Linear transformations with characteristic subspaces that are not hyperinvariant
aa r X i v : . [ m a t h . R A ] J a n Linear transformations with characteristicsubspaces that are not hyperinvariant
Pudji AstutiFaculty of Mathematicsand Natural SciencesInstitut Teknologi BandungBandung 40132Indonesia Harald K. WimmerMathematisches InstitutUniversit¨at W¨urzburg97074 W¨urzburgGermanyJuly 31, 2018 The work of the first author was supported by the program “Riset dan InovasiKK ITB” of the Institut Teknologi Bandung. bstractMathematical Subject Classifications (2000):
Keywords: hyperinvariant subspaces, characteristic subspaces, invariantsubspaces, Ulm invariants, characteristic hull, exponent, height.
Abstract: If f is an endomorphism of a finite dimensional vector space over a field K then an invariant subspace X ⊆ V is called hyperinvariant (respectively,characteristic) if X is invariant under all endomorphisms (respectively, auto-morphisms) that commute with f . According to Shoda (Math. Zeit. 31, 611–624, 1930) only if | K | = 2 then there exist endomorphisms f with invariantsubspaces that are characteristic but not hyperinvariant. In this paper we ob-tain a description of the set of all characteristic non-hyperinvariant subspacesfor nilpotent maps f with exactly two unrepeated elementary divisors. Address for Correspondence:
H. WimmerMathematisches InstitutUniversit¨at W¨urzburgAm Hubland97074 W¨urzburgGermany e -mail: [email protected] F ax: +49 931 8 88 46 11 e -mail: [email protected] Introduction
Let K be a field, V an n -dimensional vector space over K and f : V → V a K -linear map. A subspace X ⊆ V is said to be hyperinvariant (under f ) [9,p. 305] if it remains invariant under all endomorphisms of V that commutewith f . If X is an f -invariant subspace of V and if X is invariant under allautomorphisms of V that commute with f , then [1] we say that X is char-acteristic (with respect to f ). Let Inv( V, f ), Hinv(
V, f ), and Chinv(
V, f ) besets of invariant, hyperinvariant and characteristic subspaces of V , respec-tively. These sets are lattices, andHinv( V, f ) ⊆ Chinv(
V, f ) ⊆ Inv(
V, f ) . If the characteristic polynomial of f splits over K (such that all eigenvaluesof f are in K ) then one can restrict the study of hyperinvariant and ofcharacteristic subspaces to the case where f has only one eigenvalue, and tothe case where f is nilpotent. Thus, throughout this paper we shall assume f n = 0. Let Σ( λ ) = diag(1 , . . . , , λ t , . . . , λ t m ) ∈ K n × n [ λ ] be the Smithnormal form of f such that t + · · · + t m = n . We say that an elementarydivisor λ r is unrepeated if it appears exactly once in Σ( λ ).The structure of the lattice Hinv( V, f ) is well understood ([11], [6], [12],[9, p. 306]). We point out that Hinv(
V, f ) is the sublattice of Inv(
V, f )generated by Ker f k , Im f k , k = 0 , , . . . , n. It is known ([13], [10, p. 63/64], [1]) that each characteristic subspace ishyperinvariant if | K | >
2. Hence, only if V is a vector space over the field K = GF (2) one may find K -endomorphisms f of V with characteristicsubspaces that are not hyperinvariant. A necessary and sufficient conditionfor the existence of such mappings f is due to Shoda (see also [3, Theorem 9,p. 510] and [10, p. 63/64]). It involves unrepeated elementary divisors of f . Theorem 1.1. [13, Satz 5, p. 619]
Let V be a finite dimensional vector spaceover the field K = GF (2) and let f : V → V be nilpotent. The followingstatements are equivalent. (i) There exists a characteristic subspace of V that is not hyperinvariant. (ii) For some numbers R and S with R + 1 < S the map f has exactly oneelementary divisor λ R and exactly one of the form λ S . Provided that f satisfies condition (ii) of Shoda’s theorem how can oneconstruct all characteristic subspaces of V that are not hyperinvariant? For1he moment the answer to that question is open. In this paper we assumethat f has exactly one pair of unrepeated elementary divisors. In that casewe show how to construct the family of characteristic and non-hyperinvariantsubspaces associated to f . For that purpose we prove rather general resultson the structure of characteristic non-hyperinvariant subspaces and we clarifythe role of unrepeated elementary divisors of f . We note that our study canbe interpreted in the setting of module theory. In the context of abeliangroup theory [8] one would deal with characteristic subgroups of p -groupsthat are not fully invariant.We first discuss an example, which displays features of characteristic sub-spaces that will become important later, and we introduce concepts that willallow us to state Theorem 1.3 at the end of this section. The inequality R + 1 < S in Theorem 1.1 is valid for ( R, S ) = (1 , f with elementary divisors λ and λ .We first introduce some notation, in particular we define the concepts ofexponent and height. We set V [ f j ] = Ker f j , j ≥
0. Thus, f n = 0 implies V = V [ f n ]. Define ι = id V and f = ι . Let x ∈ V . The smallest nonnegativeinteger ℓ with f ℓ x = 0 is called the exponent of x . We write e( x ) = ℓ . Anonzero vector x is said to have height q if x ∈ f q V and x / ∈ f q +1 V . In thiscase we write h( x ) = q . We set h(0) = −∞ . The n -tuple H ( x ) = (cid:0) h( x ) , h( f x ) , . . . , h( f n − x ) (cid:1) is the indicator [8, p. 3] or Ulm sequence [10] of x . Thus, if e( x ) = k then H ( x ) = (h( x ) , . . . , h( f k − x ) , ∞ , . . . , ∞ ). We say that H ( x ) has a gap at j , if1 ≤ j < e( x ) and h ( f j x ) > h ( f j − x ). Let End( V, f ) be the algebra ofall endomorphisms of V that commute with f . The group of automorphismsof V that commute with f will be denoted by Aut( V, f ). Let h x i = span { f i x, i ≥ } = { c x + c f x + · · · + c n − f n − x ; c i ∈ K, i = 0 , , . . . , n − } be the f -cyclic subspace generated by x . If B ⊆ V we define h B i = P b ∈ B h b i and B c = h αb ; b ∈ B, α ∈ Aut(
V, f ) i . We call B c the characteristic hull of B . Clearly, if α ∈ Aut(
V, f ) then α ( f j x ) = f j ( αx ) for all x ∈ V . Hence it is obvious thate( αx ) = e( x ) and h( αx ) = h( x ) for all x ∈ V, α ∈ Aut(
V, f ) . (1.1)2et e = (1 , , . . . , T , . . . , e m = (0 , . . . , , T be the unit vectors of K m ,and let N m denote the lower triangular nilpotent m × m Jordan block.
