Semigroups of Linear Transformations with Invariant Subspaces

Transcript

1 International Journal of Algebra, Vol 6, 2012, no 8, Semigroups of Linear Transformations with Invariant Subspaces Preeyanuch Honyam Department of Mathematics, Faculty of Science Chiang Mai University, Chiang Mai 50200, Thailand preeyanuch Jintana Sanwong Department of Mathematics, Faculty of Science Chiang Mai University, Chiang Mai 50200, Thailand Material Science Research Center Faculty of Science, Chiang Mai University Abstract Let V be a vector space and let T V denote the semigroup under composition of all linear transformations from V into itself For a fixed subspace W of V, let SV,W be the subsemigroup of T V, consisting of all linear transformations on V which leave a subspace W of V invariant The purpose of this paper is to characterize Green s relations and ideals on SV,W and prove that SV,W is never isomorphic to T U for any vector space U when W is a non-zero proper subspace of V Mathematics Subject Classification: 20M12, 20M20, 15A04 Keywords: linear transformation semigroups, Green s relations, ideals 1 Introduction Let T X denote the semigroup under composition of all transformations from X into itself For a fixed nonempty subset Y of X, let SX, Y ={α T X :Yα Y } Then SX, Y is a semigroup of total transformations of X which leave a subset Y of X invariant Magill [4] introduced and studied the semigroup SX, Y

2 376 P Honyam and J Sanwong in 1966 Later in 2006 Nenthein and Kemprasit [5] characterized its regular elements Recently, we described Green s relations and ideals on SX, Y see [2] for details For a vector space V, let T V be the semigroup under composition of all linear transformations from V into itself It is known that T V is a regular semigroup [1, Volume 1, exercise 226] Here, we consider the subsemigroup of T V, analogous to SX, Y, defined by SV,W ={α T V :Wα W }, where W is a subspace of V and Wα denotes the range of W under α Note that id V and 0, the identity map and the zero map on V respectively, belong to SV,W In addition, SV,W contains LV,W ={α T V :Vα W } as an ideal of SV,W If {0} W V, then SV,W is not a 0-simple semigroup since id V SV,W\LV,W [5, page 9] In 2006, Nenthein and Kemprasit [5] defined the semigroup SV,W and characterized the regular elements of SV,W The authors proved that for α SV,W, α is a regular element of SV,W if and only if Vα W = Wα, and showed that, SV,W is regular if and only if either W = V or W = {0} Here, in Section 2, we prove that SV,W is never isomorphic to T U for any vector space U when W is a non-zero proper subspace of V We also show that SV,W is almost never isomorphic to SX, Y In Section 3, we characterize Green s relations on SV,W, and we describe its ideals in Section 4 In this paper, the vector space V we consider, dim V can be finite or infinite The cardinality of a set A is denoted by A Also X = A B means X is a disjoint union of A and B 2 Isomorphism of SV,W For a vector space V, let T V denote the semigroup under composition of all linear transformations from V into itself For a fixed subspace W of V, let SV,W ={α T V :Wα W }, where Wα denotes the range of W under α Then SV,W is a subsemigroup of T V, which contains id V and 0, the identity map and the zero map on V respectively If W = {0} or W = V, we have SV,W =T V which is a regular semigroup In general SV,W is not a regular subsemigroup of T V The following lemma characterizes the regular elements of SV,W, and gives a necessary and sufficient condition for SV,W to be regular

