Maps on idempotent matrices over division rings

Transcription

1 Maps on idempotent matrices over division rings Peter Šemrl Department of Mathematics University of Ljubljana Jadranska 19 SI-1000 Ljubljana Slovenia July 27, Math. Subj. Class. 06A99, 15A30. Abstract Let D be an arbitrary division ring and P n(d) the set of all n n idempotent matrices over D. Under some mild conditions we give a complete description of maps on P n(d) that preserve either commutativity, or order, or orthogonality. We give examples showing that our assumptions cannot be relaxed much further. As an application we will prove a quaternionic analogue of Ovchinnikov s result that is important in quantum mechanics. Other applications of our theorems include results on automorphisms of operator and matrix semigroups, local automorphisms, linear preserver problems and geometry of matrices and Grassmanians. 1 Introduction Let D be a division ring and n a positive integer. We denote by M n (D) the set of all n n matrices over D. The symbol E ij, 1 i, j n, will be used for a matrix having all entries zero except the (i, j)-entry which is equal to 1. By P n (D) we denote the set of all n n idempotent matrices, P n (D) = {P M n (D) : P 2 = P }. Let 1 k n. Then Pn k (D) and Pn k (D) denote the set of all n n idempotent matrices of rank k, and the set of all n n idempotents of rank at most k, respectively. Supported in part by a grant from the Ministry of Science of Slovenia 1

2 We will call a division ring D an EAS-division ring if every nonzero endomorphism τ : D D is automatically surjective. The field of real numbers and the field of rational numbers are well-known to be EAS-fields while this is not true for the complex field. It turns out that the division ring of quaternions is an EAS-division ring as well. It is well-known that the set of all n n idempotent matrices is a poset (partially ordered set) with P Q if P Q = QP = P. Recall that an automorphism of the poset is a bijective map preserving order in both directions. Automorphisms of the poset of real or complex n n idempotent matrices were characterized by Ovchinnikov [18] who treated also the infinite-dimensional case because of applications in physics (see the review MR 95a:46093). In the finitedimensional case we can get Ovchinnikov s result under the weaker assumption of preserving order in one direction only [21]. A substantial improvement for idempotent matrices over EAS-fields was obtained in [23] where all injective maps on the poset of n n idempotents preserving order in one direction only were characterized. The proof was based on an improved version of the fundamental theorem of affine geometry for affine spaces over such fields. Using a different approach we will extend this result to idempotent matrices over any EAS-division ring. Our aim when developing this new approach was to prove such a characterization for injective order preserving maps on idempotent matrices over an arbitrary division ring. Much to our surprise our method gives as a byproduct a counterexample showing that this characterization is not valid even in the complex case. It should be mentioned here that Ovchinnikov s result recently proved to be useful also in the study of quantum mechanical invariance transformations. Molnár [17] used it to considerably improve the classical Wigner s unitaryantiunitary theorem. Motivated by applications in quantum mechanics he asked whether a quaternioinic analogue of Ovchinnikov s result holds true. The positive answer to this question will be one of the simple consequences of our finite-dimensional improvement of Ovchinnikov s theorem. The study of symmetry transformations in quantum mechanics is closely related to the problem of characterizing maps on idempotents or projections preserving orthogonality [16, 17, 24]. In [26] it was shown that in the infinitedimensional case the problem of characterizing automorphisms of the poset of idempotents is equivalent to the problem of characterizing orthogonality preserving bijective maps on idempotent operators. The solutions of both problems follow from a more general result on bijective maps on idempotents preserving commutativity in both directions. In this paper we will study besides order preserving maps also related structural problems for orthogonality or commutativity preserving maps on idempotent matrices. We will characterize such maps under mild assumptions (much weaker than those needed for the corresponding results in the infinite-dimensional case) and give counterexamples showing that our assumptions cannot be relaxed much further. 2

3 To be more precise we recall that a map φ : P n (D) P n (D) preserves order if φ(p ) φ(q) whenever P Q, P, Q P n (D). A map φ : P n (D) P n (D) preserves orthogonality if φ(p ) φ(q) whenever P Q, P, Q P n (D). Here, P Q if and only if P Q = QP = 0. And finally, a map φ : P n (D) P n (D) preserves commutativity if φ(p )φ(q) = φ(q)φ(p ) whenever P Q = QP, P, Q P n (D). A map φ : P n (D) P n (D) preserves order in both directions if for every pair P, Q P n (D) we have φ(p ) φ(q) if and only if P Q. In a similar way we define maps preserving orthogonality (commutativity) in both directions. If A = [a ij ] M n (D) is any matrix and σ : D D an automorphism or an anti-automorphism (bijective additive map satisfying σ(λµ) = σ(µ)σ(λ), λ, µ D) of division ring D, then we denote by A σ the matrix obtained from A by applying σ entrywise, A σ = [a ij ] σ = [σ(a ij )]. Clearly, for every invertible matrix T M n (D) and every automorphism σ of D the map φ : P n (D) P n (D) defined by φ(p ) = T P σ T 1, P P n (D), (1) is a bijective map preserving order, orthogonality, and commutativity in both directions. Similarly, for every invertible matrix T M n (D) and every antiautomorphism τ of D the map φ : P n (D) P n (D) defined by φ(p ) = T t (P τ )T 1, P P n (D), (2) is a bijective map preserving order, orthogonality, and commutativity in both directions. Here, t A denotes the transpose of a matrix A. Every map of the form (1) or (2) will be called a standard map on P n (D). If σ and τ in (1) and (2) are assumed to be a nonzero (not necessarily bijective) endomomorphism and anti-endomomorphism of D, respectively, then the map φ is an injective map preserving order, orthogonality, and commutativity in both directions. We will call such maps almost standard maps on P n (D). The problem here is, of course, whether the converse holds true. That is, if we assume that φ : P n (D) P n (D) is a bijective (injective) map preserving order in both directions, does it follow that φ is an (almost) standard map? And the same question can be posed with the assumption of preserving order in both directions replaced by either the assumption of preserving commutativity in both directions, or the assumption of preserving orthogonality in both directions. Further, we can ask if we can get such a conclusion under a weaker assumption of preserving any of these relations in one direction only. Do we really need the assumption of bijectivity or even injectivity? Let us mention here that in the case of commutativity preserving maps we must slightly modify our questions if we want to have affirmative answers. Namely, assume that φ : P n (D) P n (D) is any map which sends every idempotent either into itself, or into its orthocomplement, that is, φ(p ) {P, I P } for every P P n (D). Clearly, every such map preserves commutativity in both 3

