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Subspace-hypercyclic abelian linear semigroups
Journal Pre-proof Subspace-hypercyclic abelian linear semigroups
Salah Herzi, Habib Marzougui
PII:
S0022-247X(20)30122-0
DOI:
https://doi.org/10.1016/j.jmaa.2020.123960
Reference:
YJMAA 123960
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
17 October 2019
Please cite this article as: S. Herzi, H. Marzougui, Subspace-hypercyclic abelian linear semigroups, J. Math. Anal. Appl. (2020), 123960, doi: https://doi.org/10.1016/j.jmaa.2020.123960.
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SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS SALAH HERZI AND HABIB MARZOUGUI Abstract. We consider subspace-hypercyclicity for abelian semigroups G of matrices on Kn (n ≥ 1), K = R or C. We say that G is subspacehypercyclic for a non-zero subspace M of Kn , if there exists x ∈ Kn such that G(x) ∩ M is dense in M . We give an effective way of checking that a given abelian semigroup is subspace-hypercyclic. We construct a nontrivial example of abelian semigroup which has an orbit dense relative to a certain straight line but is not subspace-hypercyclic for this straight line. Additionally, we construct, for every n ≥ 2, an example of abelian semigroup generated by two matrices which is subspace hypercyclic but not hypercylic. We determine the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn . On the other hand, we prove that Gk is subspace-hypercyclic for every k ∈ Np0 whenever G is a subspace-hypercyclic abelian semigroup of matrices on Cn .
1. Introduction Let K = R or C, Mn (K) be the set of all square matrices over K of order n ≥ 1 and let GL(n, K) be the group of invertible matrices of Mn (K). Let G be an abelian sub-semigroup of Mn (K). By a sub-semigroup of Mn (K), we mean a subset of Mn (K) which is stable under multiplication and contains the identity matrix. For a vector v ∈ Kn , we consider the orbit of G through v: G(v) = {Av : A ∈ G} ⊂ Kn . A subset E ⊂ Kn is called G-invariant if A(E) ⊂ E for any A ∈ G. The orbit G(v) ⊂ Kn is dense in Kn if G(v) = Kn , where E denotes the closure of a subset E ⊂ Kn . The semigroup G is called hypercyclic if there exists a vector v ∈ Kn such that G(v) is dense in Kn . In this case, v is called hypercyclic vector for G. We refer the reader to the books [5] and to [9] and papers [1], [2], [3], [7], [8], [13] for a thorough account on hypercyclicity. Here and throughout the paper, N and Z denote the sets of non-negative integers and integers, respectively, N = {0, 1, 2, . . . } while N0 = N\{0} denotes the set of positive integers. In [11] B.F. Madore and R.A. Mart´ınez-Avenda˜ no consider for a single operator A on a Banach space X the concept of the density of orbits in a nontrivial subspace M of X instead of X, this notion is called subspacehypercyclicity. They proved in particular that no operator on Kn , n ∈ N0 , 2000 Mathematics Subject Classification. 47A16. Key words and phrases. hypercyclic, subspace-hypercyclic, semigroup, abelian, somewhere dense, dense orbit. 1
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SALAH HERZI AND HABIB MARZOUGUI
can be subspace-hypercyclic (see [11], theorem 4.9). Moreover, they showed that there are operators on a separable Hilbert space which are subspacehypercyclic but not hypercyclic. However, they asked whether hypercyclicity of an operator on a Banach space implies subspace-hypercyclicity. Bamerni et al. [4] provide an affirmative answer to the above question. More examples and results of subspace-hypercyclic operators can be found in ([15], [11], [12]). Here we are interested to introduce this new notion for abelian semigroups of matrices on Kn , n ∈ N0 because phenomena of hypercyclicity for those semigroups can arise in Kn (see e.g. [2], [3]), and so are of their subspacehypercyclicity by [4]. Definition 1.1. We say that G is subspace-hypercyclic for a non-zero subspace M of Kn , if there exists x ∈ Kn such that G(x) ∩ M is dense in M . In this case, x is called subspace-hypercyclic vector for G. We say that G is somewhere subspace-hypercyclic if there exists x ∈ Kn and a non-zero subspace M of Kn such that G(x) ∩ M is somewhere dense in M . Notice that subspace-hypercyclicity is reduced to that of hypercyclicity whenever M = Kn or n = 1. For an abelian semigroup G generated by A1 , . . . , Ap of Mn (C), we denote k by Gk the semigroup generated by Ak11 , . . . , App defined by m k 1 k1 Gk = {Am . . . . Ap p p : m1 , . . . , mp ∈ N}, where k = (k1 , . . . , kp ) ∈ Np0 . 1 The main purpose of this paper is to provide firstly, for any abelian subsemigroup of Mn (C) an effective way of checking that a given semigroup is subspace-hypercyclic. In particular, we show that no matrix on Cn can be subspace-hypercyclic. Also, we show that there are subspace-hypercyclic semigroups that are not hypercyclic (see proposition 3.1). We also prove that the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn for a subspace M of dimension k is k + 1. Secondly, we show that there are semigroups on Rn that have somewhere dense orbits in a subspace M that are not everywhere dense in M (see proposition 6.3). On the other hand, we show that Gk is subspace-hypercyclic for every k ∈ Np0 whenever G is subspace-hypercyclic finitely abelian sub-semigroup of Mn (C). To state our main results, we need to introduce the following notations and definitions. Let n ∈ N0 be fixed. By a partition of n we mean a r ni = n. finite sequence of positive integers η = (n1 , . . . , nr ) such that i=1
The number r will be called the length of the partition. Given a partition η = (n1 , . . . , nr ), we define the following: • Kη (C) := Tn1 (C) ⊕ · · · ⊕ Tnr (C), where Tm (C) (m = 1, 2, . . . , n) is the set of lower-triangular matrices over C with only one eigenvalue. Obviously Kη (C) is a sub-semigroup of Mn (C). In particular: If r = 1, Kη (C) = Tn (C). If r = n, Kη (C) is the semigroup of diagonal
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
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matrices on Cn . • Kη∗ (C) := Kη (C) ∩ GL(n, C), it is a sub-semigroup of GL(n, C). • B0 = {e1 , . . . , en } the canonical basis of Cn . The following “normal form of G” allows us to deduce the results for an arbitrary semigroup: Proposition 1.2 ([3], proposition 2.4). Let G be an abelian sub-semigroup of Mn (C), n ∈ N0 . Then there exists a P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η of n. Let G be an abelian sub-semigroup of Kη (C), for some partition η = (n1 , . . . , nr ) of n. For each A = diag(A1 , . . . , Ar ) ∈ G and for each 1 ≤ j ≤ r, we denote by • Gj = {Aj : A ∈ G} • σ(Gj ) = ∪ σ(Aj ) : the spectrum of Gj i.e. the set of eigenvalues of all A∈G
elements Aj of Gj . If G is an abelian sub-semigroup of Mn (C) and P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), we denote by • Gj = (P −1 GP )j and σ(Gj ) = σ((P −1 GP )j ). Our principal results can now be stated as follows: Theorem 1.3. Let G be an abelian sub-semigroup of Mn (C), n ∈ N0 , and let P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η = (n1 , . . . , nr ) of n. Then the following assertions are equivalent: (i) G is subspace-hypercyclic. (ii) σ(Gj ) is dense in C, for some j ∈ {1, . . . , r}. (iii) G is subspace-hypercyclic for the straight line CP en1 +···+nj , for some j ∈ {1, . . . , r}. Corollary 1.4. Let G be an abelian sub-semigroup of Mn (C) generated by p matrices A1 , . . . , Ap (p ≥ 1) and let P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η = (n1 , . . . , nr ) of n. Let σ(Ai ) = {λi,j : 1 ≤ j ≤ r} denote the set of eigenvalues of Ai , 1 ≤ i ≤ p. Then the following assertions are equivalent: (i) G is subspace-hypercyclic. kp 1 . . . λp,j : k1 , . . . , kp ∈ N} is dense in C, for some j ∈ {1, . . . , r}. (ii) {λk1,j (iii) G is subspace-hypercyclic for the straight line CP en1 +···+nj , for some j ∈ {1, . . . , r}. Corollary 1.5 ([11], theorem 4.9). For n ∈ N0 , there is no subspacehypercyclic matrix on Cn . On the other hand, we determine the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn .
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Theorem 1.6. Let M be a subspace of Cn . The minimum number of diagonalizable matrices generating a subspace-hypercyclic abelian semigroup of Mn (C) for a subspace M is dim(M ) + 1. Theorem 1.6 extends partially corollary 2.2 obtained for M = Cn corresponding to hypercyclicity. In the following theorem, we deal with answer negatively whether, given a subspace M of Cn and x ∈ M , does M ⊂ G(x) imply that G(x) ∩ M = M i.e. G is subspace-hypercyclic for M ? Theorem 1.7. For every n ≥ 2, there exists a hypercyclic abelian semigroup G of matrices of Mn (C) with a hypercyclic vector x, but not subspacehypercyclic for Cx. Feldman showed (corollary 5.8, [8]) that if a finitely generated abelian semigroup G of matrices over C is hypercyclic, then for every k = (k1 , . . . kp ) ∈ Np0 , Gk is hypercyclic. Here we ask if similar result holds for subspacehypercyclicity. We prove the following theorem: Theorem 1.8. Let G be a finitely generated abelian semigroup of Mn (C). If G is subspace-hypercyclic, then for every k = (k1 , . . . kp ) ∈ Np0 , Gk is subspace-hypercyclic. Moreover, if G is subspace-hypercyclic for a subspace M then Gk is subspace-hypercyclic for a straight line in M . 2. Some results on hypercyclicity and subspace-hypercyclicity First we introduce the following notations. • In the identity matrix on Cn . For a row vector v ∈ Cn , we denote by v T the transpose of v. We also have that • uη = [eη,1 , . . . , eη,r ]T ∈ Cn , where eη,k = [1, 0, . . . , 0]T ∈ Cnk , k = 1, . . . , r. (k)
(k)
(k)
• vη = [(vη )1 , . . . , (vη )r ] ∈ Cn , where for every j = 1, . . . , r, if j = k 0 ∈ Cnj (k) (vη )j = if j = k eη,k (k)
• If r = n, then uη = [1, . . . , 1]T and vη k-position and 0’s on the other positions.