3 Linear transformations with invariant subspaces 377 Lemma 1 [5] The following statements hold for the semigroup SV,W 1 For α SV,W, α is a regular element of SV,W if and only if Vα W = Wα 2 SV,W is regular if and only if either W = V or W = {0} From [2, Theorem 1], we see that SX, Y is isomorphic to T Z if and only if X = Y and Y = Z Here, the following corollary shows that SV,W is almost never isomorphic to T U for any vector space U Corollary 1 If {0} W V, then SV,W is not isomorphic to T U for any vector space U Proof Suppose that {0} W V By Lemma 1, we have SV,W is not regular, thus SV,W T U For convenience, we adopt the convention introduced in Clifford and Preston [1, Volume 2, page 241]: namely, if α T X then we write Xi a i and take as understood that the subscript i belongs to some unmentioned index set I, the abbreviation {a i } denotes {a i : i I}, and that X {a i } and a i α 1 = X i Similarly, we can use this notation for elements in T V To construct a map α T V, we first choose a basis {e i } for V and a subset {a i } of V, and then let e i a i for each i I and extend this map linearly to V To shorten this process, we simply say, given {e i } and {a i } within the context, then for each α T V, we can write ei a i A subspace U of V generated by a linearly independent subset {e i } of V is denoted by e i and when we write U = e i, we mean that the set {e i } is a basis of U, and we have dim U = I For each α T V, the kernel and the range of α denoted by ker α and Vα, respectively We note that SV,W always contains a zero element, the zero map But, for SX, Y we have the following lemma Lemma 2 SX, Y has a zero element if and only if Y =1

4 378 P Honyam and J Sanwong Proof Assume that Y = 1 Let Y = {a} and define X SX, Y a Then α βα for all β SX, Y and therefore α is a zero element of SX, Y Conversely, assume that SX, Y has a zero element, say α Suppose that Y > 1 Let b, c Y be such that b c and define X b and γ = X c Then β,γ SX, Y and β γ Since α is a zero element, we have α αγ But αβ and γ = αγ Thus α αγ = γ which is a contradiction Therefore, Y =1 Thus, if Y > 1 then SX, Y has no zero elements following corollary So we have the Corollary 2 Let Y be a fixed nonempty subset of X If Y > 1, then SX, Y is not isomorphic to SV,W for any subspace W of a vector space V 3 Green s relations on SV,W It is well-known that αlβ in T V if and only if Vα= Vβ; and αrβ in T V if and only if ker kerβ see [1, Volume 1, exercise 226] But, for SV,W we have the following lemmas Lemma 3 Let α, β SV,W Then γβ for some γ SV,W if and only if Vα Vβ and Wα Wβ Consequently, αlβ if and only if Vα= Vβ and W Wβ Proof Assume that γβ for some γ SV,W Then V Vγ Vγβ Vβ and W WγWγβ Wβ since γ SV,W Conversely, suppose that Vα Vβ and Wα Wβ Then W α W W β W where α W,β W T W Hence α W = δβ W for some δ T W Write W = w i and V = w i v j Sow i w i δβ Now, for each v j, there exists some v j V such that v j v j β since Vα Vβ Thus we extend δ T W toγ T V by γ = wi v j w i δ v j Then γ SV,W and γβ as required