4 directions. Every such map will be called an orthopermutation. Obviously, an orthopermutation is bijective if and only if for every P P n (D) we have either or φ(p ) = P and φ(i P ) = I P, φ(p ) = I P and φ(i P ) = P. So, in the case of commutativity preserving maps we ask under which conditions such maps must be almost standard maps composed with an orthopermutation. The purpose of this paper is to give best possible answers to these questions. For each of the above questions we will either show that it has the affirmative answer, or give a counterexample showing that the answer is negative. When the answer is negative one can further ask what are all degenerate maps (maps that are not of the expected standard form) satisfying given assumptions. In all such cases we will either answer this question, or give examples showing that the structure of such degenerate maps cannot be described in a nice way. We have already mentioned that results concerning order or orthogonality preserving maps on idempotents are important because of applications in physics. Other applications include results on automorphisms of operator semigroups (see [23, Section 8] and [24, Corollary 1.3]) and linear and non-linear preservers (see [23, Section 8]). The main motivation for our study of maps on idempotents comes from some open problems concerning geometry of matrices. The study of geometry of matrices was initiated by Hua in the forties [5] - [12]. This topic is closely related to the geometry of Grassmann spaces studied also by Chow and Dieudonné [1, 2, 3]. An interested reader can find more information on this research area in the recent book by Wan [27]. For some recent results on geometry of matrices published after the appearance of this book we refer to [13], [19], [20], and [23]. In [23] we showed that the fundamental theorem of geometry of square matrices can be deduced from results of the Ovchinnikov s type. Applications of our results in this direction will be presented in our forthcoming paper. The proofs are rather long and will be therefore divided into several steps. Another reason to divide them into several steps is that some of them will be useful in our forthcoming paper considering geometry of matrices. These auxilary results will be presented in the next section where we will also introduce the notation. Already in this section we will give some examples showing that these statements cannot be further improved. More examples showing the optimality of our main results will be given in the third section. The next section will be devoted to the formulation of our main results and the proofs will be given in the fifth section. The last section will be devoted to the proof of the quaternionic analogue of Ovchinnikov s theorem. 4

5 2 Notation and preliminary results Let us recall first the definition of the rank of an n n matrix A with entries in a division ring D. We will denote by D n the set of all 1 n matrices and consider it always as a left vector space over D. Correspondingly, we have the right vector space of all n 1 matrices t D n. We first take the left vector subspace of D n generated by the rows of A (the row space of A) and define the row rank of A to be the dimension of this subspace. The column rank of A is the dimension of the right vector space generated by the columns of A. This space is called the column space of A. These two ranks are equal for every matrix over D and this common value is called the rank of a matrix. If rank A = r then there exist invertible matrices T, S M n (D) such that [ ] Ir 0 T AS =. (3) 0 0 Here, I r is the r r identity matrix and the zeroes stand for zero matrices of the appropriate size. Rank satisfies the triangle inequality, that is, rank (A + B) rank A + rank B for every pair A, B M n (D) [14, p.46, Exercise 2]. Note that in general rank A need not be equal to rank t A. However, if τ : D D is a nonzero anti-endomorphism of D, then rank A = rank t (A τ ). Let a D n and t b t D n be arbitrary nonzero vectors. Then t ba = ( t b)a is a matrix of rank one and every matrix of rank one can be written in this form. We will frequently use the obvious fact that a rank one matrix t ba is an idempotent if and only if a t b = 1. The elements of the standard basis of the left vector space D n and the elements of the standard basis of the right vector space t D n will be denoted by e 1,..., e n and t e 1,..., t e n, respectively. Thus, E ij = t e i e j, 1 i, j n. Assume that A, B M n (D). Since the multiplication in D is not necessarily commutative we do not have t (AB) = t B t A in general. But if τ is an antiendomorphism of D then t [(AB) τ ] = t (B τ ) t (A τ ). As usual we will identify n n matrices with linear operators acting on D n. Namely, each n n matrix A gives rise to a linear operator defined by x xa, x D n. Then the rank of the matrix A is the dimension of the image Im A of the corresponding operator A. The kernel of an operator A, Ker A = {x D n : xa = 0}, is the set of all vectors x D n satisfying x( t y) = 0 for every t y from the column space of A. Note that n = rank A + dim Ker A. In the sequel we shall need the following fact that is well-known for idempotent matrices over fields and can be also generalized to idempotent matrices over division rings [14, p.62, Exercise 1]. Assume that P 1,..., P k P n (D) are pairwise orthogonal. Denote by r i the rank of P i. Then there exists an invertible matrix A M n (D) such that for each i, 1 i k, we have AP i A 1 = diag (0,..., 0, 1,..., 1, 0,..., 0) 5

6 where diag (0,..., 0, 1,..., 1, 0,..., 0) is the diagonal matrix in which all the diagonal entries are zero except those in (r r i 1 +1)st to (r r i )th rows. Let P, Q P n (D). If P Q then clearly, Q P is an idempotent orthogonal to P. Thus, by the previous paragraph, we have P Q, P 0, Q I, and P Q if and only if there exist an invertible A M n (D) and positive integers r 1, r 2 such that AP A 1 = I r and AQA 1 = I r I r and 0 < r 1 < r 1 + r 2 < n. In particular, if we identify matrices with linear operators, then the image of P is a subspace of the image of Q, while the kernel of Q is a subspace of the kernel of P. We have seen that the relations and are closely related. Even more is true. The partial order can be easily characterized with the orthogonality relation. Indeed, for an arbitrary subset S P n (D) we define S to be the set of all R P n (D) satisfying R P for every P S. If S = {P } then we write shortly {P } = P. It is not difficult to see that for every pair P, Q P n (D) we have P Q if and only if Q P. This suggests that the study of order preserving maps on P n (D) is in close connection with the study of orthogonality preserving maps on P n (D). If P Q or P Q, then clearly, P and Q commute. So, we will consider also a related problem of characterizing commutativity preserving maps. Let us prove now some technical lemmas that will be used in the proofs of our main results. In order to formulate the first one we need one more definition. For two rank one idempotents t xy, t uv Pn(D) 1 we write t xy t uv if and only if y and v are linearly dependent in D n or t x and t u are linearly dependent in t D n. In other words, for two rank one idempotent operators P and Q we have P Q if and only if they have the same image or the same kernel. So, if P Q and P Q, then there exists an invertible T M n (D) such that either T P T 1 = E 11 and T QT 1 = E 11 + E 12, or T P T 1 = E 11 and T QT 1 = E 11 +E 21. Indeed, after applying a similarity transformation we may assume that P = E 11 = t e 1 e 1. Because P Q we have either Q = t e 1 x for some x D n, or Q = t ue 1 for some t u t D n. We will consider only the first case. Then, since P Q and x t e 1 = 1 we have 1 q q 1n Q = , where [ q q 1n ] 6