Example 1.2. [10, p. 63/64]) Let K = GF (2). Consider V = K and let f : V → V be K -linear with elementary divisors λ and λ . With respect tothe basis { e i } , i = 1 , . . . ,
4, the map f is given by f x = N x with N = diag( N , N ) = , and V = h e i ⊕ h e i with e( e ) = 1 and e( e ) = 3. Then α ∈ Aut(
V, f ) if andonly if αx = Ax and A ∈ K × is a nonsingular matrix satisfying AN = N A ,that is (see e.g. [14, p. 28]), A = ν ω κ µ ω , κ, µ, ν, ω ∈ K. Define z = e + e . We show that the characteristic subspace X = h z i c is nothyperinvariant. We have αz = ( e + κe ) + ( e + ωe ), and therefore X = h z i c = h e + f e i c =span { e + e , e + e + e } = { , e + e , e + e + e , e } . Let π = diag(1 , , ,
0) be the orthogonal projection on Ke . Then π ∈ End(
V, f ). We have π z = e , but e / ∈ X . Therefore X is a characteristicsubspace that is not hyperinvariant. From H ( z ) = (0 , , ∞ , ∞ ) we see that H ( z ) has a gap at j = 1. (cid:3) To make the connection with Kaplansky’s exposition of Shoda’s theorem[10, p. 63/64] we define the numbers d ( f, r ) = dim (cid:0) V [ f ] ∩ f r − V / V [ f ] ∩ f r V (cid:1) , r = 1 , , . . . , n. In accordance with the terminology of abelian p -groups [7, p. 154] or p -modules [10, p.27] we call d ( f, r ) the ( r − Ulm invariant of f . Then d ( f, r ) is equal to the number of entries λ r in the Smith form of f , and d ( f, r ) = 1 means that λ r is an unrepeated elementary divisor. In the follow-ing it may be convenient to write d ( r ) instead of d ( f, r ). We recall (see e.g.[4]) that methods or concepts of abelian group theory must be translated to3odules over principal ideal domains and then specialized to K [ λ ]-modulesbefore they can be applied to linear algebra.With regard to Theorem 1.3 below we note additional definitions. Sup-pose dim Ker f = m . Let λ t , . . . , λ t m be the elementary divisors of f suchthat t + · · · + t m = dim V . Then V can be decomposed into a direct sum of f -cyclic subspaces h u i i such that V = h u i ⊕ · · · ⊕ h u m i and e( u i ) = t i , i = 1 , . . . , m. (1.2)If (1.2) holds and if the elements in U are ordered by nondecreasing exponentssuch that e( u ) ≤ · · · ≤ e( u m )then we call U = ( u , . . . , u m ) a generator tuple of V (with respect to f ). Thetuple ( t m , . . . , t ) of exponents - written in nonincreasing order - is known as Segre characteristic of f . The set of generator tuples of V will be denotedby U . We call u ∈ V a generator of V (see also [7, p.4]) if u ∈ U for some U ∈ U . In other words, u ∈ V is a generator if and only if u = 0 and V = h u i ⊕ V for some V ∈ Inv(
V, f ) . (1.3)If f has only two elementary divisors then part (ii) of the following the-orem gives a description of the set Chinv( V, f ) \ Hinv(
V, f ). Theorem 1.3.
Assume | K | = 2 . Suppose λ R and λ S are unrepeated elemen-tary divisors of f and R + 1 < S . Let u and v be corresponding generatorsof V such that e( u ) = R and e( v ) = S . (i) A subspace X = h f R − s u + f S − q v i c (1.4) is characteristic and not hyperinvariant if the integers s, q satisfy < s ≤ R, s < q, R − s < S − q. (1.5)(ii) Suppose V = h u i ⊕ h v i . Then an invariant subspace X ⊆ V is charac-teristic and not hyperinvariant if and only if X is of the form (1.4) and s, q satisfy (1.5) . The proof of Theorem 1.3(ii) will be given in Section 2, where two propo-sitions will be proved, one dealing with sufficiency and the other one withnecessity of condition (1.5). In Section 3 we split the space V into twocomplementary invariant subspaces E and G such that the unrepeated ele-mentary divisors of f are those of f | E and the repeated ones are those of f | G .It will be shown that a characteristic subspace X is hyperinvariant in V ifand only if X ∩ E is hyperinvariant in E . An application of that approach isa proof of Theorem 1.3(i). In Section 4 we extend Theorem 1.3(ii) assumingthat f has only two unrepeated elementary divisors.4 The case of two elementary divisors
In this section we give a proof of Theorem 1.3(ii). It is based on auxiliaryresults on hyperinvariant subspaces and on images of generators under f -commuting automorphisms of V . Suppose dim Ker f = m . Let U = ( u , . . . , u m ) ∈ U be a generator tuplewith e( u i ) = t i , i = 1 , . . . , m , such that0 < t ≤ · · · ≤ t m . Set ~t = ( t , . . . , t m ) and t = 0. Let L ( ~t ) be the set of m -tuples ~r =( r , . . . , r m ) ∈ Z m satisfying0 ≤ r ≤ · · · ≤ r m and 0 ≤ t − r ≤ · · · ≤ t m − r m . (2.1)We write ~r (cid:22) ~s if ~r = ( r i ) mi =1 , ~s = ( s i ) mi =1 ∈ L ( ~t ) and r i ≤ s i , i = 1 , . . . , m .Then (cid:0) L ( ~t ) , (cid:22) (cid:1) is a lattice. The following theorem is due to Fillmore, Herreroand Longstaff [6]. We refer to [9] for a proof. Theorem 2.1.