= ek : vector with 1 on the
Consider the matrix exponential map exp : Mn (C) → GL(n, C), set exp(M ) = eM . Finally, for a semigroup G of Mn (K), we denote by G∗ = G ∩ GL(n, K). Recall the following theorem which characterizes the hypercyclicity of any abelian semigroup of matrices on Cn . Theorem 2.1 ([3], theorem 1.1). Let G be an abelian sub-semigroup of Kη (C) for some partition η with length r, 1 ≤ r ≤ n. Assume that G∗
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
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is finitely generated by p matrices (p ≥ 1) and let B1 , . . . , Bp ∈ Kη (C) such that eB1 , . . . , eBp generate G∗ . Then G is hypercyclic if and only if p r (k) NBk uη + 2iπZvη is dense in Cn . k=1
k=1
Corollary 2.2 ([3]). For n ∈ N0 , the minimum number of matrices of Mn (C) that generate a hypercyclic abelian semigroup is n + 1. In particular: Corollary 2.3 ([6]). For n ∈ N0 , there is no hypercyclic matrix on Cn . For the particular case r = n, we obtain the following. Theorem 2.4 ([3], theorem 1.1). Let G be an abelian sub-semigroup of diagonal matrices of Mn (C). Assume that G∗ is finitely generated by p matrices (p ≥ 1) and let B1 , . . . , Bp ∈ Kη (C) such that eB1 , . . . , eBp generate p n NBk uη + 2iπZek is dense in G∗ . Then G is hypercyclic if and only if Cn .
k=1
k=1
Corollary 2.5 ([3], corollary 1.8). For n ∈ N0 , the minimum number of diagonalizable matrices of Mn (C) that generate a hypercyclic abelian semigroup is n + 1. Theorem 2.6 ([4]). Let A be a dense subset of a Banach space X with dim(X) > 1. Then there is a nontrivial closed subspace M of X such that A ∩ M is dense in M . By a nontrivial subspace M of X, we mean that M is non-zero and distinct from X. Remark 2.7. Notice that the theorem 2.6 is not true for dim(X) = 1. As a consequence: Corollary 2.8. Let G be an abelian semi-group of Mn (K), n ≥ 2. If G is hypercyclic, then it is subspace-hypercyclic for a nontrivial subspace M of Kn . Proposition 2.9. Let G be an abelian semi-group of Mn (K), n ≥ 2. If G is subspace-hypercyclic, then it is subspace-hypercyclic for a straight line Δ = Kz, z ∈ Kn . Proof. Let x ∈ Kn such that G(x) ∩ M = M , where M is a subspace of Kn . If dimM = 1, the proposition 2.9 is obvious. So assume that dimM > 1. According to theorem 2.6, there is a nontrivial subspace M1 of M such that G(x)∩M ∩M1 = G(x)∩M1 is dense in M1 . Now, if M1 is a straight line, the result follows, otherwise the theorem 2.6 is applied again and so on until we get a straight line Δ = Kz (z ∈ Kn ) such that G(x) ∩ Δ is dense in Δ.
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Lemma 2.10. Let G be a finitely generated abelian sub-semigroup of Mn (K). If G is subspace-hypercyclic for a non-zero subspace M of Kn , then there exists y ∈ M such that G(y) ∩ M is dense in M . Proof. Let A1 , . . . , Ap generate G and let x ∈ Cn such that G(x)∩M is dense in M . Wewill prove that there exists y ∈ M such that G(x)∩M = G(y)∩M . n Set E1 = n1 ∈ N : ∃(n2 , . . . , np ) ∈ Np−1 ; An1 1 An2 2 . . . Ap p x ∈ M . Since G(x) ∩ M = ∅, E1 = ∅. So let k1 = min(E1 ). Now set n
Ek1 = {n2 ∈ N : ∃n1 ≥ k1 , ∃(n3 , . . . , np ) ∈ Np−2 ; An1 1 An2 2 . . . Ap p x ∈ M }. n
Then Ek1 = ∅ (since ∃(n2 , . . . , np ) ∈ Np−1 such that Ak11 An2 2 . . . Ap p x ∈ M ). So set k2 = min(Ek1 ). In this way, we construct by induction, n
Ek1 ,k2 = {n3 ∈ N : ∃n1 ≥ k1 , ∃n2 ≥ k2 , ∃(n4 , . . . , np ) ∈ Np−3 ; An1 1 An2 2 . . . Ap p x ∈ M }, n
. . . , Ek1 ,...,kp−1 = {np ∈ N : ∃n1 ≥ k1 , . . . , ∃np−1 ≥ kp−1 ; An1 1 . . . Ap p x ∈ M }, with k3 = min(Ek1 ,k2 ), . . . , kp = min(Ek1 ,...,kp−1 ). In particular, we see that k
y := Ak11 . . . App x ∈ M . Observe that G(x) ∩ M
n
= {An1 1 . . . Ap p x : (n1 , . . . , np ) ∈ Np } ∩ M n
= {An1 1 . . . Ap p x : n1 ≥ k1 , . . . , np ≥ kp } ∩ M = G(y) ∩ M, In result, G(y) ∩ M is dense in M .
3. Subspace-hypercyclic abelian semigroups Proof of theorem 1.3. We may assume, by taking P −1 GP if necessary that, that G ⊂ Kη (C). Indeed, G := P −1 GP is a semigroup of Kη (C) and the property G(x) ∩ M = M with x ∈ Cn is equivalent to that of G (x ) ∩ M = M , where M = P −1 (M ) and x = P −1 (x) ∈ Cn . Furthermore dim(M ) = dim(M ). (i) ⇒ (ii): Suppose that G is subspace hypercyclic for a subspace M of Cn of dimension k. Then there is x = [x1 , . . . , xn ]T ∈ Cn such that G(x) ∩ M is dense in M . Let B = {y1 , y2 , . . . , yk } be a basis of M with yi = [y1,i , . . . , yn,i ]T . Since dim(M ) = k, the n − k + 1 first rows of B are not all zero. So there is 1 ≤ i ≤ n such that Li = (yi,1 , . . . , yi,k ) is the first non-zero row of B. One can then assume that yi,1 = 1; yi,2 = · · · = yi,k = 0. Let 1 ≤ j ≤ r be such that 0 i=1
j−1 i=1
ni < i ≤
j
ni with the convention
k=1
ni = 0. If y ∈ G(x) ∩ M then y = [0, . . . 0, di , . . . , dn ]T = Ax, where
A = diag(B1 , . . . , Br ) ∈ G. So the ith -row in A corresponds to a row in Bj . As for 1 ≤ l ≤ r, Bl has only one eigenvalue λl , then di = λl xi . Now for every
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
c1 , . . . , ck ∈ C, y =
k
7
ci yi ∈ M and its ith -coordinate in Cn is ci . There-
i=1
fore, from G(x) ∩ M = M and above, we deduce that {λj xi : λj ∈ σ(Gj )} is dense in C. In particular xi = 0 and hence, σ(Gj ) is dense in C. (ii) ⇒ (iii) Set Δ = Cen1 +···+nj . Then G is subspace-hypercyclic for Δ. Indeed, en1 +···+nj is an eigenvector of any element of G, so G(en1 +···+nj ) ∩ Δ = G(en1 +···+nj ) = λj en1 +···+nj : λj ∈ σ(Gj ) which is dense in Δ by the hypothesis (ii). (iii) ⇒ (i) is clear.
Proof of corollary 1.4. We may assume, by taking P −1 GP if necessary that, that G ⎛ ⊂ Kη (C). For each ⎞ i ∈ {1, . . . , p}, Ai = diag(Ti,1 , . . . , Ti,p ), 0 λi,j 0 ⎜ ⎟ where Ti,j = ⎝ ∗ . . . 0 ⎠, j ∈ {1, . . . , r}. As any element of G can ∗
∗
λi,j k k1 form A1 . . . App , for some k1 , . . . , kp ∈ N, then for each kp 1 2 = {λk1,j λk2,j . . . λp,j : k1 , . . . , kp ∈ N}. Hence, corollary 1.4
be written of the 1 ≤ j ≤ r, σ(Gj ) follows from theorem 1.3.
Proof of corollary 2.3. This follows from corollary 1.4, (ii) applied for the particular case p = 1. Notice that corollary 2.3 can also be proven by the following argument. For a single matrix A ∈ Mn (K) and x ∈ Kn , it is known that one of the following assertions holds: lim Ak x = 0 or lim Ak x = +∞ or k→+∞
k→+∞
∃ c1 , c2 > 0 such that c1 ≤ Ak x ≤ c2 for all k ∈ N, where is any norm on Cn . In particular the orbit of x by A is not dense in M for any nontrivial subspace M of Kn . Contrarily to a single matrix, subspace-hypercyclicity can occur for semigroups: Proposition 3.1. For every n ≥ 2, there exist two matrices of Mn (C) generating an abelian semi-group G which is subspace-hypercyclic, but not hypercyclic. Proof. Let a, b ∈ C such that {ak bn : k, n ∈ N} is dense in C (see e.g. [8], proposition 3.1). Let G be the abelian semigroup generated by A = diag(a, In−1 ) and B = diag(b, In−1 ), where In−1 is the identity matrix on Cn−1 . Then G is a subspace-hypercyclic for the subspace Ce1 since G(e1 ) ∩ Ce1 = G(e1 ) = Ce1 . The fact that G is not hypercyclic follows from corollary 2.2 since n ≥ 2. Remark 3.2.
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(1) In the proof of proposition 3.1, the fact that G is subspace-hypercyclic for the subspace Ce1 also follows from corollary 1.4, since G is an abelian semigroup of Kη (C), where η = (n1 , n2 ), n1 = 1, n2 = n − 1, j = 1, en1 = e1 . (2) The proposition 3.1 shows that the semigroup G is subspace-hypercyclic for Ce1 , but not for its orthogonal complement (Ce1 )⊥ . As a consequence of proposition 3.1 and corollary 2.3, we get: Corollary 3.3. The minimal number of matrices generating a subspacehypercyclic abelian semigroup over C is 2. In the following, we give an explicit parameter family (Gθ )θ∈R of abelian semigroups which are subspace-hypercyclic if and only if θ is irrational. Example 3.4. Let p, q ≥ 2 two coprime numbers and θ ∈ R. For every n ≥ 3, let Gθ be the abelian semigroup generated by A1 = diag (T1 , In−2 ), A2 = diag (T2 , In−2 ) and A3 = diag (T3 , In−2 ), where T1 , T2 , T3 ∈ T2 (C) with only one eigenvalue p, 1q and e2iπθ respectively. Then: (i) Gθ is subspace-hypercyclic for Ce2 , but not hypercyclic whenever θ is irrational. (ii) Gθ is not subspace-hypercyclic whenever θ is rational. Proof. We have that Gθ ⊂ Kη (C) with η = (n1 , n2 ), n1 = 2, n2 = n − 2 and k r = 2. By ([8], corollary 4.2), the set { qpm : k, m ∈ N} is dense R+ . k
(i) If θ is irrational, then the set { qpm e2iπsθ : k, s, m ∈ N} is dense C. By corollary 1.4, Gθ is subspace-hypercyclic for Ce2 (since j = 1 and en1 = e2 ). Moreover, Gθ is not hypercyclic by corollary 2.2 since n ≥ 3. k (ii) If θ is rational, then the set { qpm e2iπsθ : k, s, m ∈ N} is nowhere dense C. Therefore, from corollary 1.4, Gθ is not subspace-hypercyclic. The question whether there exists a subspace-hypercyclic abelian semigroup G for a nontrivial subspace M of Cn which is not G-invariant is answered positively as follows. Proposition 3.5. For every n ≥ 2, there is an abelian semigroup G of Mn (C) which is subspace-hypercyclic for a subspace M , but M is not Ginvariant. Proof. Let a, b ∈ C such {ak bn : k, n ∈ N} is dense C. Set M = Ce1 and let G be the semigroup generated by A1 = diag (T1 , In−2 ), A2 = diag (T2 , In−2 ) and A3 = diag (T3 , In−2 ), where
1 0 a 0 b 0 , T2 = , T3 = T1 = 1 1 −a a −b b We have that An1 1 An2 2 An3 3 (e1 ) = an2 bn3 [1, n1 − (n2 + n3 ), 0, . . . , 0]T .