7 is a nonzero vector in D n 1. M n 1 (D) such that It follows that there exists an invertible S [ q 12 q q 1n ] S = [ ]. Hence, [ ] [ ] S 1 E 11 = E 0 S 11 and [ ] [ ] S 1 Q = E 0 S 11 + E 12, as desired. Our first auxilary result is an improvement of [18, Lemma 3.2]. Lemma 2.1 Let D be a division ring, n an integer 3, and P, Q P 1 n(d). Then the following are equivalent: P Q, the set {R P 2 n(d) : P R and Q R} has at most one element. Moreover, if E is a division ring, E D, P, Q P 1 n(e), P Q, and if R P 2 n(d) is a rank two idempotent satisfying P R and Q R, then all the entries of R belong to E, that is, R P 2 n(e). Proof. Assume first that P Q and that R is a rank two idempotent satisfying P R and Q R. Then the two-dimensional image of R contains the image of P as well as the image of Q. These are linearly independent onedimensional subspaces and consequently, the image of R is the direct sum of the images of P and Q. Similarly, the kernel of R, which is of codimension two in D n, has to be the intersection of the kernels of P and Q, which are both of codimension one. If R 1 is another idempotent of rank two satisfying P R 1 and Q R 1, then again, the image of R 1 must be the direct sum of the images of P and Q and the kernel of R 1 has to be the intersection of the kernels of P and Q. Thus, R = R 1, as desired. To prove the other direction, assume that P Q. Then, after applying an appropriate similarity, we may assume that either P = E 11 and Q = E 11 + E 12, or P = E 11 and Q = E 11 + E 21. Let us consider just the first case. Observing that P R λ and Q R λ for every λ D, where R λ = E 11 + E 22 + λe 32, we complete the proof of the first part of our statement. To prove the second part, we assume that E is a division ring, E D, P = t xy, Q = t uv Pn(E), 1 P Q, and R Pn(D) 2 is a rank two idempotent satisfying P R and Q R. Here, we may assume that y, v E n and t x, t u t E n. Then we know that the vectors y and v form a basis of the image of R. Moreover, we can find linearly independent vectors z 3,..., z n E n such that z t j x = z t j u = 0, j = 3,..., n. These vectors form a 7

8 basis of the kernel of R, and because D n is the direct sum of the image and the kernel of R, the n-tuple of vectors y, v, z 3,..., z n is a basis of D n. The operator R maps y and v into themselves and the vectors z 3,..., z n into the zero vector. It follows that the matrix R is similar to E 11 + E 22. The similarity is induced by a matrix B whose rows are y, v, z 3,..., z n. Thus, all the entries of both B and B 1 belong to E, and then the same must be true for the matrix R. Remark. We have proved that if P and Q are two rank one idempotents such that P Q then either there is exactly one rank two idempotent R satisfying P R and Q R, or there is no such R. Let us just show that both possibilities can occur. Indeed, if P = E 11 and Q = E 22 then R = E 11 + E 22 is an idempotent of rank two such that both P and Q are below R. The possibility that the set {R Pn(D) 2 : P R and Q R} is empty can occur already in the simplest case when n = 3 and D is the field of real or complex numbers. If we set P = and Q = then a straightforward computation shows that the only idempotent R satisfying P R and Q R is the 3 3 identity matrix. In the following few lemmas we will always assume that D is a division ring, n an integer 3, and φ : P n (D) P n (D) an injective order preserving map. Lemma 2.2 For every P P n (D) we have rank φ(p ) = rank P. In particular, φ(0) = 0 and φ(i) = I. Proof. Let rank P = r. Then we can find idempotents P 0, P 1,..., P n with P 0 = 0 P 1... P n 1 P n = I, rank P k = k, k = 0, 1,..., n, and P = P r. From φ(0) φ(p 1 )... φ(p n ) and injectivity of φ we conclude that rank φ(p ) = r, as desired. For a nonzero x D n and a nonzero t y t D n we denote by P R(x) and P L( t y) the subsets of P 1 n(d) defined by P R(x) = { t ux : t u t D n, x t u = 1} and P L( t y) = { t yv : v D n, v t y = 1}. Clearly, if t ux, t wx P R(x) for some t u, t w t D n, then either t u = t w, or t u and t w are linearly independent. Moreover, if nonzero vectors x 1 and x 2 are linearly dependent then P R(x 1 ) = P R(x 2 ). 8

9 Lemma 2.3 For every nonzero t y t D n either there exists a nonzero x D n such that φ(p L( t y)) P R(x), or there exists a nonzero t w t D n such that φ(p L( t y)) P L( t w). Proof. Take u v D n such that u t y = v t y = 1. Then, by Lemma 2.1, the set {R P 2 n(d) : t yu R and t yv R} has at least two elements. By injectivity, the set of rank two idempotents majorizing φ( t yu) and φ( t yv) has more than one element. Thus, applying Lemma 2.1 again we have either φ( t yu) = t ax and φ( t yv) = t bx for some nonzero x D n and linearly independent t a, t b t D n with x t a = x t b = 1, or φ( t yu) = t wc and φ( t yv) = t wd for some nonzero t w t D n and linearly independent c, d D n with c t w = d t w = 1. We will consider only the first case. Let z D n \ {u, v} be any vector such that z t y = 1. Then, by Lemma 2.1, φ( t yz) φ( t yu) and φ( t yz) φ( t yv). From the first relation we get φ( t yz) = t ae or φ( t yz) = t fx for some e D n or some t f t D n, while the second relation yields that φ( t yz) = t bg or φ( t yz) = t hx for some g D n or some t h t D n. As t a and t b are linearly independent, we have necessarily φ( t yz) P R(x). Hence, φ(p L( t y)) P R(x), as desired. Lemma 2.4 Assume that there exists a nonzero t y t D n such that φ(p L( t y)) P R(x) for some nonzero x D n. Then for every nonzero t w t D n we have φ(p L( t w)) P R(u) for some u D n. Proof. Assume on the contrary that there exists a nonzero t w t D n such that φ(p L( t w)) P L( t z) for some nonzero t z t D n. Because the intersection P R(x) P L( t z) contains at most one element, the vectors t w and t y must be linearly independent. Since n 3 we can find linearly independent vectors a, b D n satisfying a t y = b t y = a t w = b t w = 1. Denote P 1 = t ya, P 2 = t yb, Q 1 = t wa, and Q 2 = t wb. assumptions we have φ(p 1 ) = t cx and φ(p 2 ) = t dx According to our 9