Let f : V → V be nilpotent. (i) If ~r ∈ L ( ~t ) , then W ( ~r ) = f r V ∩ V [ f t − r ] + · · · + f r m V ∩ V [ f t m − r m ] is a hyperinvariant subspace. Conversely, each W ∈ Hinv(
V, f ) is ofthe form W = W ( ~r ) for some ~r ∈ L ( ~t ) . (ii) If ~r ∈ L ( ~t ) then W ( ~r ) = f r h u i ⊕ · · · ⊕ f r m h u m i . (iii) The mapping ~r W ( ~r ) is a lattice isomorphism from (cid:0) L ( ~t ) , (cid:22) (cid:1) onto (Hinv( V, f ) , ⊇ ) .For a given ~t the number of hyperinvariant subspaces is n H ( ~t ) = Y mi =1 (1 + t i − t i − ) . (2.2)Let X H denote the largest hyperinvariant subspace contained in a char-acteristic subspace X . Using a generator tuple U = ( u , . . . , u m ) ∈ U onecan give an explicit description of X H . Let x ∈ V be decomposed as x = x + · · · + x m , x i ∈ h u i i , i = 1 , . . . , m, (2.3)5nd let π j : V → V be the projections defined by π j x = x j , i = 1 , . . . , m . If X ⊆ V then π j x ∈ X for all x ∈ X is equivalent to π j X = X ∩ h u j i . (2.4)The following theorem shows that (2.4) holds if e( u j ) = t and λ t is a repeatedelementary divisor. Theorem 2.2. [2, Lemma 4.5, Lemma 4.2, Theorem 4.3]
Suppose X is acharacteristic subspace of V . Let U = ( u , . . . , u m ) ∈ U . (i) If d ( t j ) > then π j X = X ∩ h u j i . (ii) The subspace X is hyperinvariant if and only if π j X = X ∩ h u j i , j = 1 , . . . , m, (2.5) or equivalently, X = ⊕ mi =1 (cid:0) X ∩ h u i i (cid:1) . (2.6)(iii) The subspace X H = ⊕ mi =1 (cid:0) X ∩ h u i i (cid:1) (2.7) is the largest hyperinvariant subspace contained in X . In a characteristic subspace X elements outside of X H are of specialinterest (if they exist). Lemma 2.3.
Let X be a characteristic subspace. (i) If W is a proper subspace of X then X = h X \ W i c . (ii) If X is not hyperinvariant then X = h X \ X H i c and h X \ X H i c = h X \ X H i . (2.8) Proof. (i) From X \ W ⊆ X and X c = X follows h X \ W i c ⊆ X . Conversely,if x ∈ X then either x ∈ X \ W or x ∈ W . In the first case it is obviousthat x ∈ h X \ W i c . Suppose x ∈ W . Choose an element z ∈ X \ W . Then x + z ∈ X \ W . Thus z ∈ h X \ W i c and x + z ∈ h X \ W i c , and therefore x ∈ h X \ W i c . Hence X ⊆ h X \ W i c , which completes the proof.(ii) Because of X % X H we can choose W = X H , and obtain X = h X \ X H i c . Let us show that α ( X \ X H ) = X \ X H for all α ∈ Aut(
V, f ) . (2.9)Since X H is hyperinvariant and X is characteristic we have α ( X H ) = X H and α ( X ) = X . Consider x ∈ X \ X H . Suppose αx ∈ X H . Then x ∈ α − ( X H ) = X H , which is a contradiction. It is obvious that (2.9) is equivalentto (2.8). 6 .2 Images under automorphisms If x ∈ V and α ∈ Aut(
V, f ) then it follows from (1.1) that H ( αx ) = H ( x ).We note a converse result due to Baer. Theorem 2.4. (See [10], [8, p. 4])
Let x, y ∈ V . Then H ( x ) = H ( y ) if andonly if y = αx for some α ∈ Aut(
V, f ) . We shall use Baer’s theorem in Lemma 2.7 to determine the set { αu ; α ∈ Aut(
V, f ) } for generators u of V . With regard to the proof of Lemma 2.7 weput together basic facts on exponent and height.If x ∈ V , x = 0, and e( x ) = k , then e( f j x ) = k − j , j = 0 , , . . . , k −
1. Theheight of f j x satisfies the inequality h( f j x ) ≥ j + h( x ), j = 0 , , . . . , k − x , . . . , x m ∈ V thenh( x + · · · + x m ) ≥ min { h( x i ); 1 ≤ i ≤ m } . (2.10)In general, the inequality (2.10) is strict. Consider Example 1.2 with x = e + e , x = e + e , and h( x ) = h( x ) = 0 and h( x + x ) = 1. We haveequality in (2.10) if the vectors x i satisfy the assumption of the followinglemma. Lemma 2.5.
Let U = ( u , . . . , u m ) ∈ U . If x = P mi =1 x i , x i ∈ h u i i , i = 1 , . . . , m , x = 0 , then h( x ) = min { h( x i ); 1 ≤ i ≤ m, x i = 0 } (2.11) and e( x ) = max { e( x i ); 1 ≤ i ≤ m, x i = 0 } . (2.12) Proof.
To prove the identity (2.11) we set ˜ q = min { h( x i ); x i = 0 } and q =h( x ). Then q ≥ ˜ q . On the other hand we have x = f q y , y = P y i , y i ∈ h u i i .Hence x i = f q y i for all i , and therefore ˜ q ≥ q .With regard to (2.12) we set ˜ ℓ = max { e( x i ); x i = 0 } and ℓ = e( x ). Then0 = f ℓ x = P mi =1 f ℓ x i implies f ℓ x i = 0 for all i . Hence ˜ ℓ ≤ ℓ . On the otherhand 0 = f ℓ − x = X mi =1 f ℓ − x i implies f ℓ − x j = 0 for some j , 1 ≤ j ≤ k , and therefore ˜ ℓ ≥ ℓ .We remark that the preceding lemma can be deduced from results onmarked subspaces in [5]. 7 emma 2.6. Suppose λ t is an elementary divisor of f . Then u is a generatorof V with e( u ) = t if and only if f t u = 0 and h( f j u ) = j, j = 0 , , . . . , t − . (2.13) Proof.
