof. Similar to Lemma 2.10.We next characterize scalar multiples of rank-one idempotents in termsof Ω • A ∩ Ω A • A , i.e., in terms of the zeros of polynomial p . This is a chiefLemma in the proof of Theorem 2.1. Lemma 2.12.
Fix A ∈ M n ( F ) . Under the assumptions of Theorem 2.1, precisely one of the following three possibilities occurs for the set Ω • A ∩ Ω A • A :(i). Ω • A ∩ Ω A • A = { } .(ii). Ω • A ∩ Ω A • A contains a square-zero matrix of rank-one.(iii). Ω • A ∩ Ω A • A = F P where P is a rank-one idempotent.Proof. Obviously, the listed three possibilities are exclusive. It, hence, re-mains to see that at least one of them does occur. We will rely on the factthat X ∈ Ω • A ∩ Ω A • A is equivalent to Eq. (5) and Eq. (6), simultaneously.Now, with the help of similarity we may assume A = C n ( λ ) ⊕ · · · ⊕ C n r ( λ r ) is already in its Jordan block-diagonal form, where λ , . . . , λ r arepairwise distinct eigenvalues of A , and an n i × n i matrix C n i ( λ i ) is a sum ofall Jordan blocks that correspond to eigenvalue λ i . We may decompose X =10 X ij (cid:1) ≤ i,j, ≤ r accordingly. Then, A s = C n ( λ ) s ⊕ · · · ⊕ C n r ( λ r ) s , for s =1 , , . . . , n , are also block-diagonal, so the ( i, j )-th block of Eqs. (5)–(6) read X ij C n j ( λ j ) k − + (cid:0) β C n i ( λ i ) X ij C n j ( λ j ) k − + · · · + β k C n i ( λ i ) k − X ij (cid:1) = 0(7)respectively, ξ k C n i ( λ i ) k − X ij + (cid:0) ˜ β k − C n i ( λ i ) k − X ij C n j ( λ j ) + · · · + ˜ β X ij C n j ( λ j ) k − (cid:1) = 0(8)Obviously, Sp (cid:0) C n i ( λ i ) s (cid:1) = { λ si } for any integer s . In view of Lemma 2.6 weconsequently introduce two polynomials p • A ( λ, µ ) := µ k − + ( β λµ k − + · · · + β k λ k − ) ∈ F [ λ, µ ]as well as its counterpart p A • A ( λ, µ ) := ξ k λ k − + ( ˜ β k − λ k − µ + · · · + ˜ β µ k − ) , and proceed in five steps: Step 1 . Assume first that for no pair ( λ i , λ j ) ∈ Sp( A ) × Sp( A ) we havesimultaneously p • A ( λ i , λ j ) = 0 = p A • A ( λ i , λ j ).In this case, we note that the left-hand sides of each of the Eqs. (7)–(8)define an elementary operator. Therefore, Lemma 2.6, with L := C n i ( λ i )and M := C n j ( λ j ) implies that their spectra are equal to { p • A ( λ i , λ j ) } , and { p A • A ( λ i , λ j ) } , respectively. At least one does not contain zero, and there-fore, the corresponding elementary operator is invertible. The correspondingequation, on the other hand, implies that X ij = 0. Hence, all blocks of X arezero. This clearly demonstrates Ω • A ∩ Ω A • A = { } , and we have condition (i)satisfied. Step 2 . Suppose next p • A ( λ i , λ j ) = 0 = p A • A ( λ i , λ j ) for distinct λ i , λ j . Forsimplicity, we assume ( i, j ) = (1 , X with all, but the (1 , p ( X, A, . . . , A ), but the (1 , ,
2) block equals to (cid:0) p ( X, A, . . . , A ) (cid:1) = X C n ( λ ) k − ++ (cid:0) β C n ( λ ) X C n ( λ ) k − + · · · + β k C n ( λ ) k − X (cid:1) Now, write C n ( λ ) = λ Id n + N and C n ( λ ) = λ Id n + N , for someupper-triangular nilpotents N , N . Next, consider an n × n matrix ˆ X ,11ith all entries, but the upper-right one, equal to zero. It is easy to see thatˆ X N = 0 = N ˆ X , which in turn, implies that the right side of the aboveequation simplifies into ˆ X λ k − +( β λ X λ k − + · · · + β k λ k − X ) = ( λ k − + β λ λ k − + · · · + β k λ k − ) ˆ X = p • A ( λ , λ ) ˆ X = 0. Consequently, if a block-matrix X ∈ M n ( F ) has its (1 , X while the other blocksare zero then it is a rank-one nilpotent, and p ( X, A, . . . , A ) = 0. Similarly, byEq. (8), we also infer p ( A, . . . , A, X , A, . . . , A ) = 0. Therefore, such X ∈ Ω • A ∩ Ω A • A , which guaranties the condition (ii). Step 3 . Suppose we are not under the conditions of Step 2. We, thus,consider the case p • A ( λ i , λ i ) = 0 = p A • A ( λ i , λ i ) for some i .Clearly, p • A ( λ i , λ i ) = λ k − i (1 + ξ ) where ξ := β + · · · + β k . However, ξ = −
1, so p • A ( λ i , λ i ) = 0 implies λ i = 0. We now consider two options,regarding the dimension of the corresponding block C n i ( λ i ) = C n i (0). Step 4 . Suppose λ i = 0 and the corresponding block C n i ( λ i ) has dimen-sion n i ≥
2. For simplicity, assume i = 1, so λ = 0.Now, the first block of p ( X, A, . . . , A ) equals X C n (0) k − + (cid:0) β C n (0) X C n (0) k − + · · · + β k C n (0) k − X (cid:1) Note that C n (0) is an upper-triangular nilpotent. Since n ≥
2, a straight-forward computations show that E n C n (0) k − + (cid:0) β C n (0) E n C n (0) k − + · · · + β k C n (0) k − E n (cid:1) = 0. Therefore, X := E n ∈ Ω • A . Similarly, we canalso show that E n ∈ Ω A • A and we are in the condition (ii) again. Step 5 . Finally, suppose we are not under the conditions of Step 2 andthere exists i such that λ i = 0 with the corresponding block C n i ( λ i ) = C n i (0)of dimension n i = 1. Again, for simplicity i = 1, so that A = 0 ⊕ C ,where C := C n ( λ ) ⊕ · · · ⊕ C n r ( λ r ) is an ( n − × ( n −
1) matrix. Recallthat λ , . . . , λ r = 0, so C is invertible.It is straightforward to see that F E ⊆ Ω • A ∩ Ω A • A in this case. Let usshow that also F E ⊇ Ω • A ∩ Ω A • A .Retaining the block structure of X , the Eqs. (7)–(8) simplify for theblocks in the first row/column into X j C n j ( λ j ) k − = 0, respectively, into ξ k C n j ( λ j ) k − X j = 0. Since C n j ( λ j ) are invertible for j = 2 , . . . , r , andsince ξ k = 0, we get X j = 0 = X j whenever j ≥ X st for s, t ≥
2. Now, if s = t , it is impossible tohave simultaneously p • A ( λ s , λ t ) = 0 = p A • A ( λ s , λ t ), by conditions of Step 2.This remains true if s = t ≥
2, for otherwise p • A ( λ s , λ s ) = 0, which wouldwrongly imply λ s = 0. Then, however, we may copy the arguments fromStep 1 to deduce X st = 0. Therefore, the only possible nonzero block of X X , and so X = αE for some scalar α . Therefore, Ω • A ∩ Ω A • A = F E .This gives the case (iii) with P = E .Recall that I is the set of rank-one idempotents in M n ( F ). Corollary 2.13.