5 Linear transformations with invariant subspaces 379 Lemma 4 Let α, β SV,W Then βγ for some γ SV,W if and only if ker β ker α and Wβ 1 Wα 1 Consequently, αrβ if and only if ker kerβ and Wα 1 = Wβ 1 Proof Assume that βγ for some γ SV,W If a ker β, then a aβγ =aβγ =0γ = 0 and hence ker β ker α If b Wβ 1, then bβ W and so b bβγ =bβγ Wγ W Thus b Wα 1 implies Wβ 1 Wα 1 To prove the converse, we assume that ker β ker α and Wβ 1 Wα 1 Let Vβ= v i ThusV = v i u j Since v i Vβ, there exists z i V such that v i = z i β So we define γ T V by γ = vi u j z i α 0 Then γ is well-defined since ker β ker α Now, we prove that γ SV,W Since W is a subspace of V, we can write W = w k where {w k } {v i,u j }If w k = u j, then w k γ = u j γ =0 W Ifw k = v i, then there exists z i V such that w k = v i = z i βthusz i w k β 1 Wβ 1 Wα 1 Sow k γ = z i α W Also, we have vβγ =vβγ = vα for all v V by the definition of γ Theorem 1 Let α, β SV,W Then αdβ if and only if dimw dimwβ, dimvα W /W dimvβ W /W β and dimvα/vα W = dimvβ/vβ W Proof We first note that for any α SV,W, dimw dimw/w ker α, dimvα W /W dimwα 1 /Wαα 1 and dimvα/vα W = dimv/wα 1 Assume that αdβ Then αlγrβ for some γ SV,W So V Vγ and W Wγ by Lemma 3, and ker γ =kerβ and Wγ 1 = Wβ 1 by Lemma 4 Thus dimw dimwγ = dimw/w ker γ = dimw/w ker dimwβ, and dimvα/vα W = dimvγ/vγ W = dimv/wγ 1 = dimv/wβ 1 = dimvβ/vβ W From ker γ =kerβ, we have Wγγ 1 =Wββ 1 So dimvα W /W dimvγ W /W γ = dimwγ 1 /Wγγ 1 = dimwβ 1 /Wββ 1 = dimvβ W /W β Conversely suppose that the conditions hold Assume that dimwα= I = dimwβ, dimvα W /W J = dimvβ W /W β and dimvα/vα W = K = dimvβ/vβ W Then we can write W w i α and W w i β ; Vα W /W u jα + Wα and Vβ W /W u jβ +Wβ ; Vα/Vα W = v k α+vα W and Vβ/Vβ W = v k β + Vβ W Since W w iα and Vα W /W u j α + Wα, we have Vα W = w i α u j α Thus Vα= w i α u j α v k α since Vα/Vα W = v k α+vα W Similarly, we get Vβ= w iβ u jβ v k β

6 380 P Honyam and J Sanwong Let ker z r and ker z t So V = z r w i u j v k and V = z t w i u j v k Thus we can write Define zr w i u j v k 0 w i α u j α v k α and zt w i u j v k 0 w i β u j β v k β μ = zt w i u j v k 0 w i α u j α v k α Then Wμ = w i α W and hence μ SV,W Also, we have Wμ = w i α = Wα and Vμ= w i α u j α v k α = Vα So αlμ by Lemma 3 Also, we have ker μ = z t =kerβ and Wμ 1 = z t w i u j = Wβ 1, thus μrβ by Lemma 4 and therefore αdβ Theorem 2 Let α, β SV,W Then λβμ for some λ, μ SV,W if and only if dimvα dimvβ, dimwα dimwβ and dimvα/vα W dimvβ/vβ W Consequently, αjβ if and only if dimvα= dimvβ, dimw dimwβ and dimvα/vα W = dimvβ/vβ W Proof Assume that λβμ for some λ, μ SV,W Then Vα= Vλβμ= Vλβμ Vβμ and W Wλβμ =Wλβμ Wβμ Since dimvβμ dimvβ and dimwβμ dimwβ, so we have dimvα dimvβ and dimwα dimwβ Also, dimvα/vα W = dimvλβμ/vλβμ W = dimv λβμ/v λβμ W dimvβμ/vβμ W dimvβ/vβ W, since there exists a map from Vβ/Vβ W ontovβμ/vβμ W Conversely, assume that the conditions hold From dimwα dimwβ, we can write W w i α and W w i β w lβ Since Wα is a subspace of Vα W and Wβ is a subspace of Vβ W, write Vα W = w i α u j α and Vβ W = w iβ w l β u mβ And from dimvα/vα W dimvβ/vβ W, we can write Vα/Vα W = v k α + Vα W and Vβ/Vβ W = v k α + Vβ W v nβ + Vβ W Then V w i α u j α v k α and V w i β w l β u m β v n β v k β Let ker z r and ker z t SoV = z r w i u j v k and V = z t w i w l u m v n v k Then zr w i u j v k 0 w i α u j α v k α