Suppose u is a generator and V = h u i ⊕ V and e( u ) = t. (2.14)Let h( f t − u ) = ( t − τ , τ ≥
0. Then f t − u = f t − τ ˜ w for some ˜ w = w + w with w ∈ h u i , w ∈ V . Then h u i ∩ V = 0 implies f t − u = f t − τ w , and weobtain τ = 0. Hence h( f t − u ) = t −
1, which is equivalent to (2.13).Now suppose u ∈ V satisfies (2.13). Let v ∈ V be a generator corre-sponding to λ t such that V = h v i ⊕ W and e( v ) = t . We have shownbefore that H ( v ) = (0 , , . . . , t − , ∞ , . . . , ∞ ). Hence H ( u ) = H ( v ), andtherefore Theorem 2.4 implies u = αv for some α ∈ Aut(
V, f ). Then V = α ( h v i ⊕ W ) = h u i ⊕ αW shows that u is also a generator.A consequence of Lemma 2.6 is the following observation. Suppose u isa generator of V and w ∈ h u i and h( w ) = 0. Then h w i = h u i . Lemma 2.7.
Suppose λ t is an unrepeated elementary divisor of f . Let U =( u , . . . , u m ) ∈ U , and e( u p ) = t . If u ∈ V then the following statements areequivalent. (i) The vector u is a generator of V with e( u ) = t . (ii) There exists an α ∈ Aut(
V, f ) such that u = αu p . (iii) We have u = y + g with y = cu p + f v, c = 0 , v ∈ h u p i ,g ∈ h u i ; i = p i [ f t ] . (2.15) Proof. If u and u ρ are generators of V then we have e( u ) = e ( u ρ ) if and onlyif H ( u ) = H ( u ρ ). Hence it follows from Theorem 2.4 and Lemma 2.6 that(i) and (ii) are equivalent.(i) ⇒ (iii) Let u be decomposed such that u = y + g and y ∈ h u p i and g ∈ h u i ; i = p i . Then (2.12) implies e( g ) ≤ e( u ) = t , that is g ∈ V [ f t ]. Letus show that h( y ) = 0, or equivalently y = cu p + f v, c = 0 , v ∈ h u p i . (2.16)8e have g = g < + g > with g < ∈ h u i ; i < p i , and g > ∈ h u i ; i > p i . If i < p then e( u i ) < e( u p ) = t . Hence f t − g < = 0, and f t − u = f t − y + f t − g > . (2.17)If i > p then e( u i ) > t . Therefore h u i i [ f t ] = f e( u i ) − t h u i i ⊆ f V. Hence h( g > ) ≥ f t − g > ) ≥ t . Now suppose h( y ) = 0. Then h( y ) ≥ f t − y ) ≥ t . Therefore (2.17) implies h( f t − u ) ≥ t . This is a contra-diction to the assumption that u is a generator with h( f t − u ) = t −
1. Henceh( y ) = 0.(iii) ⇒ (i) Assume (2.15). Then (2.16) implies that y is a generator withe( y ) = t . Thus h( y ) = 0 and h( f t − y ) = t −
1. Moreover, we have e( g ) ≤ t .Hence e( u ) = max { e( y ) , e( g ) } = t . From (2.11) follows t − ≤ h( f t − u ) = min { h( f t − y ) , h( f t − g ) } ≤ h( f t − y ) = t − . Hence h( f t − u ) = t −
1. Then Lemma 2.6 completes the proof.Lemma 2.7 will be used in the proof of Proposition 2.11. We note thatthe assumption | K | = 2 implies that the vector y in (2.15) is of the form y = u ρ + f v, v ∈ h u ρ i . (2.18)If u is a generator with e( u ) = t then there is no gap in the indicatorsequence H ( f j u ) = ( j, j + 1 , . . . , t − , ∞ , . . . , ∞ ). We mention withoutproof that h x i c is hyperinvariant if and only if H ( x ) has no gap. We onlyneed the following special case of that result. Lemma 2.8.
Let λ t be an unrepeated elementary divisor of f and u a gen-erator of V with e( u ) = t . Then h f j u i c = Im f j ∩ Ker f t − j , j = 0 , . . . , t, (2.19) and h f j u i c is hyperinvariant.Proof. Let U = ( u , . . . , u m ) ∈ U such that e( u p ) = t and u p = u . ThenKer f t = h u p i M (cid:0) ⊕ ≤ i ≤ m ; i = p h u i i [ f t ] (cid:1) and h u p i = (cid:10) cu p + f v ; c = 0 , v ∈ h u p i (cid:11) . HenceKer f t = h cu p + f v + g ; c = 0 , v ∈ h u p i , g ∈ h u i ; i = p i [ f t ] i = h αu p ; α ∈ Aut(
V, f ) i = h u p i c .
9f 0 < j < t then h f j u p i c = f j h u p i c = f j Ker f t = f j Ker f t − j = Im f j ∩ Ker f t − j . A general theorem that contains the following lemma can be found in [8,Lemma 65.4, p. 4].
Lemma 2.9.
Suppose V = h u i ⊕ h u i and e( u ) < e( u ) . If x ∈ V thenthere exists an automorphism α ∈ Aut(
V, f ) such that αx = f k u + f k u (2.20) for some k , k ∈ N .Proof. Let x = x + x , x i ∈ h u i i , i = 1 ,
2. Suppose x = 0 and x = 0.If h( x i ) = k i then x i = f k i ˜ u i , ˜ u i ∈ h u i i , h(˜ u i ) = 0, i = 1 ,
2. Therefore h ˜ u i i = h u i i , i = 1 ,
2, such that (˜ u , ˜ u ) ∈ U . Then α : (˜ u , ˜ u ) ( u , u ) ∈ Aut(
V, f ) yields (2.20).
In this section we assume dim Ker f = 2 such that V = h u i ⊕ h u i . Weprove two propositions. They provide the complete description of the setChinv( V, f ) \ Hinv(
V, f ) in Theorem 1.3(ii). The following notation will beconvenient. If we write α : ( u , u ) (ˆ u , ˆ u ) ∈ Aut(
V, f )we assume (ˆ u , ˆ u ) ∈ U , and α denotes the automorphism in Aut( V, f ) de-fined by ( αu , αu ) = (ˆ u , ˆ u ). Proposition 2.10.