Let conditions of Theorem 2.1 be satisfied. Then(i) Φ( I ) ⊆ F I .(ii) If Φ( X ) ∈ I then X = αY for certain α ∈ F \ { } and Y ∈ I .Proof. Step 1 . Pick a rank-one idempotent P . Then, A := Id − P is anidempotent of rank ( n − A ∈ D . The direct calculations show thatΩ • A ∩ Ω A • A = Ω • (Id − P ) ∩ Ω (Id − P ) • (Id − P ) consists precisely of those matrices X which satisfy XA + βAXA + β k AX = 0 = ξ k AX + ˜ βAXA + ˜ β XA ; ( β := β + · · · + β k − ) . By Lemma 2.7, X is orthogonal to idempotent A = Id − P . Hence, X = λP ,and so, Ω • A ∩ Ω A • A = F P . Step 2 . Since Φ preserves zeros of p , the first step implies Φ( P ) ∈ Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) . By Corollary 2.9, Φ( P ) = 0, so Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) = { } . By Lemma 2.12 it follows that Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) either is equal toa scalar multiple of an idempotent or contains a rank-one nilpotent. Weassume erroneously the later, i.e., that there exists a rank-one nilpotent Y ∈ Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) . By the surjectivity of Φ it follows that Y = Φ( X )for some X ∈ D . Since Φ preserves zeros of p strongly we have X ∈ (Ω • A ∩ Ω A • A ) ∩ D = F P ∩ D .By Corollary 2.9, X = 0. On the other hand, Y = 0 since Y is a nilpotentof rank-one. Hence, p ( Y, . . . , Y ) = 0, so also 0 = p ( X, . . . , X ) = (1 + ξ ) X k .This is clearly a contradiction since X ∈ F P . Hence, Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) contains no rank-one nilpotents. Therefore, Φ( P ) ∈ Ω • Φ( A ) ∩ Ω Φ( A ) • Φ( A ) = F Q , for some rank-one idempotent Q . This proves (i). Step 3 . Conversely, if Φ( X ) is an idempotent of rank-one, then F Φ( X ) =Ω • B ∩ Ω B • B for B := Id − Φ( X ) ∈ D . Since Φ is surjective and preserves zerosof p strongly, we can prove that X ∈ F P for some rank-one idempotent P in a similar way, as in the proof of Item (i). Remark 2.14.
Since the polynomial p is homogeneous, the assumptions andthe conclusion of Theorem 2.1 will not be affected if we replace Φ by a map-ping ˆΦ : A δ ( A ) · Φ( A ) , where δ : D → F \{ } is a scalar function.We may define δ in such a way that ˆΦ preserves rank-one idempotents (i.e., ˆΦ( I ) ⊆ I ), while ˆΦ( A ) = Φ( A ) for any other matrix from D .The redefined ˆΦ may not be surjective, but we clearly have I ⊆ Im ˆΦ .
13e are ready now to prove the main result.
Proof of Theorem 2.1.
We proceed in several steps.
Step 1 . The transformation ˆΦ preserves orthogonality among rank-oneidempotents. Indeed, suppose
P, Q are two orthogonal rank-one idempotents. Then Q ∈ Ω • P ∩ Ω P • P . Therefore, ˆΦ( Q ) ∈ Ω • ˆΦ( P ) ∩ Ω ˆΦ( P ) • ˆΦ( P ) , so that X ˆΦ( P ) + β ˆΦ( P ) X ˆΦ( P ) + β k ˆΦ( P ) X = 0 = ˆΦ( P ) X + ˆ β ˆΦ( P ) X ˆΦ( P ) + ˆ β X ˆΦ( P ) for X := ˆΦ( Q ). The orthogonality between X = ˆΦ( Q ) and ˆΦ( P ) now followsfrom Lemma 2.7. Step 2.
Let us show that ˆΦ is injective transformation on the set ofrank-one idempotents. We assume erroneously that ˆΦ( P ) = ˆΦ( P ) for some distinct rank-one idempotents P = x f tr1 and P = x f tr2 . Then either x , x are lin-early independent or f , f are linearly independent or both. Assume that x , x are. Then, we can construct a rank-one idempotent Q such that p ( Q, P , . . . , P ) = QP + βP QP + β k P Q = 0, but p ( Q, P , . . . , P ) = QP + βP QP + β k P Q = 0.Indeed, we choose any nonzero y with f tr1 y = 0 = f tr2 y . Suppose first y = µ x + µ x . Then, µ = 0, since otherwise y = µ x , contradicting f tr1 x = 1, f tr1 y = 0. Now, as x , x are linearly independent, we may choose g such that g tr x = 0 and g tr x = 1 /µ . Then g tr y = 1. Now, if y , x , x are linearly independent, we may choose g such that g tr x = 0, g tr x = 1, g tr y = 1. In both cases, Q := yg tr is the required idempotent.For the chosen Q we have0 = p ( ˆΦ( Q ) , ˆΦ( P ) , . . . , ˆΦ( P )) = p ( ˆΦ( Q ) , ˆΦ( P ) , . . . , ˆΦ( P )) = 0 , a contradiction.On the other hand, if f , f are independent, we can similarly find Q with p ( P , Q, . . . , Q ) = 0, but p ( P , Q, . . . , Q ) = 0. As before, this leads to acontradiction. Indeed, ˆΦ( P ) = ˆΦ( P ). Step 3.