7 Linear transformations with invariant subspaces 381 and zt w i w l u m v n v k 0 w iβ w l β u mβ v nβ v k β Since dimv α dimv β, we have I + J + K I + L + M + N + K To find λ, μ SV,W such that λβμ, we follow the ideas of [2, Theorem 5] Now, we consider two cases: Case 1 : J L + M + N Let L M N = P Q where P = J Then we can write {w l } {u m} {v n} = {s p } {s q } and rewrite β as follow: zt w i s p s q v k 0 w iβ s p β s q β v k β Since J = P, there is a bijection ϕ : J P Now define zr w λ = i u j v k 0 w i s jϕ v k Then Wλ = w i W and so λ SV,W From V w iβ s p β s q β v k β, we let V = w i β s pβ s q β v k β z s and define μ = w i β s jϕ β v k β {s qβ,z s} w i α u j α v k α 0 So μ SV,W and λβμ Case 2 : J > L + M + N Then dimv β is infinite for if dimv β is finite, then dimvα= I + J + K > I + L + M + N + K = dimvβ which is a contradiction Hence J I or J K are infinite cardinals If J I is an infinite cardinal, then write I = P Q where P = I, Q = J Thus we can write {w i } = {s p} {s q } and rewrite β as follow: zt s p s q w l u m v n v k 0 s p β s q β w l β u mβ v nβ v k β Since I = P and J = Q, there are bijections ϕ : I P and ψ : J Q And from Vβ= s p β s q β w l β u m β v n β v kβ, we can write V = s p β s q β w l β u mβ v nβ v k β z t Then define λ and μ as follows: zr w λ = i u j v k 0 s iϕ s jψ v k, and μ = siϕ β s jψ β v k β {w l β,u m β,v n β,z t } w i α u j α v k α 0

8 382 P Honyam and J Sanwong So, we see that λ, μ SV,W and λβμ For the case J K is an infinite cardinal, we write K = G H where G = J, H = K Write {v k } = {t g} {t h } and rewrite β as follow: zt w i w l u m v n t g t h 0 w i β w l β u m β v n β t gβ t h β As before, we can define λ, μ SV,W such that λβμ The following example shows that in general D J on SV,W Example 1 Let V = w, w k,u 1,u 2 and W = w, w k, where K is infinite Choose k 0 K and let K = K\{k 0 } Then we define u1 u 2 {w, w k0 } w k u1 {u and 2,w,w k0 } w k 0 w w k0 w k 0 w k0 w k Hence α, β SV,W and V w, w k, V w k ; W w k = Wβ and Vα W = w, w k, Vβ W = w k So dimv K = dimv β, dimw K = dimwβ and dimv α/v α W = 0 = dimvβ/vβ W, thus αjβ Since dimvα W /W 1 0 = dimvβ W /W β, we have α and β are not D-related on SV,W Even dimw is finite, we still have D J on SV,W Example 2 Let V = w 1,w 2,w 3,u,u k and W = w 1,w 2,w 3, where K is infinite Then we define {u, w1 } w 2 w 3 u k w1 w and 2 w 3 u u k 0 w 1 w 3 u k 0 w 1 w 3 w 2 u k Then α, β SV,W and V w 1,w 3,u k, V w 1,w 2,w 3,u k ; W w 1,w 3 = Wβ and Vα W = w 1,w 3, Vβ W = w 1,w 2,w 3 Thus dimvα= K = dimvβ, dimw 2 = dimwβ and dimvα/vα W = K = dimvβ/vβ W, so αjβ But dimvα W /W 0 1 = dimvβ W /W β, we have α and β are not D-related on SV,W Theorem 3 D = J on SV,W if and only if W = V or W = {0} or dim V is finite Proof If W = V or W = {0}, then SV,W =T V and thus D = J by [1, Volume 1, exercise 226] Now, assume that dim V is finite Let α, β SV,W be such that αjβ So by Theorem 2, dimv dimvβ, dimw dimwβ and dimvα/vα W = dimvβ/vβ W Then we can write W w i α, W w i β and Vα/Vα W = u kα + Vα W, Vβ/Vβ W = u kβ + Vβ W Let Vα W /W v jα + Wα and Vβ W /W v l β + Wβ Thus V w iα v j α u k α and