Let | K | = 2 . Suppose V = h u i ⊕ h u i and e( u ) = R , e( u ) = S , and R + 1 < S . If X is a characteristic non-hyperinvariantsubspace of V then X = h f R − s u + f S − q u i c with < s < q and ≤ R − s < S − q, (2.21) and X H = h f R − s +1 u , f S − q +1 u i = Im f R − s +1 ∩ Ker f q − (2.22) is the largest hyperinvariant subspace contained in X . roof. Theorem 2.2(iii) and Theorem 2.1(ii) imply X H = ( X ∩ h u i ) ⊕ ( X ∩ h u i ) = W ( ~r ) = h f r u i ⊕ h f r u i (2.23)for some pair ~r = ( r , r ) satisfying r ≤ R , r ≤ S , and0 ≤ r ≤ r and 0 ≤ R − r ≤ S − r . (2.24)Since X is not hyperinvariant we have X % X H . Let x ∈ X \ X H . ByLemma 2.9 there exists an automorphism α ∈ Aut(
V, f ) be such that αx = z = f µ u + f µ u ∈ X \ X H . (2.25)Suppose µ ≥ r . Then f µ u ∈ h f r u i ⊆ X , and because z ∈ X also f µ u = z − f µ u ∈ X . Hence f µ u ∈ X ∩ h u i and f µ u ∈ X ∩ h u i , andwe would obtain z ∈ X H . Similarly, it is impossible that µ ≥ r . Therefore µ < r , µ < r . (2.26)We shall see that µ + 1 = r , µ + 1 = r . (2.27)If R = 1 then µ = 0, and µ + 1 = r = 1. If R > f u = 0, and β : ( u , u ) ( u + f u , u ) ∈ Aut(
V, f )yields βz = ( f µ u + f µ +1 u ) + f µ u = z + f µ +1 u ∈ X. Since X is characteristic and z ∈ X , we have βz ∈ X . Hence f µ +1 u ∈ X ∩ h u i = h f r u i , and we obtain µ + 1 ≥ r . A similar argument yields, µ + 1 ≥ r . Hence(2.26) implies the relations (2.27). Thus z = f µ u + f µ u , µ = r − , µ = r − . Hence there exists a unique vector z ∈ X \ X H with a representation (2.25).Then (2.9) implies X \ X H = { αz ; α ∈ Aut(
V, f ) } and (2.8) yields X = h z i c = h f µ u + f µ u i c . (2.28)According to the definitions of s and t we have 0 ≤ µ ≤ µ and 0 ≤ R − µ ≤ S − µ . Hence it remains to show that µ = R, µ = µ and R − µ = S − µ . (2.29)11uppose R = µ . Then z = f µ u , and Lemma 2.8 implies X = h f µ u i c ∈ Hinv(
V, f ). Suppose µ = µ . Then z = f µ ( u + u ). Using γ : ( u , u ) ( u , u + u ) ∈ Aut(
V, f )we obtain γ − z = f µ u ∈ X . Hence Lemma 2.8 implies X = h f µ u i c ∈ Hinv(
V, f ). Suppose R − µ = S − µ . Then S − ( µ − µ ) = R and z = f µ ( u + f µ − µ u ) = f µ ( u + f S − R u ) . Therefore σ : ( u , u ) ( u + f S − R u , u ) ∈ Aut(
V, f ) yields σ − z = f µ u .Then X = h f µ u i c ∈ Hinv(
V, f ). Hence the inequalities (2.29) are valid.From (2.23) and (2.27) follows X H = h f µ +1 u , f µ +1 u i . Moreover (2.29)implies h f µ +1 u , f µ +1 u i = Im f µ +1 ∩ Ker f S − ( µ +1) .We obtain (2.21) and (2.22) if we set µ = R − s and µ = S − q . Proposition 2.11.
Assume | K | = 2 . Suppose V = h u i ⊕ h u i , and e( u ) = R , e( u ) = S such that R + 1 < S . Let s, q be integers satisfying < s < q, ≤ R − s < S − q. (2.30) Then the subspace X = h f R − s u + f S − q u i c is characteristic and not hyperinvariant, and X = h f R − s u + f S − q u , f R − s +1 u , f S − q +1 u i . (2.31) We have dim X = s + q − . (2.32) If s > then f | X has the elementary divisors λ q and λ s − . If s = 1 then X = h f R − s u + f S − q u i and the corresponding elementary divisor is λ q .Proof. Define z = f R − s u + f S − q u . Then X = h z i c . Set ˜ q = R + ( q − s ) and˜ z = u + f S − ˜ q u and ˜ X = h ˜ z i c . Then ( R − s ) + ( S − ˜ q ) = S − q . Therefore z = f R − s ˜ z and X = f R − s ˜ X , and(2.30) is equivalent to R < ˜ q, < S − ˜ q. (2.33)Let us first deal with the height-zero space ˜ X = h ˜ z i c and then pass to X = h z i c . Let α ∈ Aut(
V, f ). We determine α ˜ z using Lemma 2.7. Recall12hat | K | = 2 implies (2.18), that is we have y = u ρ + f v , v ∈ h u ρ i in (2.15).If αu = x + x , x i ∈ h u i i , i = 1 ,
2, then x = u + f v , v ∈ h u i and x ∈ h u i ; i = 1 i [ f R ] = h u i [ f R ] = f S − R h u i . Similarly, αu = y + y , y i ∈ h u i i , i = 1 ,
2, and y = u + f w , w ∈ h u i and y ∈ h u i ; i = 2 i [ f S ] = h u i [ f S ] = h u i . Then α ˜ z = ( u + f v + x ) + ( f S − ˜ q u + f S − ˜ q +1 w + f S − ˜ q y ) =˜ z + ( f v + f S − ˜ q y ) + ( x + f S − ˜ q +1 w ) . (2.34)From S − ˜ q > f S − ˜ q y ∈ f h u i . From ˜ q > R follows S − R ≥ S − ˜ q +1,and therefore x ∈ f S − ˜ q +1 h u i . Set ˆ v = f v + f S − ˜ q y and ˆ w = x + f S − ˜ q +1 w .Then α ˜ z = ˜ z + ˆ v + ˆ w, and ˆ v ∈ f h u i , ˆ w ∈ f S − ˜ q +1 h u i . (2.35)Define ˜ L = h ˜ z, f S − ˜ q +1 u i . Because of f ˜ z = f u + f S − ˜ q +1 u we obtain ˜ L = h ˜ z, f u , f S − ˜ q +1 u i . Then (2.35) implies h ˜ z i c ⊆ ˜ L . If β : ( u , u ) ( u , u + f u ) ∈ Aut(