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
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Therefore, G(e1 ) ∩ M = {an2 bn3 e1 : n2 , n3 ∈ N} which is dense in M . So G is subspace-hypercyclic for M , but clearly M is not G-invariant. Remark 3.6. By applying corollary 1.4 to the example in proposition 3.5, we see that η = (n1 , n2 ), n1 = 2, n2 = n − 2, j = 1, en1 = e2 . It follows that G is subspace hypercyclic for Ce2 , however, Ce2 is G-invariant. In ([11], theorem 4.5) B.F. Madore and R.A. Mart´ınez-Avenda˜ no proved that if an operator T on a Hilbert space X is subspace-hypercyclic, then its spectrum σ(T ) meets the unit circle S 1 i.e. σ(T ) ∩ S 1 = ∅. This result fails for abelian semigroup. Indeed, first we define the spectrum of an abelian semigroup G of Mn (C) generated by p matrices Ai , 1 ≤ i ≤ p, by σ(G) := ∪ σ(Ai ). Now consider A1 = aIn , A2 = bIn , where a, b ∈ C 1≤i≤k
such {ak bn : k, n ∈ N} is dense C. Then σ(G) = {a, b} and by theorem 1.3, the semigroup G generated by A1 and A2 is subspace-hypercyclic for Cen . However, σ(G) ∩ S 1 = ∅. 4. Minimal generators for subspace-hypercyclic semigroup The aim of this section is to prove theorem 1.6. • Firstly, we will show that there exists an abelian semigroup of Mn (C) generated by k + 1 diagonalizable matrices which is subspace-hypercyclic for Ek = Ck × {0}n−k . Following corollary 2.5, there exists a hypercyclic abelian semigroup G1 of Mk (C) generated by k + 1 diagonalizable matrices D1 , . . . , Dk+1 . Let u1 be a hypercyclic vector for G1 and set Ai = diag (Di , In−k ) ∈ Mn (C), for all 1 ≤ i ≤ k + 1, where In−k is the identity matrix of Cn−k . Let G be the semigroup generated by the Ai , 1 ≤ i ≤ k+1 and let u = [u1 , 0, . . . , 0]T ∈ Cn . As G1 (u1 ) = Ck and G(u) ∩ Ek = G(u) = [G1 (u1 ), 0n−k ]T , it follows that G(u) ∩ Ek = Ek . Hence, G is subspace-hypercyclic for Ek . • Secondly, let G be an abelian semigroup of Mn (C) generated by p diagonalizable matrices Ai , 1 ≤ i ≤ p. Assume that G is subspace-hypercyclic for a subspace M of Cn of dimension k. By lemma 2.10, there exists x1 ∈ M such that G(x1 ) ∩ M = M . Claim 1. One can assume that the Ai , 1 ≤ i ≤ p are diagonal. Indeed, following the proposition 1.2, there exists a P ∈ GL(n, C) such that G := P −1 GP is a semigroup of diagonal matrices. From G(x1 ) ∩ M = M , it follows that G (x1 ) ∩ M = M , where x1 = P −1 (x1 ) and M = P −1 (M ). Observe that dim(M ) = dim(M ) and x1 ∈ M . This proves the claim. Assume that x1 has the form x1 = [x1,1 , x2,1 , . . . , xn,1 ]T ∈ M . Let y = [y1 , y2 , . . . , yn ]T ∈ M , there is a sequence of diagonal matrices (Aj )j ∈ G
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SALAH HERZI AND HABIB MARZOUGUI
such that y = lim Aj x1 with Aj x1 ∈ M . Write Aj = diag(d1,j , . . . , dn,j ), j→+∞
then we have y = lim [d1,j x1,1 , d2,j x2,1 , . . . , dn,j xn,1 ]T . j→+∞
Hence, if xi,1 = 0 for some 1 ≤ i ≤ n, then yi = 0. Let B = {x1 , . . . , xk } formed by extending x1 to a basis of M , where xj = [x1,j , x2,j , . . . , xn,j ]T , 1 ≤ j ≤ k and let mj = min{i ∈ {1, . . . , n} : xi,j = 0}. By above, we have that m1 ≤ mj , for every j ∈ {1, . . . , k}. It follows that xj = [0, . . . , 0, xmj ,j , . . . , xn,j ]T and xmj ,j = 0, for all 1 ≤ j ≤ k. Furthermore k ≤ n − m1 + 1 since dim(M ) = k. Upon reordering, and taking appropriate linear combinations of the xj , 1 ≤ j ≤ k, we can assume that m1 < m2 < m3 < · · · < mk ≤ n. We let N = n − m1 + 1. Denote by G(m1 ) the semigroup of MN (C) generated by (m1 )
Aj
(m1 )
= diag(dm1 ,j , . . . , dn,j ), x1
= [xm1 ,1 , . . . , xn,1 ]T ∈ CN
and (m1 )
xj
= [0, . . . , 0, xmj ,j , . . . , xn,j ]T ∈ CN , for 2 ≤ j ≤ k.
We let M (m1 ) = {y = [ym1 , . . . , yn ]T ∈ CN : [0, . . . , 0, ym1 , . . . , yn ]T ∈ M }. It is a subspace of CN . Observe that (m1 )
G(m1 ) (x1
) ∩ M (m1 ) = M (m1 ) .
Without loss of generality, we can write for short (m1 )
G(m1 ) = G, x1 (m1 )
xj
= x1 , with x1,1 = 0,
= xj = [0, . . . 0, xmj ,j , . . . , xn,j ]T , for 2 ≤ j ≤ k
and M (m1 ) = M. Then 1 ≤ k ≤ N and we have G(x1 ) ∩ M = M . Set {ik+1 , . . . , iN } = {m1 , . . . , n}\{m1 , . . . , mk } so that ik+1 < · · · < iN . Let P, Q ∈ GL(N, C) defined by if 1 ≤ j ≤ k P (emj ) = xj if k + 1 ≤ j ≤ N P (eij ) = eij and
Q(ej ) = emj Q(ej ) = eij
if 1 ≤ j ≤ k if k + 1 ≤ j ≤ N
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
11
Set G = (P Q)−1 G(P Q). Then G is an abelian semigroup of MN (C) formed of diagonalizable matrices. As P Q(ej ) = xj , for 1 ≤ j ≤ k, then P Q(Ek ) = M and so G (e1 ) ∩ Ek = Ek .
A1 0 with A1 ∈ Claim 2. Every A ∈ G has the form A = A2 A3 Mk (C), A2 ∈ MN −k,k (C), A3 ∈ MN −k (C). Indeed, let A = (P Q)−1 A(P Q) = (ap,q )1≤p,q≤N ∈ G , where A = (ap,q )1≤p,q≤N ∈ G. As A is diagonal, then ap,q = 0 whenever p = q. Therefore, we have that ∀(p, q) ∈ {1, . . . , k} × {k + 1, . . . , N }, ap,q = eTp (P Q)−1 AP Qeq = eTp Q−1 P −1 Aeiq = aiq ,iq eTp Q−1 eiq = aiq ,iq eTp eq = 0 This proves the claim. Claim 3. p ≥ k + 1.
A1 0 ∈ G and by Indeed, denote by = : = A2 A3 e1 = [1, 0, . . . , 0]T ∈ Ck . Observe that G1 is an abelian semigroup of Mk (C). We have that G (e1 ) ∩ Ek ⊂ G1 (e1 ) × {0}n−k ⊂ Ek . Therefore, from G (e1 ) ∩ Ek = Ek , we get G1 (e1 ) = Ck . In result, G1 is hypercyclic. As the number of generators of G1 is less than that of G and this later is equal to the number p of generators of G, so by corollary 2.2, p ≥ k + 1. G1
A1
A
5. On semigroups with orbits dense relative to nontrivial subspaces Given a subspace M of Cn and x ∈ X, we see that if G(x) ∩ M = M then G(x) is dense in M ; that is M ⊂ G(x). Conversely, does the density of G(x) in M imply that G is subspace-hypercyclic for M ? In this section, we prove theorem 1.7 which answer negatively the above question. We will construct n + 1 diagonal matrices A1 , . . . , An+1 on Cn generating a hypercyclic abelian semigroup G with a hypercyclic vector x but G is not subspace-hypercyclic for M = Cx. For this, let recall the following multidimensional real version of Kronecker’s theorem (see e.g. [13], lemma 2.2; see also [10], theorem 442). Kronecker’s Theorem ([13], lemma 2.2) Let α1 , . . . , αn be negative real numbers such that 1, α1 , . . . , αn are linearly independent over Q. Then the
12
SALAH HERZI AND HABIB MARZOUGUI
set
Nn + N[α1 , . . . , αn ]T := [s1 , . . . , sn ]T + k[α1 , . . . , αn ]T : k, s1 , . . . , sn ∈ N
is dense in Rn . By identifying Cn with R2n in the natural way, we deduce the complex version as follows. Theorem 5.1 (Kronecker’s theorem: complex version). Let α1 , . . . , αn ; β1 , . . . , βn be negative real numbers such that 1, α1 , . . . , αn , β1 , . . . , βn are linearly independent over Q. Then Nn + iNn + N[α1 + iβ1 , . . . , αn + iβn ]T is dense in Cn . Lemma 5.2. Let n ∈ N0 . Then there exist n + 1 vectors u1 , . . . , un+1 of Cn such that n+1 n Nuk + 2iπZek k=1
k=1
is dense in Cn . Proof. Let α1 , . . . , αn , β1 , . . . , βn be negative real numbers such that the numbers 1, α1 , . . . , αn , β1 , . . . , βn are linearly independent over Q. Set u = [α1 + iβ1 , . . . , αn + iβn ]T and define
−2πek 2iπu
uk : = Write H :=
n+1 k=1
Nuk +
if 1 ≤ k ≤ n if k = n + 1 n
2iπZek
k=1
Then we have H=
n
− 2πNek + N2iπu +
k=1
n
2iπZek
k=1
= 2iπH where H =
n k=1
iNek + Nu +
n k=1
Zek = Zn + iNn + Nu. By theorem 5.1, we
see that H is dense in Cn , and thus so is H. This proves the lemma.
Proof of theorem 1.7. Let u1 , . . . , un+1 of Cn are those of from the proof (k) (k) of Lemma 5.2. Write uk = [x1,1 , . . . , xn,1 ]T . Let B1 , . . . , Bn+1 be defined by (k) (k) (k) Bk = diag x1,1 , x2,1 , . . . , xn,1 , 1 ≤ k ≤ n + 1. Let G be the abelian sub-semigroup of Kη∗ (C) generated by eB1 , . . . , eBn+1 .