9 Linear transformations with invariant subspaces 383 V w i β v l β u kβ So I + J + K = dimv dimvβ= I + L + K is finite since dim V is finite and thus J = L, that is dimvα W /W dimvβ W /W β Therefore, αdβ and hence D = J Conversely, assume that D = J on SV,W, and suppose on contrary that dim V is infinite and {0} W V We consider two cases: Case 1 : dim W is finite Let V = w 1,,w n,u,u k and W = w 1,,w n n 1, where K is infinite Define α, β SV,W as follows: {u, w1 } w 2 w n u k 0 w 2 w n u k and w1 w 2 w n u u k 0 w 2 w n w 1 u k Thus V w 2,,w n,u k, V w 1,,w n,u k ; W w 2,,w n = Wβ and Vα W = w 2,,w n, Vβ W = w 1,,w n Hence dimvα= K = dimvβ, dimwα=n 1 = dimwβ and dimvα/vα W = K = dimvβ/vβ W, so αjβ Since dimvα W /W 0 1= dimvβ W /W β, we have α and β are not D-related on SV,W by Theorem 1 and this contradicts the assumption Case 2 : dim W is infinite Let V = w, w k,u,u i and W = w, w k, where K is infinite Choose k 0 K and let K = K\{k 0 } Define α, β SV,W as follows: ui {w, w k0 } w k u 0 w k0 w k w ui {u, w, w and k0 } w k 0 w k0 w k Hence Vα= w, w k, V w k ; W w k = Wβ and Vα W = w, w k, Vβ W = w k Thus dimv K = dimvβ, dimw K = dimwβ and dimvα/vα W = 0 = dimvβ/vβ W, so αjβ Since dimvα W /W 1 0 = dimvβ W /W β, we have α and β are not D-related on SV,W by Theorem 1 and this leads to a contradiction 4 Ideals of SV,W Let p be any cardinal number and let p = min{q : q>p} Note that p always exists since the cardinals are well-ordered and when p is finite we have p = p + 1 = the successor of p To describe ideals of SV,W for any vector space V and any subspace W of V, we let dim V = a, dim W = b and dim V/W = c

10 384 P Honyam and J Sanwong In addition, for each cardinals r, s, t such that 1 r a, 1 s b and 1 t c, define Sr, s, t ={α SV,W : dimvα <r,dimwα <sand Then if r = a,s= b and t = c, we have dimvα/vα W <t} Sr, s, t =Sa,b,c =SV,W And if r = b,s = b and t = 1, then Sr, s, t =Sb,b, 1 = {α SV,W : dimvα <b, dimwα <b and dimvα/vα W < 1} = LV,W = {α T V :Vα W } which is an ideal of SV,W [5, page 9] We observe that: if W = V, then dim V = a = dim W and dim V/W =0 Thus, Sr, r, 1 = {α SV,W : dimvα <r} = {α T V : dimvα < r} which is an ideal of T V Theorem 4 The set Sr, s, t is an ideal of SV,W Proof Let α Sr, s, t and λ, μ SV,W Then dimvα <r, dimwα < s and dimvα/vα W <t By using the same proof as given in Theorem 2, we have dimvλαμ dimvα <r, dimwλαμ dimwα <sand dimvλαμ/vλαμ W dimvα/vα W <t Hence λαμ Sr, s, t Therefore, Sr, s, t is an ideal of SV,W We note that: if r u, s v and t w, then Sr, s, t Su, v, w The following example shows that there is an ideal in SV,W which is not of the form Sr, s, t and the set of ideals of SV,W does not form a chain under the set inclusion Example 3 Let V = w 1,w 2,u 1,u 2 and W = w 1,w 2 Then dim V = 4, dim W = 2 and dim V/W = 2 Since S3, 3, 1 and S4, 2, 2 are ideals of SV,W, we have S3, 3, 1 S4, 2, 2 is also an ideal of SV,W Suppose that S3, 3, 1 S4, 2, 2 = Sl, m, n for some 1 l 5, 1 m 3 and 1 n 3 Let w1 w 2 u 1 u 2 0 w 1 w 2 u 2 Then dimv 3, dimw 1 and dimvα/vα W = 1 and hence α S4, 2, 2 If l<4orn<2, then we have α S4, 2, 2\Sl, m, n If m<3, there is w1 {w 2,u 1 } u 2 S3, 3, 1\Sl, m, n w 1 w 2 0

