International Journal of Algebra, Vol. 6, 2012, no. 27, 1319 - 1324 A Note on Semigroups of Linear Transformations with Invariant Subspaces Pei Huisheng Deparment of Mathematics Xinyang Normal University Xinyang Henan, 464000, P. R. China [email protected] Abstract Let V be a vector space over a field and T (V ) the semigroup of all linear transformations from V into V . For a fixed subspace W of V , denote S(V,W)={α ∈ T (V ):Wα⊆ W }. Then S(V,W) is a subsemigroup of T (V ). In this note, we show that the semigroup S(V,W) is always abundant. Keywords: semigroups; Green’s-starred relations; abundant semigroups; vector spaces; linear transformations 1 Introduction Let V be a vector space over a field and T (V ) the semigroup (under composi- tion) consisting of all linear transformations from V into V . Let W be a fixed subspace of V . In [4], R. P. Sullivan studied a kind of linear transformation semigroups involved with the subspace W , T (V,W)={α ∈ T (V ):Vα⊆ W } She described Green’s relations and ideals for the semigroup T (V,W). In [6] and [7], the authors consider another kind of subsemigroups of T (V ) involved with a fixed subspace W , that is, S(V,W)={α ∈ T (V ):Wα ⊆ W }. In [6], Green’s relations and ideals on S(V,W) are characterized. In [7], the regular elements in S(V,W) are described and it is proved that S(V,W)is regular if and only if W = V or W = {0}. 1320 Pei Huisheng Let S be a semigroup. We say that a, b ∈ S are L∗-related (R∗-related) if they are L-related (R-related) in a semigroup T such that S is a subsemigroup of T . The relations L∗ and R∗ are equivalence relations on S. A semigroup S is called abundant if every L∗-class and every R∗-class of S contains an idempotent. It is well-known that regular semigroups are abundant. However, the converse is not true. For example, Umar showed in [5] that the semigroups S−(X) of order-decreasing transformations on a totally ordered finite set X is abundant but not regular. Also, letting T (X) be the full transformation semigroup on a set X, in [3], the authors find out some equivalences E on X for which the semigroup TE(X)={α ∈ T (X):(xα, yα) ∈ E for all (x, y) ∈ E} is abundant but not regular. In this note, we first describe the relations L∗ and R∗ on S(V,W). Then we show that the semigroups S(V,W) is always abundant. For convenience, we adopt the convention used in [4]. We write a subset {ei : i ∈ I} of V as {ei}, letting the subscript denote an (unspecified) index set I. The subspace U of V generated by a linearly independent subset {ei} is denoted by ei and we write dimU = |I|. Often it is necessary to construct some α ∈ T (V ) by first choosing a basis {ei} of V and some {ui}⊆V , and then letting eiα = ui for each i ∈ I and extending this action by linearity to the whole of V . To abbreviate matters, we simply say that given {ei} and {ui} within context, then α ∈ T (V ) is defined by letting ei α = . ui For undefined notation and concepts one can consult [1]. 2 Main results The following Lemma 1 and Lemma 2 come from [2], and Lemma 3 comes from [1, p.63, Exercise 19]. Lemma 1 Let S be a semigroup and α, β ∈ S. Then the following statements are equivalent: (1) (α, β) ∈R∗. (2) For all σ, γ ∈ S1,σα= γα if and only if σβ = γβ. Lemma 2 Let S be a semigroup and α, β ∈ S. Then the following statements are equivalent: (1) (α, β) ∈L∗. (2) For all σ, γ ∈ S1,ασ = αγ if and only if βσ = βγ. Semigroups of linear transformations 1321 Let V be a vector space and α ∈ T (V ). Denote kerα = {v ∈ V : vα =0}. Lemma 3 Let α, β ∈ T (V ). Then (1) (α, β) ∈Lif and only if Vα= Vβ; (2) (α, β) ∈Rif and only if kerα=kerβ. First we need the next observation. Let α ∈ S(V,W) and {zi} be a basis for Vα∩ W . Extend this, respectively, to a basis {zi,zj} for W , to a basis {zi,zl} for Vα. Then we have Lemma 4 {zi,zj,zl} is linearly independent. Proof. Let zi = ziα, zl = zlα for some zi,zl ∈ V . Assume kizi + kjzj + klzl =0 for some scalars ki,kj,kl. Then kjzj = − kizi − klzl − k zα − k zα = i i l l =(− kizi − klzl)α which implies that kjzj ∈Vα∩ W . Therefore, kjzj can be expressed by {zi}. Suppose kjzj = hizi for some scalars hi. Then, since{zi,zj} is linearly independent, we have kj = 0 for each j. Furthermore, kizi + klzl = 0. Notice that {zi,zl} is linearly independent, it must be the case that ki = 0 for each i and kl = 0 for each l. Consequently, {zi,zj,zl} is linearly independent. 