10 for some linearly independent t c, t d t D n with x t c = x t d = 1, and φ(q 1 ) = t ze and φ(q 2 ) = t zf for some linearly independent e, f D n with e t z = f t z = 1. We further know that φ(p 1 ) φ(q 1 ) and φ(p 2 ) φ(q 2 ). The injectivity assumption implies that φ(p 1 ) φ(q 1 ), and consequently, t c and t z are linearly independent, or x and e are linearly independent. We have to show that both possibilities yield a contradiction. We will consider only the first case as the second case can be treated in almost the same way. Because t c and t z are linearly independent we get from φ(p 1 ) φ(q 1 ) that x and e are linearly dependent. Replacing t z by t zλ, and e and f by λ 1 e and λ 1 f, respectively, we may assume that φ(q 1 ) = t zx. Because x and f are linearly independent, the relation φ(p 2 ) φ(q 2 ) yields that t d and t z are linearly dependent. But then φ(p 2 ) = t zµx for some µ D with x t zµ = 1. As x t z = 1 we have µ = 1, and thus φ(p 2 ) = φ(q 1 ), a contradiction. This completes the proof. Lemma 2.5 Assume that there exists a nonzero t y t D n such that φ(p L( t y)) P R(x) for some nonzero x D n. Then either for every nonzero z D n there exists a nonzero t w t D n such that φ(p R(z)) P L( t w), or φ(pn(d)) 1 P R(x). Proof. Using exactly the same approach as in the proof of Lemma 2.4 we show that either for every nonzero z D n there exists a nonzero t w t D n such that φ(p R(z)) P L( t w), or for every nonzero z D n there exists a nonzero u D n such that φ(p R(z)) P R(u). All we have to do is to show that the second possibility implies that φ(p 1 n(d)) P R(x). In order to get this inclusion we have to show φ(p R(z)) P R(x) for every nonzero z D n. If z t y 0, then t y(z t y) 1 z P L( t y), and consequently, φ( t y(z t y) 1 z) = t ax for some t a t D n with x t a = 1. We know that φ(p R(z)) P R(u) for some nonzero u D n. Thus, φ(p R(z)) P R(x) for every z D n satisfying z t y 0. If z t y = 0, then we can find t y 1 t D n and z 1 D n such that z t y 1 0, z 1 t y 0, and z 1 t y 1 = 1. We know that φ(p L( t y 1 )) P R(x 1 ) for some nonzero x 1 D n. Moreover, by the previous step we have φ(p R(z 1 )) P R(x). As t y 1 z 1 P L( t y 1 ) P R(z 1 ) we have necessarily φ(p L( t y 1 )) P R(x). Because z t y 1 0 the above argument shows that φ(p R(z)) P R(x) in this case as well. We have shown that φ(p R(z)) P R(x) for every nonzero z D n. This completes the proof. Lemma 2.6 Assume that there exists a nonzero t y t D n such that φ(p L( t y)) P R(x) for some nonzero x D n. Then either for every pair of linearly independent n-tuples t y 1,..., t y n t D n and z 1,..., z n D n there exist linearly 10

11 independent n-tuples x 1,..., x n D n and t w 1,..., t w n t D n such that φ(p L( t y i )) P R(x i ) and φ(p R(z i )) P L( t w i ), i = 1,..., n, or φ(p 1 n(d)) P R(x). Proof. We will prove this lemma by induction on n. We start with the case when n = 3. If φ(p 1 3 (D)) P R(x), we are done. Otherwise, we know by Lemma 2.4 and Lemma 2.5 that for every nonzero t y t D 3 and every nonzero z D 3 there exist nonzero x D 3 and t w t D 3 such that φ(p L( t y)) P R(x) and φ(p R(z)) P L( t w). We will show only that if t y 1, t y 2, t y 3 t D 3 are linearly independent and if φ(p L( t y i )) P R(x i ), i = 1, 2, 3, then x 1, x 2, x 3 are linearly independent. In the same way one can then prove that if z 1, z 2, z 3 D 3 are linearly independent and if φ(p R(z i )) P L( t w i ), i = 1, 2, 3, then t w 1, t w 2, and t w 3 are linearly independent. So, assume that t y 1, t y 2, t y 3 t D 3 are linearly independent and choose nonzero x 1, x 2, x 3 D 3 such that φ(p L( t y i )) P R(x i ), i = 1, 2, 3. Let T M 3 (D) be the invertible matrix satisfying T t e 1 = t y 1, T ( t e 1 + t e 2 ) = t y 2, and T ( t e 1 + t e 2 + t e 3 ) = t y 3. Then, after replacing φ by P φ(t P T 1 ), we may assume that φ(p L( t e 1 )) P R(x 1 ), φ(p L( t e 1 + t e 2 )) P R(x 2 ), and φ(p L( t e 1 + t e 2 + t e 3 )) P R(x 3 ). We know that φ(0) = 0, φ(e 11 ) = SE 11 S 1, φ(e 11 + E 22 ) = S(E 11 + E 22 )S 1, and φ(i) = I for some invertible matrix S M 3 (D). After composing φ with the similarity transformation P S 1 P S, we may also assume that φ(0) = 0, φ(e 11 ) = E 11, φ(e 11 + E 22 ) = E 11 + E 22, and φ(i) = I. Then, of course, we also have to replace the vectors x 1, x 2, x 3 by x 1 S, x 2 S, and x 3 S, respectively. Now, we have E 11 [ ] 1 0, 0 P ([ 1 0 where P is any 2 2 idempotent. It follows that E 11 = φ(e 11 ) φ 0 P P P 2 (D). Hence, there exists an injective order-preserving map ϕ : P 2 (D) P 2 (D) such that ([ ]) [ ] φ =, P P 0 P 0 ϕ(p ) 2 (D). ]), 11

12 Because of φ(e 11 ) = E 11 and φ(p L( t e 1 )) P R(x 1 ) for some nonzero x 1 D n we have φ(p L( t e 1 )) P R(e 1 ). Further, 1 λ E 11 + E 22, λ D, and thus, φ 1 λ = η(λ) 0 0, λ D, for some injective map η : D D with η(0) = 0. For every λ D we have 1 λ , and therefore Moreover, ϕ [ 1 ([ 0 ]) ] η(λ) ϕ, λ D ([ ]) [ ] 1 0 is an idempotent of rank one. Thus 0 0 φ = a for some nonzero a D. Applying the same idea as above once more we see that φ = 1 b for some b 0. So, φ(p L( t e 1 + t e 2 )) P R(e 1 + be 2 ). Because and φ(p R(e 1 )) P L( t e 1 ) we have φ = 1 α β a

13 for some α, β D. Hence, αa = β. If α = 0, then β 0 because of injectivity, a contradiction. Thus, α 0, and consequently, β 0. We conclude that φ(p L( t e 1 + t e 2 + t e 3 )) P R(e 1 +αe 2 +βe 3 ). Now, e 1, e 1 +be 2, and e 1 +αe 2 +βe 3 are linearly independent. This completes the proof in the case n = 3. We assume now that our statement holds true for n and we want to prove it for n + 1. As before we may assume that for every nonzero t y t D n+1 and every nonzero z D n+1 there exist nonzero x D n+1 and t w t D n+1 such that φ(p L( t y)) P R(x) and φ(p R(z)) P L( t w), φ(0) = 0, φ(e 11 ) = E 11, φ(e 11 +E 22 ) = E 11 +E 22,..., φ(i) = I, and t y 1 = t e 1, t y 2 = t e 1 + t e 2,..., t y n+1 = t e t e n+1. In the same way as above we see that there exists an injective order-preserving map ϕ : P n (D) P n (D) such that φ ([ P ]) = [ ϕ(p ) ], P P n (D). Also, φ(p L( t e 1 )) P R(e 1 ) and φ(p R(e 1 )) P L( t e 1 ). Because for any choice of entries denoted by, we obtain using the same argument as before that φ = , and consequently, ϕ = Similarly, ϕ = (4)