Now we can characterize the action of ˆΦ on the set of rank-oneidempotents. Injective mappings on rank-one idempotents, which preserve orthogo-nality are classified in ˇSemrl [33, Theorem 2.3] ( this result is stated onlyfor F = C , but it was already remarked by the author that the proofs are valid forany algebraically closed field ). It follows that there exists a field homomor-phism ϕ : F → F , and invertible matrix T such that either ˆΦ( P ) = T P ϕ T − for every rank-one idempotent P , or else ˆΦ( P ) = T ( P ϕ ) tr T − for every rank-one idempotent P . 14 tep 4. Field homomorphism ϕ is surjective. It suffices to see that the restriction of ˆΦ on the set of rank-one idempo-tents, ˆΦ | I : I → I , is surjective. Let us take any F ∈ I . By Remark 2.14, I ⊆ Im ˆΦ, thus F = ˆΦ( X ) for certain X ∈ D . Corollary 2.13 shows that X = λP for some P ∈ I . Now, choose pairwise orthogonal P , . . . , P n ∈ I that are also orthogonal to P . Clearly, ( λP ) ∈ Ω • P i ∩ Ω P i • P i , so also ˆΦ( λP ) ∈ Ω • ˆΦ( P i ) ∩ Ω ˆΦ( P i ) • ˆΦ( P i ) . As in Step 1 we derive that F = ˆΦ( λP ) is orthogonalto ˆΦ( P ) , . . . , ˆΦ( P n ). On the other hand, however, ˆΦ( P ) , ˆΦ( P ) , . . . , ˆΦ( P n )are n pairwise orthogonal rank-one idempotents as well. This is possibleonly when ˆΦ( P ) = F , and the result follows. Step 5.
The conclusion of Theorem 2.1 is valid for all non-nilpotent rank-one matrices.
Similar to Step 4 it can be shown that ˆΦ( λP ) ∈ F ˆΦ( P ) for any P ∈I , i.e., there exists a transformation δ ′ : D → F \{ } such that ˆΦ( λP ) = δ ′ ( λP ) ˆΦ( P ) for all λ ∈ F , P ∈ I . Step 6. ˆΦ preserves the set of rank-one nilpotents. To see this, we choose any rank-one nilpotent N . Using similarity, wemay assume N = E . Then, we can find n − E , . . . , E nn that are also orthogonal to N . Clearly, N ∈ Ω • E ii ∩ Ω E ii • E ii , so also ˆΦ( N ) ∈ Ω • ˆΦ( E ii ) ∩ Ω ˆΦ( E ii ) • ˆΦ( E ii ) for n − E ii ), i = 3 , . . . , n .Using similarity in the image space, we may assume ˆΦ( E ii ) = E ii . Thus,we have ˆΦ( N ) ∈ Ω • E ii ∩ Ω E ii • E ii for i = 3 , . . . , n . As in Step 1 we de-rive that idempotents E ii are orthogonal on ˆΦ( N ) for i = 3 , . . . , n . Con-sequently, ˆΦ( N ) could be nonzero only at the upper left 2 × p ( N, . . . , N ) = (1 + ξ ) N k = 0. Thus also 0 = 1 / (1 + ξ ) p ( ˆΦ( N ) , . . . , ˆΦ( N )), i.e., ˆΦ( N ) k = 0. Hence, ˆΦ( N ) is a nonzero nilpotentand all its non-zero elements are concentrated in the 2 × N ) is a nilpotent of rank-one. Step 7.
The end of the proof .Finally, consider the redefined ˜Φ : A T − ˆΦ( A ) T . By Step 3 either˜Φ( P ) ≡ P ϕ for rank-one idempotents P , or else ˜Φ( P ) ≡ ( P ϕ ) tr for rank-oneidempotents P . Hence, applying Lemma 2.10 ( respectively, Lemma 2.11 ) torank-one nilpotents N := N and N := ˜Φ( N ) we obtain ˜Φ( N ) = δ ′′ ( N ) · N ϕ for certain δ ′′ : M n ( F ) → F ( respectively, ˜Φ( N ) = δ ′′ ( N ) · ( N ϕ ) tr ). Obviously,this holds for any rank-one nilpotent N .15 .2 Proof of Corollaries It will be beneficial to regard F n as the space of matrices of dimension n –by–1, i.e., column vectors . Thus, any rank–one matrix A ∈ M n ( F ) can bewritten as A = xf tr for suitably chosen x , f ∈ F n . Its trace then equalsTr A = f tr x .Now, to prove Corollary 2.3, we will rely on the folowing result due toBreˇsar and ˇSemrl [12, Thm. 2.4], which we state slightly changed, recastingit into our framework: Lemma 2.15.
Let F be an infinite field with char F = 2 , and let R , R , R ∈M n ( F ) be three matrices. Then (i) implies (ii) below.(i) The vectors R u , R u , and R u are linearly dependent for every u ∈ F n .(ii) Either R , R , R are linearly dependent, or there exist v , w , z ∈ F n such that Im R i ⊆ Lin F { v , w , z } for i = 1 , , , or there exists a rank-one idempotent P ∈ M n ( F ) such that dim Lin F { (Id − P ) R , (Id − P ) R , (Id − P ) R } = 1 . With its help, the following generalization of Lemma 2.10 can be proven:
Lemma 2.16.
Let n ≥ be an integer, let A, B ∈ M n ( F ) be nonzero ma-trices, let ϕ : F → F be a nonzero field homomorphism, and let α, β ∈ F .Assume that the following condition (i) is satisfied:(i) N AP + αP AN = 0 ⇐⇒ N ϕ BP ϕ + βP ϕ BN ϕ holds for every rank-one idempotent P and every rank-one matrix N with P N = 0 =
N P .Then there exists γ, µ ∈ F such that B = γA ϕ + µ Id .Proof. Pick any nonzero vector x ∈ F n and assume erroneously that thevector b := B x ϕ Lin F { A ϕ x ϕ , x ϕ } . Denote a := A x , and let f , . . . , f l be abasis of { a , x } ⊥ := { f ∈ F n ; f tr a = 0 = f tr x } ( here, l = n − n − x and a are linear independent or not, correspondingly ). Since the rank of a matrixequals the maximal size of its nonzero minors, the vectors f ϕ , . . . , f ϕl are alsolinearly independent. Hence, they form a basis of { a ϕ , x ϕ } ⊥ . Now, b Lin F { a ϕ , x ϕ } , so ( f ϕj ) tr b = 0 for at least one j . Consequently, there exists f = f j such that f tr x = 0 = f tr A x , and ( f ϕ ) tr B x ϕ = 0 . n > y such that x Lin F { y , A y } . (9)Indeed, write F n = Lin F { x } ⊕ M . If Ker( A | M ) = 0, then any nonzero y ∈ Ker( A | M ) satisfies Eq. (9). Assume now that Ker( A | M ) = 0 and x ∈ Lin F { y , A y } for each y ∈ M . Then A | M y = λ y x + µ y y . Since x / ∈ M , wecould deduce that λ y is a linear functional, on the space M with dim M ≥ λ y = 0 for at least one nonzero y = y ∈ M . For this y we have A y = µ y y and then Lin F { y , A y } = F y . However, x / ∈ F y ⊂ M — acontradiction.Now, by (9), we may choose a vector g with g tr y = 0 = g tr A y , but g tr x = 1. Then, P := xg tr is an idempotent of rank-one, and N := yf tr is amatrix of rank-one, and we have P N = 0 =