V, f )then β ˜ z = ˜ z + f S − ˜ q +1 u ∈ h ˜ z i c . Hence β ˜ z − ˜ z = f S − ˜ q +1 u ∈ h ˜ z i c , and therefore ˜ L ⊆ h ˜ z i c , and we obtain˜ L = h ˜ z i c = h ˜ z, f u , f S − ˜ q +1 u i . (2.36)We determine the dimension of h ˜ z i c . If R = 1 then f u = 0 and f ˜ z = f S − ˜ q +1 u . Therefore h ˜ z i c = h ˜ z i and dim h ˜ z i = e(˜ z ) = ˜ q . If R > h ˜ z i c = h ˜ z, f u i . Let x = X R − µ =0 c µ f µ u + X ˜ q − µ =0 c µ f S − ˜ q + µ u ∈ h ˜ z i . Then x ∈ h f u i if and only if c = · · · = c ˜ q − = 0, that is, x = 0. Hence h ˜ z i ∩ h f u i = 0, anddim h ˜ z i c = dim h ˜ z i + dim h f u i = ˜ q + ( R − . (2.37)13t this point we go back to X = h z i c = f R − s h ˜ z i . Then (2.36) yields(2.31). From (2.37) we obtaindim X = dim f R − s h ˜ z i +dim f R − s h f u i = [˜ q − ( R − s )]+[( R − − ( R − s )] = s + q − , which proves (2.32). If s > X = h z i ⊕ h f R − s +1 u i is a direct sumof f -cyclic subspaces of dimension q and s −
1, respectively. Hence, in thatcase, the elementary divisors of f | X are λ q and λ s − . It is easy to see thatthe case s = 1 leads to X = h z i .We show next that f R − s u / ∈ X . Suppose to the contrary that f R − s u ∈ X . Then z = f R − s u + f S − q u ∈ X would imply f S − q u ∈ X . Hence h f R − s u i ⊕ h f S − q u i ⊆ X , and therefore dim X ≥ s + q , in contradiction to(2.32). Hence f R − s u / ∈ X . Let π be the projection of V on h u i along h u i .Then π ∈ End(
V, f ). But π z = f R − s u / ∈ X . Hence the subspace X is nothyperinvariant. Example 1.2 continued.
If (
R, S ) = (1 ,
3) then ( s, q ) = (1 ,
2) is the onlysolution of (2.30). Then ( R − s, S − q ) = (0 , X = h f e + f e i c isthe only characteristic non-hyperinvariant subspace of V . According to (2.2)there are 6 hyperinvariant subspaces in V . (cid:3) According to Shoda’s theorem only unrepeated elementary divisors are rele-vant for the existence of characteristic non-hyperinvariant subspaces. In thissection we examine this fact in more detail. Let E and G be invariantsubspaces of V and assume V = E ⊕ G and d ( f, t ) = d ( f | E , t ) if d ( f, t ) = 1 and d ( f, t ) = d ( f | G , t ) if d ( f, t ) > . (3.1)Thus the unrepeated elementary divisors of f are those of f | E and the re-peated ones are those of f | G . If (3.1) holds then there exists a generator tuple U = ( u , . . . , u m ) adapted to E and G such that E = h u i ; e( u i ) = t i ; d ( t i ) = 1 i and G = h u i ; e( u i ) = t i ; d ( t i ) > i . (3.2)We shall see that a characteristic subspace X is hyperinvariant in V if andonly if X ∩ E is hyperinvariant in E . We first consider a general direct sumdecomposition of V . 14 emma 3.1. Let V and V be invariant subspaces of V such that V = V ⊕ V . (i) If X ∈ Chinv(
V, f ) then ( X ∩ V i ) ∈ Chinv( V i , f | V i ) , i = 1 , . (ii) If X ∈ Hinv(
V, f ) then ( X ∩ V i ) ∈ Hinv( V i , f | V i ) , i = 1 , . (iii) A subspace X is hyperinvariant if and only if X is characteristic and X = ( X ∩ V ) ⊕ ( X ∩ V ) and X ∩ V i ∈ Hinv( V i , f | V i ) , i = 1 , . (3.3) Proof. (i) Let x ∈ X ∩ V and g ∈ Aut( V , f | V ). To show that g ( x ) ∈ X weextend g to an automorphism g ∈ Aut(
V, f ) as follows. Let ι be the identitymap of V . Then g = g + ι ∈ Aut(
V, f ). Therefore X ∈ Chinv(
V, f ) implies g ( x ) ∈ X . On the other hand g ( x ) = g ( x ) ∈ V . Hence g ( x ) ∈ X ∩ V .(ii) Let x ∈ X ∩ V and h ∈ End( V , f | V ). If 0 is the zero map on V then h = h + 0 ∈ End(
V, f ). An argument as in part (i) shows that h ( x ) ∈ X ∩ V .(iii) Let V i = ⊕ m i ν =1 h u ( i ) ν i , i = 1 ,
2, be decomposed into cyclic subspaces, andlet U = ( u j ) mj =1 ∈ U contain the vectors u (1) ν , ν = 1 , . . . , m , and u (2) ν , ν =1 , . . . , m . Define X i = X ∩ V i , i = 1 ,
2. Suppose X is characteristic and(3.3) holds. Then Lemma 3.1(ii) and V ∩ V = 0 imply X i = ⊕ m i ν =1 ( X i ∩ h u ( i ) ν i ) = ⊕ mj =1 ( X i ∩ h u j i ) , i = 1 , . Hence X = X ⊕ X = ⊕ mj =1 (cid:16) ( X ∩ h u j i ) + ( X ∩ h u j i ) (cid:17) ⊆ ⊕ mj =1 [( X + X ) ∩ h u j i ] = ⊕ mj =1 ( X ∩ h u j i ) ⊆ X. Then X satisfies (2.6), and therefore X is hyperinvariant. Using (2.6) it isnot difficult to see that X ∈ Hinv(
V, f ) implies (3.3).We apply the preceding lemma to the decomposition V = E ⊕ G . Let π E be the projection of V on E along G , and let π G denote the complementaryprojection such that π E + π G = ι . Then π E f = f π E and π G f = f π G . Let( X ∩ E ) H | E denote the largest hyperinvariant subspace (with respect to f | E )contained in E . Lemma 3.2.