14 We can now apply the induction hypothesis on the map ϕ. By the last two equations the ϕ-image of the set of all rank one idempotents is not a subset of some P R(x). Denote S k = P n+1 (D) and P k = P n+1 (D), k = 1,..., n. Here, S k has exactly k nonzero entries in the second column and exactly the first k + 1 entries of the first column of P k are equal to 1. From S k E E k+1,k+1, (4), and the induction hypothesis we get that a k φ(s k ) = , where the entry a k in the (2, k + 1)-position is nonzero. Because φ(p R(e 1 )) P L( t e 1 ) and P k E E k+1,k+1 we have 1 w k... c k φ(p k ) = Q k = ,

15 where c k is in the (1, k + 1)-position. Moreover, Q k φ(s k ) which yields 1 w k... c k a k w k... c k = If w k = 0 then φ(p k ) = E 11 contradicting the injectivity of φ. Therefore w k a k = c k 0. Obviously, φ(p L( t y k+1 )) P R(x k+1 ) where x k+1 = e 1 + w k e c k e k+1. This completes the proof. One of the main tools in the proofs of our results will be the recently obtained non-surjective version of the fundamental theorem of projective geometry [4, Theorem 3.1]. In fact, besides [4, Theorem 3.1] we will need also the following modified version of this theorem concerning maps between projective spaces over left and right vector spaces over division rings. Let n be an integer, n 3. By P(D n ) and P( t D n ) we denote the projective spaces over left vector space D n and right vector space t D n, respectively, P(D n ) = {[x] : x D n \ {0}} and P( t D n ) = {[ t y] : t y t D n \{0}}. Here, [x] and [ t y] denote the one-dimensional left vector subspace of D n generated by x and the one-dimensional right vector subspace of t D n generated by t y, respectively. Proposition 2.7 Let D be any division ring and n an integer 3. Let ψ : P(D n ) P( t D n ) be a map such that ψ([x]) ψ([y]) + ψ([z]) whenever [x] [y] + [z], x, y, z D n \ {0}. Assume further that there exist x 1,..., x n D n \ {0} such that ψ([x 1 ]) = [ t y 1 ],..., ψ([x n ]) = [ t y n ] and t y 1,..., t y n are linearly independent in t D n. Then there exist an anti-endomorphism σ : D D and an invertible matrix T M n (D) such that ψ([x]) = [ T t (x σ ) ] for every x D n \ {0}. Proof. The proof is just a slight modification of the proof of [4, Theorem 3.1]. 15

16 Lemma 2.8 Let D be any division ring, n an integer 3, and φ : P 1 n(d) P 1 n(d) an injective map. Assume that for every pair of linearly independent n- tuples t y 1,..., t y n t D n and z 1,..., z n D n there exist linearly independent n-tuples x 1,..., x n D n and t w 1,..., t w n t D n such that φ(p L( t y i )) P R(x i ) and φ(p R(z i )) P L( t w i ), i = 1,..., n. Suppose also that for every nonzero t y, t y, t y t D n and x, x, x D n satisfying t y span { t y, t y }, φ(p L( t y )) P R(x ), φ(p L( t y )) P R(x ), and φ(p L( t y )) P R(x ) we have x span {x, x }, and similarly, for every nonzero z, z, z D n and t w, t w, t w t D n satisfying z span {z, z }, φ(p R(z )) P L( t w ), φ(p R(z )) P L( t w ), and φ(p R(z )) P L( t w ) we have t w span { t w, t w }. Then there exist nonsingular matrices T 1, T 2 M n (D) and nonzero anti-endomorphisms σ, τ : D D such that for every z D n and t y t D n with z t y = 1 we have y τ T 2 T 1 t z σ 0 and φ( t yz) = T 1 t z σ (y τ T 2 T 1 t z σ ) 1 y τ T 2. Proof. For every nonzero z D n there exists a nonzero t w t D n such that φ(p R(z)) P L( t w). The map ψ 1 : P(D n ) P( t D n ) given by ψ 1 ([z]) = [ t w], where z and t w are as above, is obviously well-defined. Our assumptions implies that ψ 1 satisfies all the assumptions of Proposition 2.7. Thus, there exists an anti-endomorphism σ : D D and an invertible matrix T 1 such that ψ 1 ([z]) = [T 1 t z σ ], z D n \ {0}. Similarly (we need an analogue of Proposition 2.7 for maps from P( t D n ) into P(D n )), there exists an anti-endomorphism τ : D D and an invertible matrix T 2 such that for every t y t D n \{0} and x D n \{0} with φ(p L( t y)) P R(x) we have [x] = [y τ T 2 ]. Now, if t yz is any idempotent of rank one, that is, z t y = 1, then φ( t yz) belongs to P R(y τ T 2 ) as well as to P L(T 1 t z σ ). It follows that φ( t yz) = T 1 t z σ α y τ T 2 for some α D. Since T 1 t z σ α y τ T 2 is an idempotent we have α y τ T 2 T 1 t z σ = 1. This clearly yields that y τ T 2 T 1 t z σ 0 and α = (y τ T 2 T 1 t z σ ) 1, as desired. We get a much nicer conclusion when D is an EAS-division ring. We start with a simple observation. Note that if D is an EAS-division ring, then every nonzero anti-endomorphism of D is automatically surjective as well. Indeed, let σ be an anti-endomorphism of D. Then σ 2 is an endomorphism which has to be surjective. The same is then true for σ. 16

17 Assume now that D is an EAS-division ring, T M n (D) an invertible matrix, and τ, σ anti-endomorphisms of D such that for every pair of vectors t y t D n and z D n satisfying z t y = 1 we have y τ T t z σ 0. Then, in particular 1 σ(z 2 ) [ ] T. σ(z n ) 0 for every z 2,..., z n D. Applying the fact that σ is surjective we easily conclude that all off diagonal entries in the first row of T must be zero. The same holds true for other rows, and so, T is a diagonal matrix with diagonal entries t 1,..., t n. Now, for every λ D we have 1 [ 1 τ(λ) τ(λ) ] T A straightforward computation then shows that t 1 = t 2. In the same way we prove that t 1 = t 2 =... = t n = t. So, if z 1,..., z n, y 1,..., y n D are arbitrary elements satisfying z 1 y z n y n = 1, then τ(y 1 )tσ(z 1 ) τ(y n )tσ(z n ) 0. Applying τ 1 on both sides of this equation and denoting τ 1 σ = η and τ 1 (t) = s we get η(z 1 )sy η(z n )sy n 0. Choosing z 1 = 1, z 2 = γ, z 3 =... = z n = 0, y 1 = 1 γδ, y 2 = δ, y 3 =... = y n = 0 we arrive at s (sγ η(γ)s)δ for every pair γ, δ D. Thus, η(γ) = sγs 1, γ D, or equivalently, σ(γ) = t 1 τ(γ)t, γ D. It follows that for every pair of vectors t y = t [ y 1... y n ] t D n and z = [ z 1... z n ] D n satisfying z t y = 1 we have y τ T t z σ = [ τ(y 1 )... τ(y n ) ] ti = τ(z 1 y z n y n )t = t. t 1 τ(z 1 )t. t 1 τ(z n )t Applying the above observation to the matrix T = T 2 T 1 and anti-endomorphisms σ, τ appearing in the conclusion of Lemma 2.8 we see that the map φ is of the form φ( t yz) = T 1 t z σ (y τ T 2 T 1 t z σ ) 1 y τ T 2 = 17