N P . Moreover,
N AP + αP AN =( f tr A x ) yg tr + α ( g tr A y ) xf tr = 0 + α · f ϕ ) tr B x ϕ ) · y ϕ ( g ϕ ) tr + β · (( g ϕ ) tr B y ϕ ) · x ϕ ( f ϕ ) tr . However, the first summand on the right is nonzero. Hence, the right sideis nonzero, since y ϕ ( g ϕ ) tr and x ϕ ( f ϕ ) tr are linearly independent matrices( namely, ( g ϕ ) tr x ϕ = 1, while ( f ϕ ) tr x ϕ = 0 ). This contradiction establishesthat B x ϕ ∈ Lin F { A ϕ x ϕ , x ϕ } for any x ∈ F n . (10)By Eq. (10), the vectors B x ϕ , x ϕ , A ϕ x ϕ are always F –linearly dependent.Let us show that even more is true: indeed the matrices B, Id , A ϕ are locallylinearly dependent, i.e., for any z ∈ F n the vectors B z , z , A ϕ z are linearlydependent. To demonstrate this, we consider a matrix Ξ( z ) := [ B z , z , A ϕ z ]with three columns. By Eq. (10), if z = x ϕ for a certain x ∈ F n , then all its3–by–3 minors must vanish.Consider any such minor. It is a polynomial q ( z , . . . , z n ) ∈ F [ z , . . . , z n ],where z = [ z , . . . , z n ] tr . By Eq. (10) this polynomial vanishes identicallywhenever the variables take the values from a subfield O := ϕ ( F ) ⊆ F ,i.e., for any values α , . . . , α n ∈ O it holds that q ( α , . . . , α n ) = 0. Now,being algebraically closed, F and hence also O = ϕ ( F ) have infinitely manyelements. It is easy to see then that then, q is a zero polynomial. For thesake of completeness we sketch the proof here. We will write q in the form q ( z , . . . , z n ) = a n ( z , . . . , z n − ) z nn + · · · + a ( z , . . . , z n − ) z n + a ( z , . . . , z n − ) . By the assumptions, this vanishes whenever z , . . . , z n ∈ O . Now, at each fixed z , . . . , z n − ∈ O , this is a polynomial in only one variable, z n . However, it is ero for infinitely many values of z n ∈ O . Hence, x q ( z , . . . , z n − , x ) is azero polynomial for each fixed ( z , . . . , z n − ) ∈ O n − . That is, all its coefficients, a i ( z , . . . , z n − ) are zero for any ( z , . . . , z n − ) ∈ O n − . It remains to show that a i ( z , . . . , z n − ) are identically zero, not only for any choice of z , . . . , z n − ∈ O ,but also for any choice of z , . . . , z n − ∈ F . Now, we may repeat the aforesaidprocedure with each a i ( z , . . . , z n − ): Write it as a i ( z , . . . , z n − ) = b im ( z , . . . , z n − ) z mn − + · · · + b i ( z , . . . , z n − )and argue as before to deduce that b j ( z , . . . , z n − ) vanishes for any choice of z , . . . , z n − ∈ O . Continuing in this way we obtain at the end certain polynomials c l ( z ) ∈ F [ z ] which are zero for any value z ∈ O . It follows that c l ( z ) is zero forinfinitely many values of z , i.e., that c l ( z ) is a zero polynomial. By the backwardinduction, we get that all coefficients a i ( z , . . . , z n − ) are zero, i.e., that q is indeeda zero polynomial. Therefore, q ( z , . . . , z n ) = 0 holds for any z , . . . , z n ∈ F . We repeat thiswith all 3–by–3 minors to deduce that rk Ξ( z ) ≤ z ∈ F n , as claimed.Consequently, ( R , R , R ) := ( B, Id , A ϕ ) are locally linearly dependent.We can now invoke Breˇsar and ˇSemrl’s theorem, see Lemma 2.15 in this text.Since R = Id and dim(Im(Id)) = n (cid:13)
3, the only three possibilities left toconsider are (a) A ϕ = λ Id, or (b) B = γ A ϕ + µ Id, or(Id − Q ) B = λ (Id − Q ) Id , (Id − Q ) A ϕ = λ ′ (Id − Q ) Id (c)for some rank-one idempotent Q . Under (a), B must also be a scalar, in viewof Eq. (10). So, under (a)–(b) we are done.Consider lastly (c). Decomposing B = (Id − Q ) B + QB , and using Eq. (c),easily reveals B = λ Id + Q ( B − λ Id) = λ Id +ˆ u ˆ v tr B for some column vectorsˆ u , ˆ v B . Likewise, A ϕ = λ ′ Id +ˆ u ˆ v tr A . By passing the appropriate scalar tothe other term (in ˆ u ˆ v tr B , ˆ u ˆ v tr A ), we may assume that at least one entry ofvector ˆ u equals 1. Then, from λ ′ Id +ˆ u ˆ v tr A = A ϕ ∈ M n ( ϕ ( F )) it follows thatˆ u , ˆ v A ∈ ϕ ( F n ), and also λ ′ ∈ ϕ ( F ). Let u , v A , and λ ′′ be such that ˆ u = u ϕ ,ˆ v A = v ϕA , and λ ′ = ϕ ( λ ′′ ). Then A ϕ = ϕ ( λ ′′ ) Id + u ϕ ( v ϕA ) tr and B = λ Id + u ϕ v tr B . Now, if v ϕA , v B are linearly dependent we are done. Assume erroneously theyare not. We first choose v such that v ϕ , v ϕA , v B are independent, and at thesame time, ( v ϕ ) tr u ϕ = 1 ( Such a vector v exists since we can enlarge v A with v , . . . , v n to a basis of F n , assuming v tr i u = 1. Then, v ϕA , v ϕ , . . . , v ϕn is still abasis, so some v ϕ := v ϕi is independent of v ϕA , v B ).By the choice of v , the vector u ϕ / ∈ { v ϕ } ⊥ . So much the more u ϕ / ∈{ v ϕ , v ϕA } ⊥ , so that Lin F { u ϕ }∩{ v ϕ , v ϕA } ⊥ = { } . We next choose w with w ϕ ∈ v ϕ , v ϕA } ⊥ \{ v B } ⊥ ( it is possible since v ϕ , v ϕA , v B are independent . ) Hence, w , u are independent, so we can choose h with h tr u = 0 and h tr w := 1. Lastly,choose nonzero s ∈ { h } ⊥ .We now form N := sv tr and P := wh tr . By its choice, w ∈ { v , v A } ⊥ soit follows P N = 0 = N P , and P = P . Moreover, N AP = 0 = P AN .By assumptive condition (i), we would have to have 0 = N ϕ BP ϕ + βP ϕ BN ϕ .However, it equals N ϕ BP ϕ + βP ϕ BN ϕ = (cid:0) ( v ϕ ) tr u ϕ (cid:1)| {z } =1 · ( v tr B w ϕ ) | {z } =0 · s ϕ ( h ϕ ) tr + β · = 0 . This contradiction finally establishes linear dependence between v ϕA and v B . Proof of Corollary 2.3.