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS (k)
Here r = n, η = (1, . . . , 1), uη = [1, . . . , 1]T and vη Bk uη = uk . Hence, n+1
NBk uη +
k=1
n
2iπZvη(k) =
k=1
n+1
Nuk +
k=1
13
= ek . We have that
n
2iπZek
k=1
=H By lemma 5.2, H = Cn and by theorem 2.1, G(uη ) = Cn . Now denote by M = Cuη the vector space generated by uη . We shall prove that G(uη ) ∩ M = M . Let A ∈ G such that Auη = λuη , for some λ ∈ C. Then (A − λIn )uη = 0 and therefore, (A − λIn )Buη = 0, for every B ∈ G. Hence, (A − λIn )(G(uη )) = {0}. Since G(uη ) = Cn , A = λIn . Write n+1
m k Bk
λ = eθ , θ ∈ C and A = ek=1 for some m1 , . . . , mn+1 ∈ N. It follows that n+1 mk Bk = θIn +2iπdiag(s1 , . . . , sn ), for some s1 , . . . , sn ∈ Z. In particular, k=1 n+1
mk Bk uη = θuη + 2iπdiag(s1 , . . . , sn )uη . Hence,
k=1 n+1
mk uk = θuη + 2iπdiag(s1 , . . . , sn )uη .
k=1
It follows that n
− 2πmk ek + mn+1 2iπu =
k=1
So
n
(θ + 2iπsk )ek
k=1
− 2πmk ek +
k=1
n
n
mn+1 2iπ(αk + iβk )ek =
k=1
n
(θ + 2iπsk )ek
k=1
Therefore, for every 1 ≤ k ≤ n, −2πmk + mn+1 2iπ(αk + iβk ) = θ + 2iπsk . This implies that −2πmk + mn+1 2iπ(αk + iβk ) = −2πm1 + mn+1 2iπ(α1 + iβ1 ) + 2iπ(sk − s1 ). Hence, mk + mn+1 βk = m1 + mn+1 β1 , for every 1 ≤ k ≤ n. In particular for k = n, we get mn+1 = 0 (since n ≥ 2 and 1, β1 , βn are linearly independent over Q). It follows that sk = s1 and mk = m1 , for every 1 ≤ k ≤ n. Therefore, −2πm1 = θ + 2iπs1 and hence, λ = e−2πm1 . We conclude that G(uη ) ∩ M ⊂ {e−2πm1 uη : m1 ∈ N} and hence, G(uη ) ∩ M ⊂ {e−2πm1 uη : m1 ∈ N} ∪ {0} = M .
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SALAH HERZI AND HABIB MARZOUGUI
Remark 5.3. Notice that in the theorem above, the subspace M = Cuη is not G-invariant. Indeed, otherwise eBk uη ∈ Cuη and hence, (k)
(k)
ex1,1 = · · · = exn,1 . As xk,1 = −2π and xj,1 = 0, for j = k, we get e−2π = 1, a contradiction. (k)
(k)
We give below another example for the particular case n = 2. Example 5.4. Let consider the example in ([3], example 7.2): Let G be the abelian semigroup generated by
2π
0 e 1 0 A1 = , A2 = , 2π 1 0 e2π
√ √ 1 √ 0 1 0 −2π( 2+i 3) A3 = , A4 = e 2iπ 1 2π(1 − i 5) 1 We have that G(e1 ) = C2 and G(e1 ) ∩ Ce1 = {e2πn1 e1 : n1 ∈ N}. Hence, G(e1 ) ∩ Ce1 = Ce1 .
6. Somewhere subspace-hypercyclicity Let M be a nontrivial subspace of Kn and x ∈ M . We address the following question for any abelian sub-semigroup G of Mn (K): Does G(x) ∩ M being somewhere dense in M imply that it is everywhere dense in M ? In the particular case where M = Cn , G is finitely generated and K = C, the answer is positive which is due to Feldman. Theorem 6.1 ([8], corollary 5.7). Let G be a finitely generated abelian semigroup of matrices on Cn . Then every somewhere dense orbit of G will be dense in Cn . • If G is infinitely generated, the answer is negative: Proposition 6.2. There exists an infinitely generated abelian semigroup G of Mn (K), a nontrivial subspace M of Kn and x ∈ Kn such that G(x) ∩ M is somewhere dense in M but G(x) ∩ M = M.
α 0 : α ∈ K, |α| ≥ 1 . Observe that G is an Proof. Let G = 0 In−1 generated; Indeed, abelian semigroup of Mn (K). In addition, G is infinitely
0 αi ; 1 ≤ i ≤ p, otherwise, G will be generated by p matrices 0 In−1 where α1 , . . . , αp ∈ K with |α1 | ≥ 1, . . . , |αp | ≥ 1. So for every α ∈ K, |α| ≥ n 1, one canwrite α = α1n1 . . . αp p , for somen1 , . . . , np ∈ N. So p log |α| = i=1 ni log |αi |. Hence, R+ = pi=1 N log |αi | and therefore, R = p i=1 Z log |αi |. This implies that R is a vector space of finite dimension over Z, a contradiction.
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
15
Now let M = Ke1 . We have that G(e1 ) ∩ M = {αe1 : α ∈ K, |α| ≥ 1} = M and clearly G(e1 ) ∩ M is somewhere dense in M since {αe1 : α ∈ K, |α| > 1} ⊂ G(e1 ) ∩ M is open in M . • If G is finitely generated and K = R, the answer is negative: Proposition 6.3. There is an abelian semigroup G of Mn (R) generated by two matrices and a subspace M of Rn such that G(e1 ) ∩ M is somewhere dense in M but G(e1 ) ∩ M = M . Proof. Let a, b ∈ R+ such {ak bn : k, n ∈ N} is dense R+ . Let G be the abelian semigroup generated by A = diag (a, In−1 ) , B = diag (b, In−1 ), where In−1 is the identity on Rn−1 and set M = Re1 . We have that G(e1 ) ∩ M = R+ e1 = M and it has non-empty interior in M . Hence, G(e1 ) ∩ M is somewhere dense in M . • If G is finitely generated and K = C, the answer is positive, whenever dim(M ) = 1: Proposition 6.4. Let G be a finitely generated abelian semigroup of Mn (C), x ∈ Cn and set M = Cx. If G(x) ∩ M is somewhere dense in M then G(x) ∩ M = M . Proof. Let Gx := {A ∈ G : Ax ∈ M }. Then Gx is an abelian sub-semigroup of G which is finitely generated (since G is abelian). In addition, G(x)∩M = Gx (x) and M is Gx -invariant (indeed if A ∈ Gx and y ∈ M , then y = αx for some α ∈ C and so A(y) = αA(x) ∈ M ). The restriction (Gx )|M is a finitely generated abelian semigroup of C. If G(x) ∩ M is somewhere dense in M , then the orbit Gx (x) = (Gx )|M (x) is somewhere dense in M and by theorem 6.1, Gx (x) is dense in M i.e. G(x) ∩ M = M . Proof of theorem 1.8: Assume that G is subspace-hypercyclic for a subspace M of Cn . Then by proposition 2.9, G is subspace-hypercyclic for a straight line Δ = Cx ⊂ M , x ∈ Cn . For every i ∈ {1, . . . , p} and ni ∈ N0 , write ni = qi ki + ri , 0 ≤ ri < ki . We have that n G(x) ∩ Δ = An1 1 . . . Ap p x : (n1 , . . . , np ) ∈ Np ∩ Δ =
=
k 1 −1
···
kp −1
r1 =0
rp =0
k 1 −1
kp −1
r1 =0
···
k r (Ak11 )q1 . . . (App )qp Ar11 . . . App x : (q1 , . . . , qp ) ∈ Np ∩ Δ r
Gk (Ar11 . . . App x) ∩ Δ
rp =0 r
Since G(x) ∩ Δ = Δ, there exist r1 , . . . , rp such that Gk (Ar11 . . . App x) ∩ Δ r is somewhere dense in Δ. So by proposition 6.4, Gk (Ar11 . . . App x) ∩ Δ = Δ and hence, Gk is subspace-hypercyclic for Δ.
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SALAH HERZI AND HABIB MARZOUGUI
Example 6.5. Let G be the abelian semigroup generated by Ai = diag (Ti , In−2 ), 1 ≤ i ≤ 4, where n ≥ 2, 1 0 1 0 2π 2π , T3 = and T1 = diag(e , e ), T2 = 2π 1 2iπ 1 √ √ 1 √ 0 . T4 = e−2π( 2+i 3) 2π(1 − i 5) 1 Then for every k = (k1 , . . . k4 ) ∈ N4 , Gk is subspace-hypercyclic for a straghtline in E2 = C2 × {0}n−2 . Indeed, following ([3], example 7.2), the abelian semigroup G1 of M2 (C) generated by Ti , 1 ≤ i ≤ 4, is hypercyclic. Let u1 ∈ C2 be a hypercyclic vector for G1 , so G1 (u1 ) = C2 . Set u = [u1 , 0, . . . , 0]T ∈ Cn . Then G(u) ∩ E2 = G(u) = [G1 (u1 ), 0n−2 ]T and therefore, G(u) ∩ E2 = E2 . Hence, G is subspace-hypercyclic for E2 . So by theorem 1.8, the example follows. Open questions. Let M be a nontrivial subspace of Cn and let G be an abelian semigroup of Mn (C). (1) If G is somewhere subspace-hypercyclic for M , does G will be subspacehypercyclic for M ? (2) If G is subspace-hypercyclic for M , does Gk is also subspace-hypercyclic for M for every k = (k1 , . . . kp ) ∈ Np0 ? (3) Can theorem 1.6 be extended to non diagonalizable matrices over C. Acknowledgements. The authors are thankful to the referee for his/her helpful remarks which improve the presentation of the paper. This work was supported by the research unit: “Dynamical systems and their applications” (UR17ES21), of Higher Education and Scientific Research, Tunisia.
References [1] Abels, H., Manoussos, A.: Topological generators of abelian Lie groups and hypercyclic finitely generated abelian semigroups of matrices, Adv. Math. 229, 1862–1872 (2012) [2] Ayadi, A., Marzougui, H.: Dense orbits for abelian subgroups of GL(n, C). Foliations 2005, pp. 47–69, World Scientific, Hackensack, NJ (2006) [3] Ayadi, A., Marzougui, H.: Hypercyclic abelian semigroups of matrices on Cn , Proc. Edinb. Math. Soc. 57, 323–338 (2013). [4] Bamerni, N., Kadets, V., Kili¸cman, A.: Hypercyclic operators are subspace hypercyclic, J. Math. Anal. Appl. 435 (2), 1812–1815 (2016). [5] Bayart, F., Matheron, E.: Dynamics of Linear Operators, Cambridge Tracts in Mathematics, 179, Cambridge University Press (2009). [6] Bourdon, P.S, Feldman, N.S. and Shapiro, J.H: Some properties of N -supercyclic operators, Studia Math. 165, 135–157 (2004) [7] Costakis, G., Hadjiloucas, D., Manoussos, A.: On the minimal number of matrices which form a locally hypercyclic, non-hypercyclic tuple, J. Math. Anal. Appl. 365, 229–237 (2010) [8] Feldman, N.S.: Hypercyclic tuples of operators and somewhere dense orbits, J. Math. Anal. Appl. 346, 82–98 (2008) [9] Grosse-Herdmann, K.G., Peris, A.: Linear Chaos, Universitext, Springer (2011).