18 t 1 τ(z 1 )t = T2 1 ti. t 1 τ(z n )t t 1 [ τ(y 1 )... τ(y n ) ] T 2 τ(z 1 ) = T2 1. [ τ(y 1 )... τ(y n ) ] T 2 = T2 1 ( t yz) τ T 2 τ(z n ) for every idempotent t yz of rank one. Thus, we have proved the following. Lemma 2.9 Let D be any EAS-division ring, n an integer 3, and φ : P 1 n(d) P 1 n(d) an injective map. Assume that for every pair of linearly independent n- tuples t y 1,..., t y n t D n and z 1,..., z n D n there exist linearly independent n-tuples x 1,..., x n D n and t w 1,..., t w n t D n such that φ(p L( t y i )) P R(x i ) and φ(p R(z i )) P L( t w i ), i = 1,..., n. Suppose also that for every nonzero t y, t y, t y t D n and x, x, x D n satisfying t y span { t y, t y }, φ(p L( t y )) P R(x ), φ(p L( t y )) P R(x ), and φ(p L( t y )) P R(x ) we have x span {x, x }, and similarly, for every nonzero z, z, z D n and t w, t w, t w t D n satisfying z span {z, z }, φ(p R(z )) P L( t w ), φ(p R(z )) P L( t w ), and φ(p R(z )) P L( t w ) we have t w span { t w, t w }. Then there exist a nonsingular matrix T M n (D) and an anti-automorphism τ : D D such that φ(p ) = T t (P τ )T 1, P P 1 n(d). However, in the general case the conclusion of Lemma 2.8 cannot be much improved. In order to see this we consider the case when D = C. There exist an endomorphism σ of the complex field and complex numbers a ij C, 1 i, j n, that are algebraically independent over the subfield σ(c) [15]. Of course, if we multiply the complex numbers a ij by nonzero rational numbers, the obtained numbers are still algebraically independent over σ(c). So, we can assume that their absolute values are as small as we want. In particular, we may chose them in such a way that the matrix B = I +[a ij ] is invertible. Define φ : P 1 n(c) P 1 n(c) by φ(p ) = 1 t tr( t P σ P σ B, B) P P 1 n(c). Here tr C denotes the trace of a matrix C. Note that for every rank one idempotent P we have tr ( t P σ (I + [a ij ])) = tr t P σ + tr( t P σ [a ij ]) = σ(tr t P ) + 1 i,j n c ija ij = 1+ 1 i,j n c ija ij for some c ij from σ(c). Thus, tr ( t P σ B) 18

19 0 for every idempotent P of rank one. Note also that for every P Pn(D) 1 the matrix 1 t tr ( t P σ P σ B B) is of rank one with trace one, and is therefore again an idempotent of rank one. Clearly, this map satisfies all the assumptions of Lemma 2.8 but does not have such a nice form as in the case of EAS-division rings. We will prove now two consequences of Lemma 2.8 and its improvement for EAS-division rings. Lemma 2.10 Let D be any division ring, n an integer 3, and φ : P 1 n(d) P 1 n(d) an injective map preserving orthogonality. Assume that for every pair of linearly independent n-tuples t y 1,..., t y n t D n and z 1,..., z n D n there exist linearly independent n-tuples x 1,..., x n D n and t w 1,..., t w n t D n such that φ(p L( t y i )) P R(x i ) and φ(p R(z i )) P L( t w i ), i = 1,..., n. Then there exist a nonsingular matrix T M n (D) and a nonzero anti-endomorphism σ : D D such that φ(p ) = T t (P σ )T 1, P P 1 n(d). Proof. Assume that nonzero t y, t y, t y t D n and x, x, x D n satisfy t y span { t y, t y }, φ(p L( t y )) P R(x ), φ(p L( t y )) P R(x ), and φ(p L( t y )) P R(x ). We want to prove that then x span {x, x }. There is nothing to prove if t y and t y are linearly dependent. So, assume that they are linearly independent. Then we can find vectors t v 3,..., t v n t D n and u 1,..., u n D n such that t y u 1, t y u 2, t v 3 u 3,..., t v n u n are pairwise orthogonal rank one idempotents. We have φ( t y u 1 ) = t s 1 x, φ( t y u 2 ) = t s 2 x, φ( t v 3 u 3 ) = t s 3 p 3,..., φ( t v n u n ) = t s n p n for some linearly independent (linear independence follows from orthogonality) t s 1,..., t s n t D n and some p 3,..., p n D n. At least one of u 1 t y and u 2 t y, say u 1 t y, is nonzero. Then φ( t y (u 1 t y ) 1 u 1 ) P R(x ) is orthogonal to t s 3 p 3,..., t s n p n, and consequently, x t s j = 0, j = 3,..., n. It follows that x belongs to the linear span of x and x. Similarly, if nonzero z, z, z D n and t w, t w, t w t D n satisfy z span {z, z }, φ(p R(z )) P L( t w ), φ(p R(z )) P L( t w ), and φ(p R(z )) P L( t w ), then t w span { t w, t w }. Now we can apply Lemma 2.8. After composing φ by a similarity transformation we may assume that φ( t yz) = t z σ (y τ T t z σ ) 1 y τ T for every rank one idempotent t yz. Let z D n and t u t D n be any pair of nonzero vectors satisfying z t u = 0. Then we can find v D n and t y t D n 19