Obviously, Φ satisfies the hypothesis of Theorem 2.1,hence also its conclusion. That is, (i) and (ii) hold for rank-one matrices.Now, we may replace Φ by a mapping X γ ( X ) T − Φ( X ) T to achievethat either Φ( X ) ≡ X ϕ holds for all rank-one matrices, or else Φ( X ) ≡ ( X ϕ ) tr holds for all rank-one matrices. It remains to see that, modulo scalarmultiplication and scalar addition, same holds for A ∈ D of rank ≥ X ) ≡ X ϕ for all X of rank-one, and let A ∈ D be ofrank ≥
2. Now, by definition, our polynomials satisfy P σ (1)=1 α σ = 1. Notethat 1 = X σ (1)=1 α σ = X σ (1)=1 σ (2)=2 α σ + X σ (1)=1 σ (3)=2 α σ + · · · + X σ (1)=1 σ ( k )=2 α σ , so at least one summand on the right is nonzero. Say, τ := P σ (1)=1 σ ( j ′ α σ = 0.Having found j ′ , we next pick any rank-one idempotent P , and any rank-onematrix N with P N = 0 =
N P . Consider now p ( N, P, . . . , P, A j ′ , P, . . . , P )with matrix A at the position j ′ . An easy argument reveals that p ( N, P, . . . , P, A j ′ , P, . . . , P ) = X σ (1)=1 σ ( j ′ α σ N AP + X σ (1)= kσ ( j ′ k − α σ P AN ++ X the rest permut. α σ · τ ( N AP + αP AN )19here α := 1 /τ P σ (1)= kσ ( j ′ k − α σ . Similarly, the value of p (Φ( N ) , Φ( P ) , . . . , Φ( P ) , Φ( A ) j ′ , Φ( P ) , . . . , Φ( P )) = p ( N ϕ , P ϕ , . . . , P ϕ , Φ( A ) j ′ , P ϕ , . . . , P ϕ )further equals τ ( N ϕ Φ( A ) P ϕ + αP ϕ Φ( A ) N ϕ ). Consequently, N AP + αP AN = 0 ⇐⇒ N ϕ Φ( A ) P ϕ + αP ϕ Φ( A ) N ϕ , so Lemma 2.16 gives Φ( A ) = γ ( A ) A ϕ + µ ( A ) Id.Assume lastly Φ( X ) ≡ ( X ϕ ) tr for all X of rank-one. Pick rank-one N, P ,with P = P and P N = 0 =
N P . Also, pick any A ∈ D and let B := Φ( A ).We deduce, similarly as before, that0 = N AP + αP AN = τ p ( N, P, . . . , P, A j ′ , P, . . . , P ) ⇐⇒⇐⇒ τ p (Φ( N ) , Φ( P ) , . . . , Φ( P ) , Φ( A ) j ′ , Φ( P ) , . . . , Φ( P ))= τ p (cid:0) ( N ϕ ) tr , ( P ϕ ) tr , . . . , ( P ϕ ) tr , B, ( P ϕ ) tr , . . . , ( P ϕ ) tr (cid:1) = ( N ϕ ) tr B ( P ϕ ) tr + α ( P ϕ ) tr B ( N ϕ ) tr (11)Choose ( N, A, P ) := ( E , E , E ). Then, P N = 0 =
N P and B := Φ( A ) =( E ϕ ) tr = E . On one hand, this gives N AP + αP AN = α E , and, on theother, ( N ϕ ) tr B ( P ϕ ) tr + α ( P ϕ ) tr B ( N ϕ ) tr = E . Consequently, the equiva-lence (11) reads 0 = αE ⇐⇒ E , and so α = 0.We may now rewrite equivalence (11) into N AP + αP AN = 0 ⇐⇒ (cid:0) N ϕ B tr P ϕ (cid:1) tr + α (cid:0) P ϕ B tr N ϕ (cid:1) tr . The right side is further equivalent to 0 = N ϕ B tr P ϕ + α P ϕ B tr N ϕ . Hence,Lemma 2.16 gives Φ( A ) tr = B tr = γ ( A ) A ϕ + µ ( A ) Id. Proof of Corollary 2.4.
Assumption char F = 2 ensures p ( x, y ) := xy + yx is a polynomial with nonvanishing sum of coefficients. Since both map-pings X X ϕ and X X tr preserve the zeros of p ( x, y ), we may re-place Φ by a mapping X (cid:0) γ ( X ) T − Φ( X ) T (cid:1) ϕ − , respectively, by X (cid:0)(cid:0) γ ( X ) T − Φ( X ) T (cid:1) tr (cid:1) ϕ − to achieve that Φ leaves fixed all rank-one matri-ces. It remains to see that Φ( A ) = γ ( A ) A holds also for diagonalizablematrices A with the spectrum { λ, − λ } . In view of Corollary 2.9 we mayassume further A = 0. 20sing a similarity S , we may write A = S (cid:0) λ Id n ⊕ ( − λ ) Id n (cid:1) S − . It iseasy to see that Ω A := { X ∈ M n ( F ); XA + AX = 0 } = S [ KL ] S − where K, L are arbitrary matrices of an appropriate size. Now, since Φ fixes rank-one matrices, we have, by the defining equation (1), X Φ( A ) + Φ( A ) X = 0at least for each rank-one X ∈ S [ KL ] S − . Obviously, the set Ω Φ( A ) of allmatrices X with X Φ( A ) + Φ( A ) X = 0 is a vector subspace of M n ( F ), soactually Ω Φ( A ) ⊇ S [ KL ] S − . We now write Φ( A ) = S [ U VW Z ] S − , and solve the identity[ U VW Z ] [ KL ] + [ KL ] [ U VW Z ] ≡ ∀ K, ∀ L ) . Straightforward calculations give V = 0 = W , and V = µ Id n , Z = − µ Id n .Thus, Φ( A ) = S diag ( µ, − µ ) S − = µλ · A . Proof of Corollary 2.5.