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[10] Hardy, G.H and Wright, E.M.: An introduction to the Theory of Numbers, Oxford University Press (2008). [11] Madore, B. F. and Mart´ınez-Avenda˜ no, R. A.: Subspace hypercyclicity, J. Math. Anal. Appl. 373 (2), 502–511 (2011). [12] Rezaei, H.: Notes on subspace-hypercyclic operators, J. Math. Anal. Appl. 397, 428– 433 (2013). [13] Shkarin, S.: Hypercyclic tuples of operator on Cn and Rn , Linear and Multilinear Alg. 60, 885–896 (2011). [14] Bourdon, P. S. and Feldman, N. S.: Somewhere dense orbits are everywhere dense, Indiana Univ. Math. J. 52 (3), 811–819 (2003). [15] Le, C. M.: On subspace-hypercyclic operators, Proc. Amer. Math. Soc. 139 (8), 2847– 2852 (2011). Salah Herzi, University of Carthage, Preparatory Engineering Institute of Bizerte, Jarzouna. 7021. Tunisia. Email address: [email protected] Habib Marzougui, University of Carthage, Faculty of Science of Bizerte, (UR17ES21), “Dynamical systems and their applications”, 7021, Jarzouna, Tunisia Email address: [email protected]
Salah Herzi, Habib Marzougui
PII:
S0022-247X(20)30122-0
DOI:
https://doi.org/10.1016/j.jmaa.2020.123960
Reference:
YJMAA 123960
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
17 October 2019
Please cite this article as: S. Herzi, H. Marzougui, Subspace-hypercyclic abelian linear semigroups, J. Math. Anal. Appl. (2020), 123960, doi: https://doi.org/10.1016/j.jmaa.2020.123960.
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SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS SALAH HERZI AND HABIB MARZOUGUI Abstract. We consider subspace-hypercyclicity for abelian semigroups G of matrices on Kn (n ≥ 1), K = R or C. We say that G is subspacehypercyclic for a non-zero subspace M of Kn , if there exists x ∈ Kn such that G(x) ∩ M is dense in M . We give an effective way of checking that a given abelian semigroup is subspace-hypercyclic. We construct a nontrivial example of abelian semigroup which has an orbit dense relative to a certain straight line but is not subspace-hypercyclic for this straight line. Additionally, we construct, for every n ≥ 2, an example of abelian semigroup generated by two matrices which is subspace hypercyclic but not hypercylic. We determine the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn . On the other hand, we prove that Gk is subspace-hypercyclic for every k ∈ Np0 whenever G is a subspace-hypercyclic abelian semigroup of matrices on Cn .
1. Introduction Let K = R or C, Mn (K) be the set of all square matrices over K of order n ≥ 1 and let GL(n, K) be the group of invertible matrices of Mn (K). Let G be an abelian sub-semigroup of Mn (K). By a sub-semigroup of Mn (K), we mean a subset of Mn (K) which is stable under multiplication and contains the identity matrix. For a vector v ∈ Kn , we consider the orbit of G through v: G(v) = {Av : A ∈ G} ⊂ Kn . A subset E ⊂ Kn is called G-invariant if A(E) ⊂ E for any A ∈ G. The orbit G(v) ⊂ Kn is dense in Kn if G(v) = Kn , where E denotes the closure of a subset E ⊂ Kn . The semigroup G is called hypercyclic if there exists a vector v ∈ Kn such that G(v) is dense in Kn . In this case, v is called hypercyclic vector for G. We refer the reader to the books [5] and to [9] and papers [1], [2], [3], [7], [8], [13] for a thorough account on hypercyclicity. Here and throughout the paper, N and Z denote the sets of non-negative integers and integers, respectively, N = {0, 1, 2, . . . } while N0 = N\{0} denotes the set of positive integers. In [11] B.F. Madore and R.A. Mart´ınez-Avenda˜ no consider for a single operator A on a Banach space X the concept of the density of orbits in a nontrivial subspace M of X instead of X, this notion is called subspacehypercyclicity. They proved in particular that no operator on Kn , n ∈ N0 , 2000 Mathematics Subject Classification. 47A16. Key words and phrases. hypercyclic, subspace-hypercyclic, semigroup, abelian, somewhere dense, dense orbit. 1
2
SALAH HERZI AND HABIB MARZOUGUI
can be subspace-hypercyclic (see [11], theorem 4.9). Moreover, they showed that there are operators on a separable Hilbert space which are subspacehypercyclic but not hypercyclic. However, they asked whether hypercyclicity of an operator on a Banach space implies subspace-hypercyclicity. Bamerni et al. [4] provide an affirmative answer to the above question. More examples and results of subspace-hypercyclic operators can be found in ([15], [11], [12]). Here we are interested to introduce this new notion for abelian semigroups of matrices on Kn , n ∈ N0 because phenomena of hypercyclicity for those semigroups can arise in Kn (see e.g. [2], [3]), and so are of their subspacehypercyclicity by [4]. Definition 1.1. We say that G is subspace-hypercyclic for a non-zero subspace M of Kn , if there exists x ∈ Kn such that G(x) ∩ M is dense in M . In this case, x is called subspace-hypercyclic vector for G. We say that G is somewhere subspace-hypercyclic if there exists x ∈ Kn and a non-zero subspace M of Kn such that G(x) ∩ M is somewhere dense in M . Notice that subspace-hypercyclicity is reduced to that of hypercyclicity whenever M = Kn or n = 1. For an abelian semigroup G generated by A1 , . . . , Ap of Mn (C), we denote k by Gk the semigroup generated by Ak11 , . . . , App defined by m k 1 k1 Gk = {Am . . . . Ap p p : m1 , . . . , mp ∈ N}, where k = (k1 , . . . , kp ) ∈ Np0 . 1 The main purpose of this paper is to provide firstly, for any abelian subsemigroup of Mn (C) an effective way of checking that a given semigroup is subspace-hypercyclic. In particular, we show that no matrix on Cn can be subspace-hypercyclic. Also, we show that there are subspace-hypercyclic semigroups that are not hypercyclic (see proposition 3.1). We also prove that the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn for a subspace M of dimension k is k + 1. Secondly, we show that there are semigroups on Rn that have somewhere dense orbits in a subspace M that are not everywhere dense in M (see proposition 6.3). On the other hand, we show that Gk is subspace-hypercyclic for every k ∈ Np0 whenever G is subspace-hypercyclic finitely abelian sub-semigroup of Mn (C). To state our main results, we need to introduce the following notations and definitions. Let n ∈ N0 be fixed. By a partition of n we mean a r ni = n. finite sequence of positive integers η = (n1 , . . . , nr ) such that i=1
The number r will be called the length of the partition. Given a partition η = (n1 , . . . , nr ), we define the following: • Kη (C) := Tn1 (C) ⊕ · · · ⊕ Tnr (C), where Tm (C) (m = 1, 2, . . . , n) is the set of lower-triangular matrices over C with only one eigenvalue. Obviously Kη (C) is a sub-semigroup of Mn (C). In particular: If r = 1, Kη (C) = Tn (C). If r = n, Kη (C) is the semigroup of diagonal
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
3
matrices on Cn . • Kη∗ (C) := Kη (C) ∩ GL(n, C), it is a sub-semigroup of GL(n, C). • B0 = {e1 , . . . , en } the canonical basis of Cn . The following “normal form of G” allows us to deduce the results for an arbitrary semigroup: Proposition 1.2 ([3], proposition 2.4). Let G be an abelian sub-semigroup of Mn (C), n ∈ N0 . Then there exists a P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η of n. Let G be an abelian sub-semigroup of Kη (C), for some partition η = (n1 , . . . , nr ) of n. For each A = diag(A1 , . . . , Ar ) ∈ G and for each 1 ≤ j ≤ r, we denote by • Gj = {Aj : A ∈ G} • σ(Gj ) = ∪ σ(Aj ) : the spectrum of Gj i.e. the set of eigenvalues of all A∈G
elements Aj of Gj . If G is an abelian sub-semigroup of Mn (C) and P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), we denote by • Gj = (P −1 GP )j and σ(Gj ) = σ((P −1 GP )j ). Our principal results can now be stated as follows: Theorem 1.3. Let G be an abelian sub-semigroup of Mn (C), n ∈ N0 , and let P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η = (n1 , . . . , nr ) of n. Then the following assertions are equivalent: (i) G is subspace-hypercyclic. (ii) σ(Gj ) is dense in C, for some j ∈ {1, . . . , r}. (iii) G is subspace-hypercyclic for the straight line CP en1 +···+nj , for some j ∈ {1, . . . , r}. Corollary 1.4. Let G be an abelian sub-semigroup of Mn (C) generated by p matrices A1 , . . . , Ap (p ≥ 1) and let P ∈ GL(n, C) such that P −1 GP ⊂ Kη (C), for some partition η = (n1 , . . . , nr ) of n. Let σ(Ai ) = {λi,j : 1 ≤ j ≤ r} denote the set of eigenvalues of Ai , 1 ≤ i ≤ p. Then the following assertions are equivalent: (i) G is subspace-hypercyclic. kp 1 . . . λp,j : k1 , . . . , kp ∈ N} is dense in C, for some j ∈ {1, . . . , r}. (ii) {λk1,j (iii) G is subspace-hypercyclic for the straight line CP en1 +···+nj , for some j ∈ {1, . . . , r}. Corollary 1.5 ([11], theorem 4.9). For n ∈ N0 , there is no subspacehypercyclic matrix on Cn . On the other hand, we determine the minimal number of diagonalizable matrices over C that generate a subspace-hypercyclic abelian semigroup on Cn .
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SALAH HERZI AND HABIB MARZOUGUI
Theorem 1.6. Let M be a subspace of Cn . The minimum number of diagonalizable matrices generating a subspace-hypercyclic abelian semigroup of Mn (C) for a subspace M is dim(M ) + 1. Theorem 1.6 extends partially corollary 2.2 obtained for M = Cn corresponding to hypercyclicity. In the following theorem, we deal with answer negatively whether, given a subspace M of Cn and x ∈ M , does M ⊂ G(x) imply that G(x) ∩ M = M i.e. G is subspace-hypercyclic for M ? Theorem 1.7. For every n ≥ 2, there exists a hypercyclic abelian semigroup G of matrices of Mn (C) with a hypercyclic vector x, but not subspacehypercyclic for Cx. Feldman showed (corollary 5.8, [8]) that if a finitely generated abelian semigroup G of matrices over C is hypercyclic, then for every k = (k1 , . . . kp ) ∈ Np0 , Gk is hypercyclic. Here we ask if similar result holds for subspacehypercyclicity. We prove the following theorem: Theorem 1.8. Let G be a finitely generated abelian semigroup of Mn (C). If G is subspace-hypercyclic, then for every k = (k1 , . . . kp ) ∈ Np0 , Gk is subspace-hypercyclic. Moreover, if G is subspace-hypercyclic for a subspace M then Gk is subspace-hypercyclic for a straight line in M . 2. Some results on hypercyclicity and subspace-hypercyclicity First we introduce the following notations. • In the identity matrix on Cn . For a row vector v ∈ Cn , we denote by v T the transpose of v. We also have that • uη = [eη,1 , . . . , eη,r ]T ∈ Cn , where eη,k = [1, 0, . . . , 0]T ∈ Cnk , k = 1, . . . , r. (k)
(k)
(k)
• vη = [(vη )1 , . . . , (vη )r ] ∈ Cn , where for every j = 1, . . . , r, if j = k 0 ∈ Cnj (k) (vη )j = if j = k eη,k (k)
• If r = n, then uη = [1, . . . , 1]T and vη k-position and 0’s on the other positions.