Let E and G be subspaces of V such that (3.1) holds. Suppose X is a characteristic subspace of V . Then π E X = X ∩ E and π G X = X ∩ G , and X = ( X ∩ E ) ⊕ ( X ∩ G ) . (3.4) Moreover, X ∩ E ∈ Chinv(
E, f | E ) and X ∩ G ∈ Hinv(
G, f | G ) , (3.5) and X H = ( X ∩ E ) H | E ⊕ ( X ∩ G ) (3.6)(ii) The subspace X is hyperinvariant in V if and only if X ∩ E is hyper-invariant in E .Proof. (i) Let U ∈ U satisfy (3.2). If x ∈ X then Theorem 2.2(i) yields π G x = (cid:18)X d ( t i ) > π i (cid:19) x = X d ( t i ) > π i x ∈ X. Hence π G X ⊆ X ∩ G , and therefore π G X = X ∩ G . Then x = π G x + π E x implies π E x ∈ X . Thus we obtain π E X = X ∩ E , and X = π E X ⊕ π G X .From Lemma 3.1(i) we conclude that X ∩ E and X ∩ G are characteristicin E , respectively in G . According to (3.1) the map f | G has only repeatedelementary divisors. Hence Theorem 1.1 implies that X ∩ G is hyperinvariantin G . The description of X H in (3.6) follows from (2.7).(ii) The subspace X ∩ G is hyperinvariant in G . We apply Lemma 3.1(iii).The assumption that X is characteristic is essential for (3.4). Example 3.3.
Let V = h u , u , u i with e( u ) = 1 , e( u ) = e( u ) = 2. Then E = h u i and G = h u , u i . Set X = h u + f u i . Then X ∩ G = 0 and X % ( X ∩ E ) ⊕ ( X ∩ G ).In the following we deal with invariant subspaces associated to subsets ofunrepeated elementary divisors of f . Let T be an invariant subspace of V such that f | T has only unrepeated elementary divisors and such that V = T ⊕ V for some V ∈ Inv (
V, f ) , and f | T and f | V have no elementary divisors in common. The subspace T can also be characterized as follows. There exists a T ∈ Inv (
V, f ) such that T ⊕ T = E, and T ⊕ G = V , E ⊕ G = V and E and G satisfy (3.1) . (3.7)16et π T be the projection on T along V and π V be the complementary pro-jection. If Y ⊆ T then Y c T denotes the characteristic hull of Y with respectto T , Y c T = (cid:10) α T y ; y ∈ Y, α T ∈ Aut(
T, f | T ) (cid:11) . A spin-off from the following lemma is a proof of Theorem 1.3(i).
Lemma 3.4.
Let T be an invariant subspace such that f | T has only unrepeatedelementary divisors and (3.7) holds. Suppose X is a characteristic subspaceof V . (i) Then X ∩ T ∈ Chinv(
T, f | T ) . If the subspace X is hyperinvariant in V then X ∩ T is hyperinvariant in T . (ii) We have
Aut(
T, f | T ) = { π T απ T ; α ∈ Aut(
V, f ) } . (3.8)(iii) If Y ⊆ T then Y c ∩ T = π T Y c = Y c T . (3.9) Proof. (i) Because of V = T ⊕ V one can apply Lemma 3.1. (ii) Let U = { u , . . . , u m } be a generator tuple of V such that a subtuple U T = { u τ , . . . , u τ q } ⊆ U is a generator tuple of T (with respect to f | T ). Set I T = { τ , . . . , τ q } . If i ∈ I T then Lemma 2.7 implies α u i = c i u i + f v i + X j ∈ I T , j = i w j + X k / ∈ I T , ≤ k ≤ m x k , where c i = 0, v i ∈ h u i i , w j ∈ h u j i [ f t i ] and x k ∈ h u k i [ f t i ]. Hence( π T α ) u i = c i u i + f v i + X j ∈ I T ,j = i w j . Then Lemma 2.5 yields e (cid:0) ( π T α ) u i (cid:1) = t i , andh( f t i − ( π T α ) u i ) = h( f t i − u i ) = t i − , and h(( π T α ) u i ) = 0. By Lemma 2.6 the element ( π T α ) u i ∈ T is a generator.Hence (cid:0) ( π T α ) u τ , . . . , ( π T α ) u τ q (cid:1) is a generator tuple of T . Then the mapgiven by π T α : u i ( π T α ) u i , i ∈ I T , is in Aut( T, f | T ). Hence π T απ T ∈ Aut(
T, f | T ). Now consider an automor-phism α T ∈ Aut(
T, f | T ). We extend α T to an automorphism ˜ α ∈ Aut(
V, f )defining˜ α : u i α T u i if i ∈ I T , and ˜ α : u i u i if i / ∈ I T . (3.10)17hen ( π T ˜ απ T ) u τ i = α T u τ i , i = 1 , . . . , q , and therefore α T = π T ˜ απ T .(iii) Let us show first that π T Y c = Y c ∩ T . If α ∈ Aut(
V, f ) then α T = π T απ T ∈ Aut(
T, f | T ). Let ˜ α ∈ Aut(
V, f ) be the extension of α T given by(3.10). If y ∈ Y ⊆ T then π T αy = π T απ T y = α T y = ˜ αy ∈ Y c . Hence π T Y c ⊆ Y c , which suffices to prove π T Y c = Y c ∩ T . Then Y c ∩ T = π T Y c h ( π T α ) y ; α ∈ Aut(
V, f ) , y ∈ Y i = h ( π T απ T ) y ; α ∈ Aut(
V, f ) , y ∈ Y i = h ( α T ) y ; α T ∈ Aut(
T, f | T ) , y ∈ Y i = Y c T . Proof of Theorem 1.3(i).