20 such that z t y = 1 = v t u and v t y = 0. It follows that t yz and t uv are orthogonal rank one idempotents, which yields that Therefore t v σ (u τ T t v σ ) 1 u τ T t z σ (y τ T t z σ ) 1 y τ T = 0. u τ T t z σ = 0 whenever z t u = 0. Taking u = e i and z = e j, i j, where the e i s form the standard basis of D n, we get from the above implication that T is a diagonal matrix. Further, the choice u = e i e j and z = e i + e j, i j, shows that all the diagonal entries of T are equal, say to λ. Hence, φ( t yz) = t z σ (y τ λ t z σ ) 1 y τ λ for every rank one idempotent t yz. Moreover, applying the above implication once more, this time with z = e 1 + µe 2 and u = µe 1 + e 2 we arrive at τ(µ)λ = λσ(µ), or equivalently, τ(µ) = λσ(µ)λ 1 for every µ D. Thus, a straightforward computation gives φ( t yz) = t ( t yz) σ for every idempotent t yz of rank one. This completes the proof. The example given after Lemma 2.9 shows that the complex case is much more complicated than the EAS-case because of the existence of nonsurjective endomorphisms. So it is not surprising that in the next result we will need the assumption that the underlying division ring D is an EAS-division ring. Lemma 2.11 Let D be any EAS-division ring, n an integer 3, and φ : P n (D) P n (D) an injective order preserving map. Assume that for every pair of linearly independent n-tuples t y 1,..., t y n t D n and z 1,..., z n D n there exist linearly independent n-tuples x 1,..., x n D n and t w 1,..., t w n t D n such that φ(p L( t y i )) P R(x i ) and φ(p R(z i )) P L( t w i ), i = 1,..., n. Then there exist a nonsingular matrix T M n (D) and an anti-automorphism τ : D D such that φ(p ) = T t (P τ )T 1, P P n (D). 20

21 Even before proving this lemma we will give a 3 3 complex example showing that the EAS-assumption on the underlying division ring is indispensable. We first choose two endomorphisms σ, τ : C C such that there exist complex numbers a, b, c, d that are algebraically independent over some subfield of C containing both σ(c) and τ(c). The existence of such endomorphisms and numbers a, b, c, d follows easily from the construction of a noncontinuous endomorphism of the complex field given in [15]. We set T = a b 0 1+bc c a d Then T 1 = 1+bc a b 0 c a d 1 We define φ : P 3 (C) P 3 (C) in three steps. First put φ(0) = 0 and φ(i) = I. Next, we define φ on the set of all rank one idempotents by φ( t yz) = 1 y σ T t z τ t z τ y σ T, t yz P 1 3 (C). In order to see that this map is well-defined on P3 1 (C) we have to check first that for every idempotent of rank one t yz we have y σ T t z τ 0. Once we prove 1 this it is straightforward to see that the matrix t y σ T t z z τ y σ T is an idempotent τ of rank one. And then we have to show that t yz = t uv P3 1 (C) implies that φ( t yz) = φ( t uv). So, assume first that z t y = 1. We have to show that y σ T t z τ 0. Write y = [ y 1 y 2 y 3 ] and z = [ z 1 z 2 z 3 ]. Assume on the contrary that y σ T t z τ = 0, that is 0 = [ σ(y 1 ) σ(y 2 ) σ(y 3 ) ] a b 0 1+bc c a 0 τ(z 1) τ(z 2 ) 0 0 d τ(z 3 ) = 1 a (a2 σ(y 1 )τ(z 1 ) + acσ(y 2 )τ(z 1 ) + abσ(y 1 )τ(z 2 ) + σ(y 2 )τ(z 2 ) +bcσ(y 2 )τ(z 2 ) + adσ(y 3 )τ(z 3 )). Now, a, b, c, d are algebraically independent over some subfield of the complex field containing both σ(c) and τ(c), and therefore σ(y 1 )τ(z 1 ) = 0 and σ(y 2 )τ(z 2 ) = 0 21

22 and σ(y 3 )τ(z 3 ) = 0, or equivalently, y 1 = 0 or z 1 = 0, and y 2 = 0 or z 2 = 0, and y 3 = 0 or z 3 = 0, contradicting our assumption that y 1 z 1 + y 2 z 2 + y 3 z 3 = 1. Next, if t yz = t uv P3 1 (C), then v = λz and u = 1 λy for some nonzero complex λ. Thus, ( ( ) ) ( ) φ( t t 1 uv) = φ λ y 1 1 (λz) = σ ( ) 1 λ yσ T τ(λ) t z τ(λ) t z τ σ y σ T τ λ = 1 y σ T t z τ t z τ y σ T = φ( t yz). So, φ is well-defined on the set of rank one idempotents. Let us show that φ is injective on P3 1 (C). Indeed, if t yz t uv, then y = [ y 1 y 2 y 3 ] and u = [ u 1 u 2 u 3 ] are linearly independent or z = [ z 1 z 2 z 3 ] and v = [ v 1 v 2 v 3 ] are linearly independent. We will consider only the second possibility. Then there exist i, j, 1 i < j 3, say i = 1 and j = 2, [ such that ] z1 z [ z 1 z 2 ] and [ v 1 v 2 ] are linearly independent, or equivalently, det 2 [ ] v 1 v 2 τ(z1 ) τ(z 0. It follows that det 2 ) 0. Thus, τ(v 1 ) τ(v 2 ) t z τ and t v τ are linearly independent, and therefore, φ( t yz) φ( t uv), as desired. Next, we define ψ : P3 1 (C) P3 1 (C) by by ψ( t yz) = 1 y τ T 1 t z σ T 1 t z σ y τ, t yz P 1 3 (C). Again, ψ is well-defined injective map on P 1 3 (C). Now we are ready to define φ on rank two idempotents. For every P P 2 3 (C) the idempotent I P belongs to P 1 3 (C). We set φ(p ) = I ψ(i P ), P P 2 3 (C). We have to show that φ is an injective order preserving map that is not an almost standard map on 3 3 idempotents. Obviously it is injective. Assume that P Q, P, Q P 3 (C). The only nontrivial case we have to consider is when rank P = 1 and rank Q = 2. So, assume P = t yz and Q = I t uv. From P Q it follows that t yz t uv and in order to see that φ(p ) φ(q) we only have to show that φ( t yz) ψ( t uv). One can obtain this orthogonality by a straightforward computation. In fact, we have t yz t uv if and only if φ( t yz) ψ( t uv). Thus, φ preserves order in both directions. It is also trivial to see that φ(e 11 ) φ(e 22 ) which further yields that φ is not an almost standard map. 22