= ek : vector with 1 on the
Consider the matrix exponential map exp : Mn (C) → GL(n, C), set exp(M ) = eM . Finally, for a semigroup G of Mn (K), we denote by G∗ = G ∩ GL(n, K). Recall the following theorem which characterizes the hypercyclicity of any abelian semigroup of matrices on Cn . Theorem 2.1 ([3], theorem 1.1). Let G be an abelian sub-semigroup of Kη (C) for some partition η with length r, 1 ≤ r ≤ n. Assume that G∗
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
5
is finitely generated by p matrices (p ≥ 1) and let B1 , . . . , Bp ∈ Kη (C) such that eB1 , . . . , eBp generate G∗ . Then G is hypercyclic if and only if p r (k) NBk uη + 2iπZvη is dense in Cn . k=1
k=1
Corollary 2.2 ([3]). For n ∈ N0 , the minimum number of matrices of Mn (C) that generate a hypercyclic abelian semigroup is n + 1. In particular: Corollary 2.3 ([6]). For n ∈ N0 , there is no hypercyclic matrix on Cn . For the particular case r = n, we obtain the following. Theorem 2.4 ([3], theorem 1.1). Let G be an abelian sub-semigroup of diagonal matrices of Mn (C). Assume that G∗ is finitely generated by p matrices (p ≥ 1) and let B1 , . . . , Bp ∈ Kη (C) such that eB1 , . . . , eBp generate p n NBk uη + 2iπZek is dense in G∗ . Then G is hypercyclic if and only if Cn .
k=1
k=1
Corollary 2.5 ([3], corollary 1.8). For n ∈ N0 , the minimum number of diagonalizable matrices of Mn (C) that generate a hypercyclic abelian semigroup is n + 1. Theorem 2.6 ([4]). Let A be a dense subset of a Banach space X with dim(X) > 1. Then there is a nontrivial closed subspace M of X such that A ∩ M is dense in M . By a nontrivial subspace M of X, we mean that M is non-zero and distinct from X. Remark 2.7. Notice that the theorem 2.6 is not true for dim(X) = 1. As a consequence: Corollary 2.8. Let G be an abelian semi-group of Mn (K), n ≥ 2. If G is hypercyclic, then it is subspace-hypercyclic for a nontrivial subspace M of Kn . Proposition 2.9. Let G be an abelian semi-group of Mn (K), n ≥ 2. If G is subspace-hypercyclic, then it is subspace-hypercyclic for a straight line Δ = Kz, z ∈ Kn . Proof. Let x ∈ Kn such that G(x) ∩ M = M , where M is a subspace of Kn . If dimM = 1, the proposition 2.9 is obvious. So assume that dimM > 1. According to theorem 2.6, there is a nontrivial subspace M1 of M such that G(x)∩M ∩M1 = G(x)∩M1 is dense in M1 . Now, if M1 is a straight line, the result follows, otherwise the theorem 2.6 is applied again and so on until we get a straight line Δ = Kz (z ∈ Kn ) such that G(x) ∩ Δ is dense in Δ.
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SALAH HERZI AND HABIB MARZOUGUI
Lemma 2.10. Let G be a finitely generated abelian sub-semigroup of Mn (K). If G is subspace-hypercyclic for a non-zero subspace M of Kn , then there exists y ∈ M such that G(y) ∩ M is dense in M . Proof. Let A1 , . . . , Ap generate G and let x ∈ Cn such that G(x)∩M is dense in M . Wewill prove that there exists y ∈ M such that G(x)∩M = G(y)∩M . n Set E1 = n1 ∈ N : ∃(n2 , . . . , np ) ∈ Np−1 ; An1 1 An2 2 . . . Ap p x ∈ M . Since G(x) ∩ M = ∅, E1 = ∅. So let k1 = min(E1 ). Now set n
Ek1 = {n2 ∈ N : ∃n1 ≥ k1 , ∃(n3 , . . . , np ) ∈ Np−2 ; An1 1 An2 2 . . . Ap p x ∈ M }. n
Then Ek1 = ∅ (since ∃(n2 , . . . , np ) ∈ Np−1 such that Ak11 An2 2 . . . Ap p x ∈ M ). So set k2 = min(Ek1 ). In this way, we construct by induction, n
Ek1 ,k2 = {n3 ∈ N : ∃n1 ≥ k1 , ∃n2 ≥ k2 , ∃(n4 , . . . , np ) ∈ Np−3 ; An1 1 An2 2 . . . Ap p x ∈ M }, n
. . . , Ek1 ,...,kp−1 = {np ∈ N : ∃n1 ≥ k1 , . . . , ∃np−1 ≥ kp−1 ; An1 1 . . . Ap p x ∈ M }, with k3 = min(Ek1 ,k2 ), . . . , kp = min(Ek1 ,...,kp−1 ). In particular, we see that k
y := Ak11 . . . App x ∈ M . Observe that G(x) ∩ M
n
= {An1 1 . . . Ap p x : (n1 , . . . , np ) ∈ Np } ∩ M n
= {An1 1 . . . Ap p x : n1 ≥ k1 , . . . , np ≥ kp } ∩ M = G(y) ∩ M, In result, G(y) ∩ M is dense in M .
3. Subspace-hypercyclic abelian semigroups Proof of theorem 1.3. We may assume, by taking P −1 GP if necessary that, that G ⊂ Kη (C). Indeed, G := P −1 GP is a semigroup of Kη (C) and the property G(x) ∩ M = M with x ∈ Cn is equivalent to that of G (x ) ∩ M = M , where M = P −1 (M ) and x = P −1 (x) ∈ Cn . Furthermore dim(M ) = dim(M ). (i) ⇒ (ii): Suppose that G is subspace hypercyclic for a subspace M of Cn of dimension k. Then there is x = [x1 , . . . , xn ]T ∈ Cn such that G(x) ∩ M is dense in M . Let B = {y1 , y2 , . . . , yk } be a basis of M with yi = [y1,i , . . . , yn,i ]T . Since dim(M ) = k, the n − k + 1 first rows of B are not all zero. So there is 1 ≤ i ≤ n such that Li = (yi,1 , . . . , yi,k ) is the first non-zero row of B. One can then assume that yi,1 = 1; yi,2 = · · · = yi,k = 0. Let 1 ≤ j ≤ r be such that 0 i=1
j−1 i=1
ni < i ≤
j
ni with the convention
k=1
ni = 0. If y ∈ G(x) ∩ M then y = [0, . . . 0, di , . . . , dn ]T = Ax, where
A = diag(B1 , . . . , Br ) ∈ G. So the ith -row in A corresponds to a row in Bj . As for 1 ≤ l ≤ r, Bl has only one eigenvalue λl , then di = λl xi . Now for every
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
c1 , . . . , ck ∈ C, y =
k
7
ci yi ∈ M and its ith -coordinate in Cn is ci . There-
i=1
fore, from G(x) ∩ M = M and above, we deduce that {λj xi : λj ∈ σ(Gj )} is dense in C. In particular xi = 0 and hence, σ(Gj ) is dense in C. (ii) ⇒ (iii) Set Δ = Cen1 +···+nj . Then G is subspace-hypercyclic for Δ. Indeed, en1 +···+nj is an eigenvector of any element of G, so G(en1 +···+nj ) ∩ Δ = G(en1 +···+nj ) = λj en1 +···+nj : λj ∈ σ(Gj ) which is dense in Δ by the hypothesis (ii). (iii) ⇒ (i) is clear.
Proof of corollary 1.4. We may assume, by taking P −1 GP if necessary that, that G ⎛ ⊂ Kη (C). For each ⎞ i ∈ {1, . . . , p}, Ai = diag(Ti,1 , . . . , Ti,p ), 0 λi,j 0 ⎜ ⎟ where Ti,j = ⎝ ∗ . . . 0 ⎠, j ∈ {1, . . . , r}. As any element of G can ∗
∗
λi,j k k1 form A1 . . . App , for some k1 , . . . , kp ∈ N, then for each kp 1 2 = {λk1,j λk2,j . . . λp,j : k1 , . . . , kp ∈ N}. Hence, corollary 1.4
be written of the 1 ≤ j ≤ r, σ(Gj ) follows from theorem 1.3.
Proof of corollary 2.3. This follows from corollary 1.4, (ii) applied for the particular case p = 1. Notice that corollary 2.3 can also be proven by the following argument. For a single matrix A ∈ Mn (K) and x ∈ Kn , it is known that one of the following assertions holds: lim Ak x = 0 or lim Ak x = +∞ or k→+∞
k→+∞
∃ c1 , c2 > 0 such that c1 ≤ Ak x ≤ c2 for all k ∈ N, where is any norm on Cn . In particular the orbit of x by A is not dense in M for any nontrivial subspace M of Kn . Contrarily to a single matrix, subspace-hypercyclicity can occur for semigroups: Proposition 3.1. For every n ≥ 2, there exist two matrices of Mn (C) generating an abelian semi-group G which is subspace-hypercyclic, but not hypercyclic. Proof. Let a, b ∈ C such that {ak bn : k, n ∈ N} is dense in C (see e.g. [8], proposition 3.1). Let G be the abelian semigroup generated by A = diag(a, In−1 ) and B = diag(b, In−1 ), where In−1 is the identity matrix on Cn−1 . Then G is a subspace-hypercyclic for the subspace Ce1 since G(e1 ) ∩ Ce1 = G(e1 ) = Ce1 . The fact that G is not hypercyclic follows from corollary 2.2 since n ≥ 2. Remark 3.2.
8
SALAH HERZI AND HABIB MARZOUGUI
(1) In the proof of proposition 3.1, the fact that G is subspace-hypercyclic for the subspace Ce1 also follows from corollary 1.4, since G is an abelian semigroup of Kη (C), where η = (n1 , n2 ), n1 = 1, n2 = n − 1, j = 1, en1 = e1 . (2) The proposition 3.1 shows that the semigroup G is subspace-hypercyclic for Ce1 , but not for its orthogonal complement (Ce1 )⊥ . As a consequence of proposition 3.1 and corollary 2.3, we get: Corollary 3.3. The minimal number of matrices generating a subspacehypercyclic abelian semigroup over C is 2. In the following, we give an explicit parameter family (Gθ )θ∈R of abelian semigroups which are subspace-hypercyclic if and only if θ is irrational. Example 3.4. Let p, q ≥ 2 two coprime numbers and θ ∈ R. For every n ≥ 3, let Gθ be the abelian semigroup generated by A1 = diag (T1 , In−2 ), A2 = diag (T2 , In−2 ) and A3 = diag (T3 , In−2 ), where T1 , T2 , T3 ∈ T2 (C) with only one eigenvalue p, 1q and e2iπθ respectively. Then: (i) Gθ is subspace-hypercyclic for Ce2 , but not hypercyclic whenever θ is irrational. (ii) Gθ is not subspace-hypercyclic whenever θ is rational. Proof. We have that Gθ ⊂ Kη (C) with η = (n1 , n2 ), n1 = 2, n2 = n − 2 and k r = 2. By ([8], corollary 4.2), the set { qpm : k, m ∈ N} is dense R+ . k
(i) If θ is irrational, then the set { qpm e2iπsθ : k, s, m ∈ N} is dense C. By corollary 1.4, Gθ is subspace-hypercyclic for Ce2 (since j = 1 and en1 = e2 ). Moreover, Gθ is not hypercyclic by corollary 2.2 since n ≥ 3. k (ii) If θ is rational, then the set { qpm e2iπsθ : k, s, m ∈ N} is nowhere dense C. Therefore, from corollary 1.4, Gθ is not subspace-hypercyclic. The question whether there exists a subspace-hypercyclic abelian semigroup G for a nontrivial subspace M of Cn which is not G-invariant is answered positively as follows. Proposition 3.5. For every n ≥ 2, there is an abelian semigroup G of Mn (C) which is subspace-hypercyclic for a subspace M , but M is not Ginvariant. Proof. Let a, b ∈ C such {ak bn : k, n ∈ N} is dense C. Set M = Ce1 and let G be the semigroup generated by A1 = diag (T1 , In−2 ), A2 = diag (T2 , In−2 ) and A3 = diag (T3 , In−2 ), where
1 0 a 0 b 0 , T2 = , T3 = T1 = 1 1 −a a −b b We have that An1 1 An2 2 An3 3 (e1 ) = an2 bn3 [1, n1 − (n2 + n3 ), 0, . . . , 0]T .