We first prove that Φ maps rank-one idempotentsinto scalar multiples of rank-one idempotents, and preserves their orthogo-nality.Indeed, let P , . . . , P n be the set of n pairwise orthogonal rank-one idem-potents. Clearly then, p ( P i , P j , . . . , P j ) = 0 = p ( P j , P i , P j , . . . , P j ) = · · · = p ( P j , . . . , P j , P i ) for all i = j . This implies a similar set of equations on theirΦ–images A i := Φ( P i ) and A j := Φ( P j ). We write them explicitly: X σ (1)=1 α σ A i A k − j + X σ (1)=2 α σ A j A i A k − j + · · · + X σ (1)= k α σ A k − j A i = 0 X σ (2)=1 α σ A i A k − j + X σ (2)=2 α σ A j A i A k − j + · · · + X σ (2)= k α σ A k − j A i = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . X σ ( k )=1 α σ A i A k − j + X σ ( k )=2 α σ A j A i A k − j + · · · + X σ ( k )= k α σ A k − j A i = 0These may be regarded as a system of k homogeneous linear equations in‘variables’ A sj A i A k − − sj . By the assumptions, the coefficient matrix is in-vertible, so the only solution is A sj A i A k − − sj = 0 for each s . In particular, A i A k − j = A k − j A i = 0. That is,Im( A k − j ) ⊆ Ker( A i ); ( i = j ) . (12)Moreover, p ( P i , . . . , P i ) = (1 + ξ ) P ki = (1 + ξ ) P i = 0 implies that 0 = p ( A i , . . . , A i ) = (1 + ξ ) A ki , for any i . 21e can now follow the arguments from [13, Lemma 2.2] of Chan, Li, andSze: Firstly, we claim that rk ( A i ) = 1 for any i . Suppose on a contrary that,say, rk ( A ) ≥
2. Then, dim Ker( A ) < n − A k − ) + . . . + Im( A k − n )) < n −
1. Hence, there exists j such thatIm( A k − j ) ⊆ X l =2 ,...,nl = j Im( A k − l ) . (13)Indeed, otherwise, by the induction, dim (cid:0) Im( A k − )+ . . . +Im( A k − n ) (cid:1) ≥ n − A j ). Thus A kj = 0 which contradicts A kj = p ( A j , . . . , A j ) =0 by the assumption (ii). Therefore, rk ( A i ) = 1 for any i .Since A ki = 0 there exist λ i ∈ F \{ } and an idempotent matrix Q i of rank-one such that A i = λ i Q i . Lastly, it follows from A i A j = λ k − j A i A k − j = 0 and A j A i = λ k − i A j A k − i = 0 that Q i are pairwise orthogonal.Thus, Φ maps orthogonal idempotents of rank-one into scalar multiples oforthogonal idempotents of rank-one. We now redefine Φ as in Remark 2.14.The rest — with the sole exception of Step 4 — follows directly the proof ofTheorem 2.1. Definition 3.1.
Let k ≥
2. A subset of nonidentical permutations Ξ ⊆ S k is called an admissible subset if the following two conditions are satisfied:(i) There exists t ∈ { , . . . , k } such that each σ ∈ Ξ fixes the first t − σ ( t ) = t ( note that t < k , otherwise, σ would have to beidentical permutation ).(ii) There exist integers w, u, v ∈ { , . . . , k } , u < v , such that σ ( w ) = v and σ ( w + 1) = u for each σ ∈ Ξ.Note that in particular, σ ( w + 1) < σ ( w ) for each σ ∈ Ξ. Example 3.2.
We give two examples of admissible sets which are the mostvisualizing on one hand and which show that there are many admissible setsamong the subsets of S k on the other hand. • Ξ := { σ } is admissible subset whenever σ is nonidentical.22 An admissible subset is also the subset of all permutations from S k thatswap 1 and 2 ( take t := 1 =: w , u := 1, v := 2 in Definition 3.1).It is not hard to see that the cardinality of an admissible set, with t = 1,i.e., which fixes no initial elements, is either ( k − k − − ( k − w and u, v .The main result of the present section is the following theorem. In con-trast to the Theorem 2.1 we do not assume that F is algebraically closed,and Φ is a strong preserver in this section. Theorem 3.3.
Let F be an arbitrary field with more than 2 elements, let n ≥ , k ≥ be integers, and let Ξ ⊂ S k be a fixed admissible subset ofpermutations. Suppose that two given homogeneous multilinear polynomials p ( x , . . . , x k ) := x · · · x k − X σ ∈ Ξ α σ x σ (1) · · · x σ ( k ) ; ( α σ ∈ F ) p ( x , . . . , x k ) := x · · · x k − X σ ∈ Ξ β σ x σ (1) · · · x σ ( k ) ; ( β σ ∈ F ) satisfy P σ ∈ Ξ α σ = 1 = P σ ∈ Ξ β σ . Then, any bijection Φ : M n ( F ) → M n ( F ) which maps the zeros of p to the zeros of p ( i.e., Φ( S p ) ⊆ S p ) preservescommutativity. Remark 3.4.
In particular if p = p the result of Theorem 3.3 also holds. Corollary 3.5.
Under the additional requirement
Φ(Id) = 0 , the conclusionof Theorem 3.3 is valid for F = Z also. The corollary below shows that the injectivity assumption on Φ can besubstituted by the requirement that Φ maps the zeros of p into the zeros of p strongly . Corollary 3.6.
In addition to the assumptions from Theorem 3.3, supposefurther that k ≥ , and that a ( possibly noninjective ) surjection Φ : M n ( F ) →M n ( F ) strongly maps the zeros of p to the zeros of p . Then, Φ( A ) = 0 implies A = 0 . Consequently, Φ preserves commutativity. The following remark provides the final forms of the transformations sat-isfying conditions of Theorem 3.3.
Remark 3.7.
A surjective and additive commutativity preservers are of theform Φ( A ) = γ T A ϕ T − + f ( A ) Id or Φ( A ) = γ T ( A ϕ ) tr T − + f ( A ) Id . Here, γ ∈ F , φ : F → F is a ring automorphism, and f : M n ( F ) → F an additivefunction; see Breˇsar [9], Petek [30], and Beidar and Lin [6]. e refer to the works by ˇSemrl [33] and Foˇsner [20] for a bijective,possibly non-additive mappings, that strongly preserve commutativity . Atleast on the subset of M n ( C ) , consisting of those matrices whose Jordanstructure has only the cells of dimension at most two, they are of the form Φ( A ) = T p A ( A ϕ ) T − or Φ( A ) = T p A (cid:0) ( A ϕ ) tr (cid:1) T − , where p A is a certainpolynomial that depends on A . Remark 3.8.
The converse of Theorem 3.3 does not hold. Namely, there aremany polynomials p = p =: p and commutativity preservers which do notpreserve the zeros of p . We refer the reader to the last section for examples. The proof will be given in a series of Lemmas.We first recall some known results about rational forms for matrices overan arbitrary field F . Definition 3.9. A companion matrix of a monic polynomial f ( x ) = x m + a m − x m − + . . . + a x + a x + a x + a , of degree m ≥
2, is the matrix C ( f ) = . . . − a . . . − a . . . − a . . . − a ... ... ... ... . . . ... ...0 0 0 0 . . . − a m − ∈ M m ( F ) . (14)If f ( x ) = x + a is of degree one we let C ( f ) := − a be the 1–by–1 matrix,i.e., a scalar.The following lemma is straightforward and well-known: Lemma 3.10.