23 Proof of Lemma The set of all rank one idempotents is invariant under φ. We will first show that the restriction of φ to P 1 n(d) satisfies the assumptions of Lemma 2.9. Assume that nonzero t y, t y, t y t D n and x, x, x D n satisfy t y span { t y, t y }, φ(p L( t y )) P R(x ), φ(p L( t y )) P R(x ), and φ(p L( t y )) P R(x ). We want to prove that then x span {x, x }. There is nothing to prove if t y and t y are linearly dependent. So, assume that they are linearly independent. Then, by our assumptions, x and x are linearly independent as well. We can find w, w, w D n such that w span { w, w }, w t y = w t y = w t y = 1, and w t y = w t y = 0. Then t y w t y w + t y w, and therefore φ( t y w ) φ( t y w + t y w ). Because φ( t y w ) P R(x ) and φ( t y w ) φ( t y w + t y w ), the image of φ( t y w + t y w ) contains x. It contains x as well. As the image of φ( t y w + t y w ) is two-dimensional, it is equal to the linear span of x and x. It follows that the image of φ( t y w ), which is the linear span of x, must be contained in the linear span of x and x. Similarly, if nonzero z, z, z D n and t w, t w, t w t D n satisfy z span {z, z }, φ(p R(z )) P L( t w ), φ(p R(z )) P L( t w ), and φ(p R(z )) P L( t w ), then t w span { t w, t w }. Now we can apply Lemma 2.9. After composing φ by an appropriate standard map we may assume that φ(p ) = P for every rank one idempotent P. Let Q be any idempotent and denote its rank by r. Then we can find pairwise orthogonal rank one idempotents P 1,..., P r such that P j Q for every j = 1,..., r. It follows that P j φ(q), j = 1,..., r. By Lemma 2.2 we have rank φ(q) = r. Hence, φ(q) = Q and we are done. We continue with results on idempotents over general division rings. Denote by D n (D) the set of all diagonal n n idempotent matrices over D. This is the set of all diagonal matrices whose diagonal entries are 0 or 1. Obviously, it has cardinality 2 n. We say that a subset S P n (D) is simultaneously diagonalizable if there exists an invertible T M n (D) such that T ST 1 D n (D). Lemma 2.12 Let D be a division ring, n an integer 3, and S a commutative subset of P n (D), that is, P Q = QP for every pair P, Q S. Then S is simultaneously diagonalizable. In particular, if S is of cardinality 2 n, then S is similar to D n (D). Proof. The proof is easy and is therefore left to the reader. Lemma 2.13 Let D be a division ring and n an integer 2. Then there exist a set S P n (D) and an invertible matrix T M n (D) such that S D n (D) = {0, I} and S = T D n (D)T 1. 23

24 Proof. Set P 1 = E 11 + E E 1n, P 2 = E 12 E E 1n + E 22 + E E 2n, P 3 = E 23 E E 2n + E 33 + E E 3n,. P n = E n 1,n + E nn. Then clearly, P j, j = 1,... n, are pairwise orthogonal idempotents of rank one. The sum j J P j is not a diagonal matrix for any subset J {1, 2,..., n} except when J = {1, 2,..., n} or when J = (here, of course, j P j is defined { to be the zero matrix). Thus, the set S = j J P j : J {1, 2,..., n}} has the desired properties. Lemma 2.14 Let D be a division ring and n an integer 3. Assume that P P n (D) is a diagonal idempotent of rank 1 or rank n 1. Then for every diagonal idempotent Q D n (D) there exists an invertible matrix T M n (D) such that P, Q T D n (D)T 1 and (D n (D) T D n (D)T 1 ) 8. Proof. We will consider only the case where P is of rank one. After applying a permutation similarity we may assume that P = E 11. Let Q be a diagonal idempotent. We have to distinguish two cases. The first one is that Q is a trivial idempotent, Q {0, I}, or Q = E 11, or Q = I E 11. We will consider only the second case when Q is a nontrival idempotent different from E 11, I E 11, since a similar idea works also in the first case. Applying yet another permutation similarity we may assume that either the first k diagonal entries of Q are 1 and all others are zero, or the first k diagonal entries of Q are zero and all others are 1. Here, k is an integer, 2 k n 1. Once again we will restrict to the first case, that is, Q = E E kk. Then, by the previous lemma, there exist invertible (k 1) (k 1) matrix T 1 and invertible (n k) (n k) matrix T 2 such that D k 1 (D) T 1 D k 1 (D)T1 1 = {0 k 1, I k 1 } and D n k (D) T 2 D n k (D)T2 1 = {0 n k, I n k }. Here, 0 j and I j denote the j j zero matrix and the j j identity matrix, respectively. Set T = T T 2 Obviously, T P T 1 = P = E 11, T QT 1 = Q = E E kk T D n (D)T 1. Moreover, if we put R 1 = E 11, R 2 = E 22 { E kk, and R 3 = E k+1,k E nn, then D n (D) T D n (D)T 1 = j J R j : J {1, 2, 3}}. This set has cardinality 8, as desired.. 24

25 Lemma 2.15 Let D be a division ring and n an integer 4. Assume that P P n (D) is a diagonal idempotent satisfying rank P {0, 1, n 1, n}. Then there exists a diagonal idempotent Q D n (D) such that for every invertible matrix S M n (D) satisfying P, Q SD n (D)S 1 we have (D n (D) SD n (D)S 1 ) > 8. Proof. As in the proof of the previous lemma there is no loss of generality in assuming that P = E E kk, 2 k n 2. Set Q = E 11 + E k+1,k+1. Let S be any invertible matrix satisfying P, Q SD n (D)S 1. Then (S 1 P S)(S 1 QS) = S 1 E 11 S is a diagonal matrix and the same must be true for S 1 (E E kk )S = S 1 P S S 1 P QS and S 1 E k+1,k+1 S = S 1 QS S 1 P QS. We can find a permutation matrix T such that (ST ) 1 E 11 (ST ) = E 11, (ST ) 1 (E E kk )(ST ) = E E kk, and (ST ) 1 E k+1,k+1 (ST ) = E k+1,k+1. It follows that the matrix ST has block diagonal form a B 0 0 ST =, 0 0 c D where a and c are nonzero elements of D, and B and D are invertible (k 1) (k 1) matrix and invertible (n k 1) (n k 1) matrix, respectively. Set R 1 = E { 11, R 2 = E E kk, R 3 = E k+1,k+1, and R 4 = E k+2,k E nn. Then j J R j : J {1, 2, 3, 4}} D n (D) ST D n (D)T 1 S 1 = D n (D) SD n (D)S 1, and thus, (D n (D) SD n (D)S 1 ) > 8. Let K be a proper subfield of the field of complex numbers C isomorphic to C and let τ : K C be an isomorphism. The following technical result will be needed for one of our counterexamples. Lemma 2.16 Let τ and K be as above and let P be a complex 4 4 idempotent matrix of rank two with at least one entry outside K. Then there exists a complex 4 4 idempotent matrix R such that Q τ R for every Q P 1 4 (K) satisfying Q P and R Q τ for every Q P 3 4 (K) satisfying P Q. Proof. Define S = {Q P4 1 (K) : Q P } and T = {Q P4 3 (K) : P Q}. If S is empty, then the choice R = 0 gives the idempotent R with the desired properties. Similarly, if S contains exactly one member S = {S} then the choice R = S τ completes the proof. In the remaining case we can choose two different idempotents S 1 and S 2 from S. If we had S 1 S 2 then by Lemma 2.1 we would have P P 4 (K), a contradiction. Hence, S 1 S 2, and after applying a similarity induced by an appropriate invertible matrix from M 4 (K) 25