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
9
Therefore, G(e1 ) ∩ M = {an2 bn3 e1 : n2 , n3 ∈ N} which is dense in M . So G is subspace-hypercyclic for M , but clearly M is not G-invariant. Remark 3.6. By applying corollary 1.4 to the example in proposition 3.5, we see that η = (n1 , n2 ), n1 = 2, n2 = n − 2, j = 1, en1 = e2 . It follows that G is subspace hypercyclic for Ce2 , however, Ce2 is G-invariant. In ([11], theorem 4.5) B.F. Madore and R.A. Mart´ınez-Avenda˜ no proved that if an operator T on a Hilbert space X is subspace-hypercyclic, then its spectrum σ(T ) meets the unit circle S 1 i.e. σ(T ) ∩ S 1 = ∅. This result fails for abelian semigroup. Indeed, first we define the spectrum of an abelian semigroup G of Mn (C) generated by p matrices Ai , 1 ≤ i ≤ p, by σ(G) := ∪ σ(Ai ). Now consider A1 = aIn , A2 = bIn , where a, b ∈ C 1≤i≤k
such {ak bn : k, n ∈ N} is dense C. Then σ(G) = {a, b} and by theorem 1.3, the semigroup G generated by A1 and A2 is subspace-hypercyclic for Cen . However, σ(G) ∩ S 1 = ∅. 4. Minimal generators for subspace-hypercyclic semigroup The aim of this section is to prove theorem 1.6. • Firstly, we will show that there exists an abelian semigroup of Mn (C) generated by k + 1 diagonalizable matrices which is subspace-hypercyclic for Ek = Ck × {0}n−k . Following corollary 2.5, there exists a hypercyclic abelian semigroup G1 of Mk (C) generated by k + 1 diagonalizable matrices D1 , . . . , Dk+1 . Let u1 be a hypercyclic vector for G1 and set Ai = diag (Di , In−k ) ∈ Mn (C), for all 1 ≤ i ≤ k + 1, where In−k is the identity matrix of Cn−k . Let G be the semigroup generated by the Ai , 1 ≤ i ≤ k+1 and let u = [u1 , 0, . . . , 0]T ∈ Cn . As G1 (u1 ) = Ck and G(u) ∩ Ek = G(u) = [G1 (u1 ), 0n−k ]T , it follows that G(u) ∩ Ek = Ek . Hence, G is subspace-hypercyclic for Ek . • Secondly, let G be an abelian semigroup of Mn (C) generated by p diagonalizable matrices Ai , 1 ≤ i ≤ p. Assume that G is subspace-hypercyclic for a subspace M of Cn of dimension k. By lemma 2.10, there exists x1 ∈ M such that G(x1 ) ∩ M = M . Claim 1. One can assume that the Ai , 1 ≤ i ≤ p are diagonal. Indeed, following the proposition 1.2, there exists a P ∈ GL(n, C) such that G := P −1 GP is a semigroup of diagonal matrices. From G(x1 ) ∩ M = M , it follows that G (x1 ) ∩ M = M , where x1 = P −1 (x1 ) and M = P −1 (M ). Observe that dim(M ) = dim(M ) and x1 ∈ M . This proves the claim. Assume that x1 has the form x1 = [x1,1 , x2,1 , . . . , xn,1 ]T ∈ M . Let y = [y1 , y2 , . . . , yn ]T ∈ M , there is a sequence of diagonal matrices (Aj )j ∈ G
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SALAH HERZI AND HABIB MARZOUGUI
such that y = lim Aj x1 with Aj x1 ∈ M . Write Aj = diag(d1,j , . . . , dn,j ), j→+∞
then we have y = lim [d1,j x1,1 , d2,j x2,1 , . . . , dn,j xn,1 ]T . j→+∞
Hence, if xi,1 = 0 for some 1 ≤ i ≤ n, then yi = 0. Let B = {x1 , . . . , xk } formed by extending x1 to a basis of M , where xj = [x1,j , x2,j , . . . , xn,j ]T , 1 ≤ j ≤ k and let mj = min{i ∈ {1, . . . , n} : xi,j = 0}. By above, we have that m1 ≤ mj , for every j ∈ {1, . . . , k}. It follows that xj = [0, . . . , 0, xmj ,j , . . . , xn,j ]T and xmj ,j = 0, for all 1 ≤ j ≤ k. Furthermore k ≤ n − m1 + 1 since dim(M ) = k. Upon reordering, and taking appropriate linear combinations of the xj , 1 ≤ j ≤ k, we can assume that m1 < m2 < m3 < · · · < mk ≤ n. We let N = n − m1 + 1. Denote by G(m1 ) the semigroup of MN (C) generated by (m1 )
Aj
(m1 )
= diag(dm1 ,j , . . . , dn,j ), x1
= [xm1 ,1 , . . . , xn,1 ]T ∈ CN
and (m1 )
xj
= [0, . . . , 0, xmj ,j , . . . , xn,j ]T ∈ CN , for 2 ≤ j ≤ k.
We let M (m1 ) = {y = [ym1 , . . . , yn ]T ∈ CN : [0, . . . , 0, ym1 , . . . , yn ]T ∈ M }. It is a subspace of CN . Observe that (m1 )
G(m1 ) (x1
) ∩ M (m1 ) = M (m1 ) .
Without loss of generality, we can write for short (m1 )
G(m1 ) = G, x1 (m1 )
xj
= x1 , with x1,1 = 0,
= xj = [0, . . . 0, xmj ,j , . . . , xn,j ]T , for 2 ≤ j ≤ k
and M (m1 ) = M. Then 1 ≤ k ≤ N and we have G(x1 ) ∩ M = M . Set {ik+1 , . . . , iN } = {m1 , . . . , n}\{m1 , . . . , mk } so that ik+1 < · · · < iN . Let P, Q ∈ GL(N, C) defined by if 1 ≤ j ≤ k P (emj ) = xj if k + 1 ≤ j ≤ N P (eij ) = eij and
Q(ej ) = emj Q(ej ) = eij
if 1 ≤ j ≤ k if k + 1 ≤ j ≤ N
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
11
Set G = (P Q)−1 G(P Q). Then G is an abelian semigroup of MN (C) formed of diagonalizable matrices. As P Q(ej ) = xj , for 1 ≤ j ≤ k, then P Q(Ek ) = M and so G (e1 ) ∩ Ek = Ek .
A1 0 with A1 ∈ Claim 2. Every A ∈ G has the form A = A2 A3 Mk (C), A2 ∈ MN −k,k (C), A3 ∈ MN −k (C). Indeed, let A = (P Q)−1 A(P Q) = (ap,q )1≤p,q≤N ∈ G , where A = (ap,q )1≤p,q≤N ∈ G. As A is diagonal, then ap,q = 0 whenever p = q. Therefore, we have that ∀(p, q) ∈ {1, . . . , k} × {k + 1, . . . , N }, ap,q = eTp (P Q)−1 AP Qeq = eTp Q−1 P −1 Aeiq = aiq ,iq eTp Q−1 eiq = aiq ,iq eTp eq = 0 This proves the claim. Claim 3. p ≥ k + 1.
A1 0 ∈ G and by Indeed, denote by = : = A2 A3 e1 = [1, 0, . . . , 0]T ∈ Ck . Observe that G1 is an abelian semigroup of Mk (C). We have that G (e1 ) ∩ Ek ⊂ G1 (e1 ) × {0}n−k ⊂ Ek . Therefore, from G (e1 ) ∩ Ek = Ek , we get G1 (e1 ) = Ck . In result, G1 is hypercyclic. As the number of generators of G1 is less than that of G and this later is equal to the number p of generators of G, so by corollary 2.2, p ≥ k + 1. G1
A1
A
5. On semigroups with orbits dense relative to nontrivial subspaces Given a subspace M of Cn and x ∈ X, we see that if G(x) ∩ M = M then G(x) is dense in M ; that is M ⊂ G(x). Conversely, does the density of G(x) in M imply that G is subspace-hypercyclic for M ? In this section, we prove theorem 1.7 which answer negatively the above question. We will construct n + 1 diagonal matrices A1 , . . . , An+1 on Cn generating a hypercyclic abelian semigroup G with a hypercyclic vector x but G is not subspace-hypercyclic for M = Cx. For this, let recall the following multidimensional real version of Kronecker’s theorem (see e.g. [13], lemma 2.2; see also [10], theorem 442). Kronecker’s Theorem ([13], lemma 2.2) Let α1 , . . . , αn be negative real numbers such that 1, α1 , . . . , αn are linearly independent over Q. Then the
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SALAH HERZI AND HABIB MARZOUGUI
set
Nn + N[α1 , . . . , αn ]T := [s1 , . . . , sn ]T + k[α1 , . . . , αn ]T : k, s1 , . . . , sn ∈ N
is dense in Rn . By identifying Cn with R2n in the natural way, we deduce the complex version as follows. Theorem 5.1 (Kronecker’s theorem: complex version). Let α1 , . . . , αn ; β1 , . . . , βn be negative real numbers such that 1, α1 , . . . , αn , β1 , . . . , βn are linearly independent over Q. Then Nn + iNn + N[α1 + iβ1 , . . . , αn + iβn ]T is dense in Cn . Lemma 5.2. Let n ∈ N0 . Then there exist n + 1 vectors u1 , . . . , un+1 of Cn such that n+1 n Nuk + 2iπZek k=1
k=1
is dense in Cn . Proof. Let α1 , . . . , αn , β1 , . . . , βn be negative real numbers such that the numbers 1, α1 , . . . , αn , β1 , . . . , βn are linearly independent over Q. Set u = [α1 + iβ1 , . . . , αn + iβn ]T and define
−2πek 2iπu
uk : = Write H :=
n+1 k=1
Nuk +
if 1 ≤ k ≤ n if k = n + 1 n
2iπZek
k=1
Then we have H=
n
− 2πNek + N2iπu +
k=1
n
2iπZek
k=1
= 2iπH where H =
n k=1
iNek + Nu +
n k=1
Zek = Zn + iNn + Nu. By theorem 5.1, we
see that H is dense in Cn , and thus so is H. This proves the lemma.
Proof of theorem 1.7. Let u1 , . . . , un+1 of Cn are those of from the proof (k) (k) of Lemma 5.2. Write uk = [x1,1 , . . . , xn,1 ]T . Let B1 , . . . , Bn+1 be defined by (k) (k) (k) Bk = diag x1,1 , x2,1 , . . . , xn,1 , 1 ≤ k ≤ n + 1. Let G be the abelian sub-semigroup of Kη∗ (C) generated by eB1 , . . . , eBn+1 .