11 Linear transformations with invariant subspaces 385 since dimv 2, dimw 2 and dimvα/vα W = 0 Both cases contradict our supposition So l 4,m 3 and n 2 Consider w1 w δ = 2 u 1 u 2, w 1 w 2 u 1 0 we have dimv 3, dimw 2 and dimv α/v α W = 1 and thus δ S4, 3, 2 but δ / S3, 3, 1 S4, 2, 2, so S3, 3, 1 S4, 2, 2 Sr, s, t for all r 4,s 3 and t 2 Since α S4, 2, 2\S3, 3, 1 and β S3, 3, 1\S4, 2, 2, we conclude that the set of ideals of SV,W does not form a chain Lemma 5 The set of ideals of SV,W forms a chain under the set inclusion if and only if V = W Proof If V = W, then SV,W =SV,V =TV and the ideals of T V are precisely the set {α T V : dimvα <r} where 1 r dim V So, we see that the set of ideals forms a chain under the set inclusion Conversely, assume that W V Let V = w, w j,u,u i and W = w, w j I or J can be empty Then dim V = I + J +2, dim W = J + 1 and dim V/W = I +1 Consider w {u, ui,w j } w 0 and u {ui,w,w j } u 0 we have α S2, 2, 1\S2, 1, 2 and β S2, 1, 2\S2, 2, 1 Thus the set of ideals of SV,W does not form a chain To obtain ideals of SV,W we need the following notation Let U be a non-zero subspace of SV,W Define KU ={α SV,W : dimvα dimvβ, dimwα dimwβ and, dimvα/vα W dimvβ/vβ W for some β U} Then we see that U KU, and U 1 U 2 implies KU 1 KU 2 Theorem 5 The ideals of SV,W are precisely the set KU for some nonzero subspace U of SV,W Proof Assume that I is an ideal of SV,W Let α KI Then dimvα dimvβ, dimwα dimwβ and dimvα/vα W dimvβ/vβ W for some β I and thus by Theorem 2 we have λβμ for some λ, μ SV,W Since β I is an ideal of SV,W, it follows that λβμ I, and that KI I Usually, we have I KI Therefore, I = KI

12 386 P Honyam and J Sanwong Conversely, we prove that KU is an ideal of SV,W Let α KU and λ, μ SV,W Then dimvα dimvβ, dimwα dimwβ and dimvα/vα W dimvβ/vβ W for some β U As before, we have dimvλαμ dimvα, dimwλαμ dimwα and dimvλαμ/vλαμ W dimvα/vα W So dimvλαμ dimvβ, dimwλαμ dimwβ and dimvλαμ/vλαμ W dimvβ/vβ W Hence λαμ KU and therefore KU is an ideal of SV,W ACKNOWLEDGEMENTS The first author would like to thank the Office of the Higher Education Commission, Thailand, for its financial support She also thanks the Graduate School, Chiang Mai University, Chiang Mai, Thailand, for its financial support that she received during the preparation of this paper The second author would like to thank the National Research University Project under the Office of the Higher Education Commission, Thailand, for its financial support References [1] AH Clifford and GB Preston, The Algebraic Theory of Semigroups, Mathematical Surveys, No 7, vol 1 and 2, American Mathematical Society, Providence, RI, USA, 1961 and 1967 [2] P Honyam and J Sanwong, Semigroups of transformations with invariant set, J Korean Math Soc, , no 2, [3] TW Hungerford, Algebra, Springer-Verlag, New York, 1974 [4] KD Magill Jr, Subsemigroups of SX, Math Japon, , [5] S Nenthein and Y Kempasit, On transformation semigroups which are BQ-semigroup, Int J Math Math Sci, vol 2006, Article ID 12757, 10 pages, 2006 doi: /IJMMS/2006/12757 Received: November, 2011