The polynomial f is a characteristic polynomial of its com-panion matrix C ( f ) . Theorem 3.11. [22, p. 144], [24, Theorem 11.20]
Any matrix A ∈ M n ( F ) is similar over F to a matrix C ( A ) = L j C ( p e j ) ⊕ . . . ⊕ L j C ( p e kj k ) , wherethe p i are distinct irreducible factors of the characteristic polynomial χ A ( x ) = Q ≤ i ≤ k ≤ j ≤ ki p i ( x ) e ij . The matrix C ( A ) is determined uniquely, up to the order ofdiagonal blocks C ( g i ) . efinition 3.12. The matrix C ( A ) described in Theorem 3.11 is called a primary rational form of A .In all statements till the end of this section we assume that conditions ofTheorem 3.3 are satisfied. Lemma 3.13. If ξ is nonzero scalar then the primary rational form of Φ( ξ Id) does not contain non-zero nilpotent blocks.Proof.
Pick a similarity P ∈ GL n ( F ) such that P − Φ( ξ Id) P equals the pri-mary rational form C (Φ( ξ Id)) of Theorem 3.11. Now, if the claim is false, atleast one block of C (Φ( ξ Id)) is a nonzero nilpotent. For simplicity, assume itis the first ( i.e.: the most upper-left ) one. Therefore, it equals Eq. (14), withzeros on the last column. Then, with E := P E P − ,Φ( ξ Id) E = P E P − , and E Φ( ξ Id) = 0 . (15)By surjectivity, E = Φ( F ) and Id = Φ( J ) for some F, J ∈ M n ( F ).Pick an integer t ∈ { , . . . , k } from the Definition 3.1 of admissible se-quence. Note that t < σ ( t ) ≤ k for each σ ∈ Ξ. Consider now the followingmatrix k -tuple (cid:0) A := J, . . . , A t − := J, A t := ( ξ Id) , A t +1 := F, . . . , A k := F (cid:1) , which lies in S p , since each σ fixes the indices { , . . . , t − } , and P α σ = 1.By the assumptions, Φ( S p ) ⊆ S p , and we haveΦ( A ) · · · Φ( A k ) = X σ ∈ Ξ (cid:0) β σ Φ( A σ (1) ) · · · Φ( A σ ( k ) ) (cid:1) , i.e., Id t − Φ( ξ Id) E k − t = X σ ∈ Ξ ( β σ Id t − E g σ Φ( ξ Id) E k − t − g σ ) , (16)where g σ := σ ( t ) − t >
0. However, the matrix E is idempotent, so E k − t = E = E g σ , and it follows from (15) that the left hand side of the equality (16)is equal to Φ( ξ Id) E = P E P − , while the right hand side is equal to 0, acontradiction. Lemma 3.14.
Suppose F = Z . Then, there exists a nonzero scalar ξ suchthat Φ( ξ Id) ∈ GL n ( F ) . roof. Since Φ is injective and the cardinality | F \{ }| ≥
2, there existsat least one nonzero scalar ξ such that Φ( ξ Id) = 0. As in the proof ofLemma 3.13, let t be fixed by Definition 3.1 let g σ := σ ( t ) − t >
0, and let J := Φ − (Id) ∈ M n ( F ). Here, we consider the following matrix k -tuple: (cid:0) A := J, . . . , A t − := J, A t := X , A t +1 := ( ξ Id) , . . . , A k := ( ξ Id) (cid:1) . Again, this k -tuple is in S p for an arbitrary matrix X . By the assumptions,Φ( S p ) ⊆ S p , and we haveId t − Φ( X )Φ( ξ Id) k − t = X σ ∈ Ξ (cid:0) β σ Id t − Φ( ξ Id) g σ Φ( X )Φ( ξ Id) k − t − g σ (cid:1) , (17)Let us assume that Φ( ξ Id) is singular. Recall Φ( ξ Id) = 0, so by Lemma 3.13,the primary rational form, C (Φ( ξ Id)), contains at least one zero block andat least one non-zero block. For simplicity, assume the first one is zero, i.e., C (Φ( ξ Id)) = ⊕ C , where C = 0 is a sum of all, but the first, blocks.By Lemma 3.13, C (Φ( ξ Id)) k − t = ⊕ C k − t = 0. Consequently, the ma-trix C (Φ( ξ Id)) k − t has a nonzero row, i.e., there exists p , 1 ≤ p ≤ n suchthat E p C (Φ( ξ Id)) k − t = 0. However, note that C (Φ( ξ Id)) E p = 0, so also C (Φ( ξ Id)) g E p = 0 for all g ∈ N \{ } , in particular for each g := g σ . Now,consider P ∈ GL n ( F ) such that Φ( ξ Id) = P C (Φ( ξ Id)) P − , and choose amatrix X with Φ( X ) = P E p P − . For such X , the left hand side of (17) isnonzero while the right hand side is zero, a contradiction. Lemma 3.15. Φ preserves commutativity.Proof. Without loss of generality we assume that ξ = 1 in Lemma 3.14, i.e.,that Φ(Id) ∈ GL n ( F ) — otherwise, the bijection Φ( ξ · xy ) would be consideredinstead of Φ. By the definition of admissible sequence, there exists an integer w such that u ≡ σ ( w + 1) < σ ( w ) ≡ v ∀ σ ∈ Ξ, we fix the smallest such index w . Consider the following matrix k -tuple:( A := Id , . . . , A w − := Id , A w := X , A w +1 := Y , A w +2 := Id , . . . , A k := Id) . (18)If XY = Y X then the k -tuple (18) is in S p . Thus,Φ(Id) w − Φ( X )Φ( Y ) Φ(Id) k − w − == X σ ∈ Ξ β σ Φ(Id) u − Φ( Y ) Φ(Id) s Φ( X ) Φ(Id) k − v , (19)where s = v − u − ≥
0. Note that by the definition of admissible se-quence u and v are independent of σ and thus the right hand side is equal26o Φ(Id) u − Φ( Y ) Φ(Id) s Φ( X ) Φ(Id) k − v . Since Φ(Id) is invertible, Eq. (19)simplifies into:Φ( X )Φ( Y ) = Φ(Id) u − w Φ( Y ) Φ(Id) s Φ( X ) Φ(Id) w − v +1 (20)whenever XY = Y X . We first claim that Φ(Id) s is a scalar matrix. Assumeon the contrary, and consider two cases: Case 1 : The primary rational form C (Φ(Id) s ) = P − Φ(Id) s P contains ablock of dimension ≥
2. Again, for the sake of simplicity, we assume this isthe first block. Therefore, the (1 , C (Φ(Id) s ) is zero. We would thenlet X = Y be such that Φ( X ) = P E P − = Φ( Y ). This contradicts (20),since the left hand side would be P E P − , while the right would be zero. Case 2 : All blocks of C (Φ(Id) s ) = P − Φ(Id) s P are one-dimensional, i.e., C (Φ(Id) s ) = diag ( d , . . . , d n ) is diagonal. Since C (Φ(Id) s ) is non-scalar, atleast two diagonal entries differ. For the sake of simplicity, assume d = d .Note that by Lemma 3.14 the matrix Φ(Id) s is invertible, so d = 0. Then,the similarity by the matrix S := h d d
11 1 i ⊕ Id n − transforms this matrix into S C (Φ(Id) s ) S − = (cid:20) d − d d + d (cid:21) ⊕ diag ( d , . . . , d n )with the zero (1 , s = λ Id = 0, andEq. (20) further simplifies into:Φ( X )Φ( Y ) = λ Φ(Id) u − w Φ( Y )Φ( X ) Φ(Id) w − v +1 (21)whenever XY = Y X . Pick any distinct indices i, j ∈ { , . . . , n } . Since n ≥ k ∈ { , . . . , n }\{ i, j } . Now, by the surjectivity, wemay choose X = Y such that Φ( X ) = Φ( Y ) = E ik + E kj , which givesΦ( X )Φ( Y ) = ( E ik + E kj ) = E ij = Φ( Y )Φ( X ). We can likewise find X = Y such that Φ( X )Φ( Y ) = E ii = E ii = Φ( Y )Φ( X ). Putting this into Eq. (21)we deduce that E ij = λ Φ(Id) u − v E ij Φ(Id) w − v +1 for any indices ( i, j ). Conse-quently, A = λ Φ(Id) u − v A Φ(Id) w − v +1 ; ∀ A ∈ M n ( F ) . With A := Id we have Φ(Id) u − v · Φ(Id) w − v +1 = 1 /λ Id. Therefore, AS =( λS ) A for each A , where S := Φ(Id) u − v . A standard procedure with A := E ij gives that S is scalar, and λ = 1. Hence, Φ(Id) w − v +1 = S − is also ascalar matrix. Since λ = 1, this further reduces Eq. (21) into the desiredΦ( X )Φ( Y ) = Φ( Y )Φ( X ) whenever XY = Y X .27 .2 Proof of Corollaries
Remark 3.16.