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS (k)
Here r = n, η = (1, . . . , 1), uη = [1, . . . , 1]T and vη Bk uη = uk . Hence, n+1
NBk uη +
k=1
n
2iπZvη(k) =
k=1
n+1
Nuk +
k=1
13
= ek . We have that
n
2iπZek
k=1
=H By lemma 5.2, H = Cn and by theorem 2.1, G(uη ) = Cn . Now denote by M = Cuη the vector space generated by uη . We shall prove that G(uη ) ∩ M = M . Let A ∈ G such that Auη = λuη , for some λ ∈ C. Then (A − λIn )uη = 0 and therefore, (A − λIn )Buη = 0, for every B ∈ G. Hence, (A − λIn )(G(uη )) = {0}. Since G(uη ) = Cn , A = λIn . Write n+1
m k Bk
λ = eθ , θ ∈ C and A = ek=1 for some m1 , . . . , mn+1 ∈ N. It follows that n+1 mk Bk = θIn +2iπdiag(s1 , . . . , sn ), for some s1 , . . . , sn ∈ Z. In particular, k=1 n+1
mk Bk uη = θuη + 2iπdiag(s1 , . . . , sn )uη . Hence,
k=1 n+1
mk uk = θuη + 2iπdiag(s1 , . . . , sn )uη .
k=1
It follows that n
− 2πmk ek + mn+1 2iπu =
k=1
So
n
(θ + 2iπsk )ek
k=1
− 2πmk ek +
k=1
n
n
mn+1 2iπ(αk + iβk )ek =
k=1
n
(θ + 2iπsk )ek
k=1
Therefore, for every 1 ≤ k ≤ n, −2πmk + mn+1 2iπ(αk + iβk ) = θ + 2iπsk . This implies that −2πmk + mn+1 2iπ(αk + iβk ) = −2πm1 + mn+1 2iπ(α1 + iβ1 ) + 2iπ(sk − s1 ). Hence, mk + mn+1 βk = m1 + mn+1 β1 , for every 1 ≤ k ≤ n. In particular for k = n, we get mn+1 = 0 (since n ≥ 2 and 1, β1 , βn are linearly independent over Q). It follows that sk = s1 and mk = m1 , for every 1 ≤ k ≤ n. Therefore, −2πm1 = θ + 2iπs1 and hence, λ = e−2πm1 . We conclude that G(uη ) ∩ M ⊂ {e−2πm1 uη : m1 ∈ N} and hence, G(uη ) ∩ M ⊂ {e−2πm1 uη : m1 ∈ N} ∪ {0} = M .
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SALAH HERZI AND HABIB MARZOUGUI
Remark 5.3. Notice that in the theorem above, the subspace M = Cuη is not G-invariant. Indeed, otherwise eBk uη ∈ Cuη and hence, (k)
(k)
ex1,1 = · · · = exn,1 . As xk,1 = −2π and xj,1 = 0, for j = k, we get e−2π = 1, a contradiction. (k)
(k)
We give below another example for the particular case n = 2. Example 5.4. Let consider the example in ([3], example 7.2): Let G be the abelian semigroup generated by
2π
0 e 1 0 A1 = , A2 = , 2π 1 0 e2π
√ √ 1 √ 0 1 0 −2π( 2+i 3) A3 = , A4 = e 2iπ 1 2π(1 − i 5) 1 We have that G(e1 ) = C2 and G(e1 ) ∩ Ce1 = {e2πn1 e1 : n1 ∈ N}. Hence, G(e1 ) ∩ Ce1 = Ce1 .
6. Somewhere subspace-hypercyclicity Let M be a nontrivial subspace of Kn and x ∈ M . We address the following question for any abelian sub-semigroup G of Mn (K): Does G(x) ∩ M being somewhere dense in M imply that it is everywhere dense in M ? In the particular case where M = Cn , G is finitely generated and K = C, the answer is positive which is due to Feldman. Theorem 6.1 ([8], corollary 5.7). Let G be a finitely generated abelian semigroup of matrices on Cn . Then every somewhere dense orbit of G will be dense in Cn . • If G is infinitely generated, the answer is negative: Proposition 6.2. There exists an infinitely generated abelian semigroup G of Mn (K), a nontrivial subspace M of Kn and x ∈ Kn such that G(x) ∩ M is somewhere dense in M but G(x) ∩ M = M.
α 0 : α ∈ K, |α| ≥ 1 . Observe that G is an Proof. Let G = 0 In−1 generated; Indeed, abelian semigroup of Mn (K). In addition, G is infinitely
0 αi ; 1 ≤ i ≤ p, otherwise, G will be generated by p matrices 0 In−1 where α1 , . . . , αp ∈ K with |α1 | ≥ 1, . . . , |αp | ≥ 1. So for every α ∈ K, |α| ≥ n 1, one canwrite α = α1n1 . . . αp p , for somen1 , . . . , np ∈ N. So p log |α| = i=1 ni log |αi |. Hence, R+ = pi=1 N log |αi | and therefore, R = p i=1 Z log |αi |. This implies that R is a vector space of finite dimension over Z, a contradiction.
ON SUBSPACE-HYPERCYCLIC ABELIAN LINEAR SEMIGROUPS
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Now let M = Ke1 . We have that G(e1 ) ∩ M = {αe1 : α ∈ K, |α| ≥ 1} = M and clearly G(e1 ) ∩ M is somewhere dense in M since {αe1 : α ∈ K, |α| > 1} ⊂ G(e1 ) ∩ M is open in M . • If G is finitely generated and K = R, the answer is negative: Proposition 6.3. There is an abelian semigroup G of Mn (R) generated by two matrices and a subspace M of Rn such that G(e1 ) ∩ M is somewhere dense in M but G(e1 ) ∩ M = M . Proof. Let a, b ∈ R+ such {ak bn : k, n ∈ N} is dense R+ . Let G be the abelian semigroup generated by A = diag (a, In−1 ) , B = diag (b, In−1 ), where In−1 is the identity on Rn−1 and set M = Re1 . We have that G(e1 ) ∩ M = R+ e1 = M and it has non-empty interior in M . Hence, G(e1 ) ∩ M is somewhere dense in M . • If G is finitely generated and K = C, the answer is positive, whenever dim(M ) = 1: Proposition 6.4. Let G be a finitely generated abelian semigroup of Mn (C), x ∈ Cn and set M = Cx. If G(x) ∩ M is somewhere dense in M then G(x) ∩ M = M . Proof. Let Gx := {A ∈ G : Ax ∈ M }. Then Gx is an abelian sub-semigroup of G which is finitely generated (since G is abelian). In addition, G(x)∩M = Gx (x) and M is Gx -invariant (indeed if A ∈ Gx and y ∈ M , then y = αx for some α ∈ C and so A(y) = αA(x) ∈ M ). The restriction (Gx )|M is a finitely generated abelian semigroup of C. If G(x) ∩ M is somewhere dense in M , then the orbit Gx (x) = (Gx )|M (x) is somewhere dense in M and by theorem 6.1, Gx (x) is dense in M i.e. G(x) ∩ M = M . Proof of theorem 1.8: Assume that G is subspace-hypercyclic for a subspace M of Cn . Then by proposition 2.9, G is subspace-hypercyclic for a straight line Δ = Cx ⊂ M , x ∈ Cn . For every i ∈ {1, . . . , p} and ni ∈ N0 , write ni = qi ki + ri , 0 ≤ ri < ki . We have that n G(x) ∩ Δ = An1 1 . . . Ap p x : (n1 , . . . , np ) ∈ Np ∩ Δ =
=
k 1 −1
···
kp −1
r1 =0
rp =0
k 1 −1
kp −1
r1 =0
···
k r (Ak11 )q1 . . . (App )qp Ar11 . . . App x : (q1 , . . . , qp ) ∈ Np ∩ Δ r
Gk (Ar11 . . . App x) ∩ Δ
rp =0 r
Since G(x) ∩ Δ = Δ, there exist r1 , . . . , rp such that Gk (Ar11 . . . App x) ∩ Δ r is somewhere dense in Δ. So by proposition 6.4, Gk (Ar11 . . . App x) ∩ Δ = Δ and hence, Gk is subspace-hypercyclic for Δ.
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SALAH HERZI AND HABIB MARZOUGUI
Example 6.5. Let G be the abelian semigroup generated by Ai = diag (Ti , In−2 ), 1 ≤ i ≤ 4, where n ≥ 2, 1 0 1 0 2π 2π , T3 = and T1 = diag(e , e ), T2 = 2π 1 2iπ 1 √ √ 1 √ 0 . T4 = e−2π( 2+i 3) 2π(1 − i 5) 1 Then for every k = (k1 , . . . k4 ) ∈ N4 , Gk is subspace-hypercyclic for a straghtline in E2 = C2 × {0}n−2 . Indeed, following ([3], example 7.2), the abelian semigroup G1 of M2 (C) generated by Ti , 1 ≤ i ≤ 4, is hypercyclic. Let u1 ∈ C2 be a hypercyclic vector for G1 , so G1 (u1 ) = C2 . Set u = [u1 , 0, . . . , 0]T ∈ Cn . Then G(u) ∩ E2 = G(u) = [G1 (u1 ), 0n−2 ]T and therefore, G(u) ∩ E2 = E2 . Hence, G is subspace-hypercyclic for E2 . So by theorem 1.8, the example follows. Open questions. Let M be a nontrivial subspace of Cn and let G be an abelian semigroup of Mn (C). (1) If G is somewhere subspace-hypercyclic for M , does G will be subspacehypercyclic for M ? (2) If G is subspace-hypercyclic for M , does Gk is also subspace-hypercyclic for M for every k = (k1 , . . . kp ) ∈ Np0 ? (3) Can theorem 1.6 be extended to non diagonalizable matrices over C. Acknowledgements. The authors are thankful to the referee for his/her helpful remarks which improve the presentation of the paper. This work was supported by the research unit: “Dynamical systems and their applications” (UR17ES21), of Higher Education and Scientific Research, Tunisia.
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[10] Hardy, G.H and Wright, E.M.: An introduction to the Theory of Numbers, Oxford University Press (2008). [11] Madore, B. F. and Mart´ınez-Avenda˜ no, R. A.: Subspace hypercyclicity, J. Math. Anal. Appl. 373 (2), 502–511 (2011). [12] Rezaei, H.: Notes on subspace-hypercyclic operators, J. Math. Anal. Appl. 397, 428– 433 (2013). [13] Shkarin, S.: Hypercyclic tuples of operator on Cn and Rn , Linear and Multilinear Alg. 60, 885–896 (2011). [14] Bourdon, P. S. and Feldman, N. S.: Somewhere dense orbits are everywhere dense, Indiana Univ. Math. J. 52 (3), 811–819 (2003). [15] Le, C. M.: On subspace-hypercyclic operators, Proc. Amer. Math. Soc. 139 (8), 2847– 2852 (2011). Salah Herzi, University of Carthage, Preparatory Engineering Institute of Bizerte, Jarzouna. 7021. Tunisia. Email address: [email protected] Habib Marzougui, University of Carthage, Faculty of Science of Bizerte, (UR17ES21), “Dynamical systems and their applications”, 7021, Jarzouna, Tunisia Email address: [email protected]