2 Theorem 1 Let α, β ∈ S(V,W). Then (α, β) ∈L∗ if and only if Vα= Vβ. Proof. If Vα = Vβ, then (α, β) ∈Lin the semigroup T (V ) by Lemma 3. Hence (α, β) ∈L∗ in S(V,W), and the sufficiency follows. Now suppose (α, β) ∈L∗.IfVα= Vβ, without loss of generality, we may assume Vα\Vβ= ∅. Take z ∈ Vα\Vβ. There are two cases to consider: z ∈ W or z ∈ W . Case 1. z ∈ W . Suppose {zi} is a basis for Vβ∩ W . Extend this to a basis {z,zi,zj} for W , to a basis {zi,zl} for Vβ, respectively. Then {z,zi,zj,zl} is linearly independent by Lemma 4. Extend it to a basis {z,zi,zj,zl,zm} for V . Define σ, γ ∈ T (V ) by letting zi zj zl zm z zi zj zl zm z σ = ,γ= . zi zj zl zm 0 zi zj zl zm z Then σ, γ ∈ S(V,W) and βσ = βγ holds. However, for each x ∈ zα−1, xασ = zσ = 0 and xαγ = zγ = z.Soασ = αγ contradicting Lemma 1. Case 2. z ∈ W . Also, suppose {zi} is a basis for Vβ∩ W . Extend this to a basis {zi,zj} for W , to a basis {zi,zl} for Vβ, respectively. Then {zi,zj,zl} is 1322 Pei Huisheng linearly independent by lemma 4. There are also two subcases: z ∈zi,zj,zl or z ∈zi,zj,zl. If the former is the case, extend {zi,zj,zl} to a basis {zi,zj,zl,zm} for V and define σ, γ ∈ T (V ) by letting zi zj zl zm zi zj zl zm σ = ,γ= . zi 0 zl zm zi zj zl zm It is routine to verify that σ, γ ∈ S(V,W) andβσ = βγholds. By hypothesis, z ∈zi,zj,zl, we can assume z = kizi + kjzj + klzlfor some scalars ki,kj,kl with kj0 = 0 for some j0 (If kj = 0 for all j then z = kizi + klzl ∈ Vβ, contradicting the hypothesis). For each x ∈ zα−1, xασ = zσ = kizi + klzl = z = zγ = xαγ. Thus, ασ = αγ, a contradiction. If the latter is the case, that is, z ∈zi,zj,zl. Extend {zi,zj,zl} to a basis {zi,zj,zl,zm,z} for V and define σ, γ ∈ S(V,W) as the same as in Case 1. Then we also have βσ = βγ and ασ = αγ, a contradiction. Consequently, it follows that Vα= Vβ and the necessity holds. 2 Theorem 2 Let α, β ∈ S(V,W). Then (α, β) ∈R∗ if and only if kerα =kerβ. Proof. The sufficiency follows immediately from Lemma 3, since kerα =kerβ implies (α, β) ∈Rin T (V ). Now suppose (α, β) ∈R∗ in S(V,W). If kerα = kerβ then, without loss of generality, we may assume there exists some z ∈ kerα\kerβ. Suppose {ul} is a basis for Wα. Extend this to a basis {ul,um} for Vα∩ W , and further extend to a basis {ul,um,un} for Vα. Choose zl ∈ −1 −1 −1 ulα ∩ W for each l, zm ∈ umα for each m and zn ∈ unα for each n. There are two possibilities: z ∈ W or z ∈ W . Case 1. z ∈ W . Take a basis {z,zi} for kerα∩W and extend this to a basis {z,zi,zj} for kerα. Then {z,zi,zl} is a basis for W while {z,zi,zj,zl,zm,zn} is a basis for V . Define σ, γ ∈ T (V ) by letting zzi zj zl zm zn zzi zj zl zm zn σ = ,γ= . 0 zi zj zl zm zn zzi zj zl zm zn One routinely verifies that σ, γ ∈ S(V,W), σα = γα and zσβ =0= zβ = zγβ. Thus σβ = γβ, a contradiction. Case 2. z ∈ W . Take a basis {zi} for kerα ∩ W and extend this to a basis {zi,z,zj} for kerα. Then {zi,zl} is a basis for W while {z,zi,zj,zl,zm,zn} is a basis for V . Define σ, γ ∈ S(V,W) as the same as in Case 1. Then we also have σα = γα and σβ = γβ, a contradiction. Consequently, we have kerα=kerβ and the proof is complete. 2 Semigroups of linear transformations 1323 Theorem 3 For each α ∈ S(V,W), there exists an idempotent β ∈ S(V,W) such that Vα= Vβ. Consequently, each L∗ class contains an idempotent.