As it was seen in the proof of Theorem 3.3, we used therequirement of injectivity just once: in the proof of Lemma 3.14. Moreover,even there it suffices to have its curtailed form, i.e., that there exists ξ ∈ F , ξ = 0 , such that Φ( ξ Id) = 0 . Thus the result of Theorem 3.3 holds underthese, even more general, conditions.Proof of Corollary 3.5. By the assumptions, Φ(Id) = 0, so the proof ofLemma 3.14 works by Remark 3.16. It can be directly checked that therest of the proof of Theorem 3.3 does not use the assumptions on the groundfield. Proof of Corollary 3.6.
Suppose Φ( A ) = 0 for some matrix A , and considerthe following matrix k -tuple (cid:0) X = Id , . . . , X w − = Id , X w = A , X w +1 = X , X w +2 = Id , . . . , X k = Id (cid:1) . This k -tuple is mapped by Φ into a k -tuple with w -th member equal toΦ( X w ) = Φ( A ) = 0. Therefore, (Φ( X ) , . . . , Φ( X k )) ∈ S p , so that also( X , . . . , X k ) = (Id , . . . , Id , X w = A, X w +1 = X, Id , . . . , Id) ∈ S p , for everychoice of X . Since σ ( w + 1) < σ ( w ) for every admissible permutation, thisfurther yields Id w − AX Id k − w − = X σ α σ Id i X Id j A Id l for some integers i, j, l . This immediately simplifies into AX = P σ α σ XA = XA for every X . Hence, A has to be scalar.It remains to show A = 0. Consider now the following k -tuple: X w := Y, X w +1 := Z , and X i := A for the rest of indices ( since k ≥
3, at least onemember equals A ). As before we deduce that this k -tuples is in S p for everychoice of Y, Z . Hence, since A is scalar, and σ ( w + 1) < σ ( w ), A k − Y Z = A k − X σ ∈ Ξ α σ ZY = A k − ZY for any choice of Y, Z . This is possible only when A k − = 0, i.e., A = 0.Therefore, nonzero scalar matrices are not annihilated by Φ, and we canfollow the proof of Theorem 3.3 to see that Φ preserves commutativity.28 Concluding remarks and examples
This section mainly contains counterexamples to show that our results cannotbe further improved without imposing additional hypothesis.
The following example shows that the inverse implication does not hold inTheorem 3.3, namely there are commutativity preserving mappings that donot preserve zeros of a fixed polynomial.
Example 4.1.
Let Φ( A ) = A + a Id for all A = [ a ij ] ∈ M n ( F ). We considerthe polynomial p ( x, y, z ) := xyz − yxz . Then the triple ( x := E , y = z := E ) is a zero of this polynomial, but its image ( E , Id + E , Id + E ) is nota zero of this polynomial. There exist ( even linear !) transformation Φ which strongly preserve com-mutativity, but have a nonzero kernel. Example 4.2.
Let us consider Φ : M n ( F ) → M n ( F ) defined by Φ( A ) := A − Tr( A ) /n Id. Then Φ is linear, Φ strongly preserves commutativity, butΦ(Id) = 0.
Corollary 2.3 is no longer true when k = 2 , i.e., the transformation Φ may not be controllable on some large subsets of D . Example 4.3.
Namely, consider p ( x, y ) := xy + yx . Pick any A ∈ Θ := { A ∈M n ( F ); 0 Sp( A ) + Sp( A ) } . Then, by Sylvester–Rosenblum Theorem( cf. the proof of Lemma 2.6 ), the elementary operator X XA + AX isinvertible, so p ( X, A ) = 0 if and only if X = 0. Hence, the restriction of Φto the subset Θ has no structure at all, i.e., Φ | Θ may arbitrarily permutethe elements of Θ, yet it still strongly preserves the zeros of p ( x, y ). Weremark that Θ is a rather large subset: when F = C it is nonempty and openin M n ( C ), by continuity of the eigenvalues. The characterization of Lemma 2.12 is no longer valid if F is not alge-braically closed. Example 4.4.
As a counterexample, choose a polynomial p ( x, y ) := xy + yx and consider A := [ −
31 0 ] ⊕ ∈ M ( R ). Then, Ω • A = Ω A • A = { (cid:2) − d cc d (cid:3) ⊕ c, d ∈ R } , and it contains no nontrivial square-zero matrices, nor itequals R P for idempotent P . The following example shows that there are field automorphism which donot preserve zeros of matrix polynomials. xample 4.5. Let us consider the polynomial p = xy − i yx and automor-phism ϕ : C → C which sends any x ∈ C to its complex conjugated element x . Then consider A := (cid:20) − (cid:21) ⊕ Id n − ∈ M n ( C ) , B := (cid:20) ii − (cid:21) ⊕ Id n − ∈ M n ( C ) . The direct computations show that the matrices
A, B are zeros of p but theirconjugated matrices A ϕ , B ϕ are not. Finally, the transposition transformation may not preserve zeros of p forsome p . Example 4.6.
Let us consider the polynomial p = xy and the matrices A = E , B = E . Then p ( A, B ) = 0, however p ( A tr , B tr ) = E E = E = 0. Acknowledgments . The authors are indebted to Professor Chi-KwongLi for inspiring conversations regarding the topic of Chapter 2.
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