We apply Lemma 3.4. Let T = h u ρ , u τ i . Set z s,q = f R − s u ρ + f S − q u τ , and Y = h z s,q i . Then Y ⊆ T , and Y c T = Y c ∩ T .It follows from Proposition 2.11 that Y c T is not hyperinvariant in T . Hence Y c = h z s,q i c is not hyperinvariant in V . (cid:3) Suppose Y ⊆ E is characteristic and non-hyperinvariant with respect to f | E How can one extend Y to a subspace X that has these properties in theentire space V ? Theorem 3.5.
Let E and G be subspaces of V such that (3.1) holds. Let Y ∈ Chinv(
E, f | E ) . If Y s is a subspace of Y and W is a subspace of G suchthat Y s + W is characteristic in V then ( Y + W ) c ∩ E = Y. (3.11) The subspace ( Y + W ) c is hyperinvariant in V only if Y is hyperinvariantin E .Proof. Set X = ( Y + W ) c . We are going to show that π E X = Y . Becauseof Y ⊆ E and Y ⊆ X we see that Y ⊆ ( X ∩ E ) ⊆ π E X. (3.12)It is obvious that ( Y + W ) c ⊆ Y c + W c . From Y, W ⊆ ( Y + W ) c follows Y c + W c ⊆ ( Y + W ) c . Hence X = Y c + W c . Therefore π E X = π E Y c + π E W c . (3.13)Since Y s + W is characteristic we have W c ⊆ ( Y s + W ) c = Y s + W . Hence Y s ⊆ E and W ⊆ G , or equivalently π E W = 0, imply π E W c ⊆ Y s ⊆ Y . Since Y is characteristic in E it follows from Lemma 3.4 that Y c E = Y = π E Y c .Hence (3.13) implies π E X ⊆ Y , and (3.12) yields Y = X ∩ E = π E X .Thus we proved (3.11). If Y is not hyperinvariant in E then it follows fromLemma 3.2 that X is not hyperinvariant in V .18 The main theorem
In this section we drop the assumption that V is generated by two vectors.Hence the map f can have more than two elementary divisors. We assumethat only two of them are unrepeated. In that case we obtain a descriptionof the family of characteristic non-hyperinvariant subspaces which extendsTheorem 1.3(ii).We first consider the case where f has no unrepeated elementary divisors.In terms of a decomposition V = E ⊕ G in (3.1) that assumption is equivalentto V = G . Lemma 4.1.
Let U = ( u , . . . , u m ) ∈ U , e( u i ) = t i , i = 1 , . . . , m . Suppose f has no unrepeated elementary divisors, that is d ( t i ) > f or all i = 1 , . . . , m. (4.1) Then X ⊆ V is hyperinvariant if and only if X = h f r u + · · · + f r m u m i c (4.2) for some ~r = ( r , . . . , r m ) satisfying (2.1) .Proof. By Shoda’s theorem the assumption (4.1) implies that each charac-teristic subspace of V is hyperinvariant. Hence, if X is of the form (4.2)then X ∈ Hinv(
V, f ). Now suppose X is hyperinvariant. According to The-orem 2.1 we have X = W ( ~r ) = h f r u i ⊕ · · · ⊕ h f r m u m i (4.3)for some ~r satisfying (2.1). Define w = f r u + · · · + f r m u m . Then w ∈ W ( ~r ).Since W ( ~r ) is hyperinvariant, it is obvious that h w i c ⊆ W ( ~r ). To prove theconverse inclusion we apply Theorem 2.2(ii) to the hyperinvariant subspace h w i c . We obtain π i w = f r i u i ∈ h w i c , i = 1 , . . . , m , and therefore W ( ~r ) ⊆h w i c . Hence X = W ( ~r ) = h w i c .The notation in Theorem 4.2 below will be the following. We write ~µ ′ = ~µ + ( ~e ρ + ~e τ ) if( µ ′ , . . . , µ ′ ρ , . . . , µ ′ τ , . . . , µ ′ m ) =( µ , . . . , µ ρ , . . . , µ τ , . . . , µ m ) + (0 , . . . , , . . . , , . . . , , (4.4)that is, µ ′ i = µ i if i = ρ, τ, and µ ′ ρ = µ ρ + 1 , µ ′ τ = µ τ + 1 . (4.5)Then ~µ ′ ∈ L ( ~t ) means µ ′ i ≤ t i , i = 1 , . . . , m, and µ ′ ≤ · · · ≤ µ ′ m , t − µ ′ ≤ · · · ≤ t m − µ ′ m . (4.6)19 heorem 4.2. Let | K | = 2 . Suppose that among the elementary divi-sors of f there are exactly two unrepeated ones, namely λ R and λ S . Let U = ( u , . . . , u m ) ∈ U and e( u ρ ) = R , e( u τ ) = S . A subspace X ⊆ V ischaracteristic and not hyperinvariant if and only if X = h f µ u + · · · + f µ m u m i c , (4.7) such that the entries µ ρ and µ τ of µ = ( µ , . . . , µ m ) satisfy ≤ µ ρ < µ τ and < R − µ ρ < S − µ τ , (4.8) and such that ~µ + ( ~e ρ + ~e τ ) ∈ L ( ~t ) .Proof. Set E = h u ρ i ⊕ h u τ i and G = h u i ; e( u i ) = t i ; d ( t i ) > i . Then V = E ⊕ G , as in (3.1). Suppose X ∈ Chinv(
V, f ) \ Hinv(
V, f ). Lemma 3.2 implies X = ( X ∩ E ) ⊕ ( X ∩ G ) . The subspace X ∩ G is hyperinvariant in G , whereas X ∩ E is characteristicbut not hyperinvariant in E . By assumption E is generated by two elements.Hence it follows from Proposition 2.10 that X ∩ E = h z i c E . Referring to (2.28)and (2.29) we obtain z = f µ ρ u ρ + f µ τ u τ for some integers µ ρ , µ τ satisfying(4.8). According to Lemma 4.1 we have X ∩ G = h w i c G for some w = X ≤ i ≤ m ; i = ρ,τ f µ i u i ∈ X ∩ G. Set ˆ z = z + w . Let us show that X = h ˆ z i c . If α E ∈ Aut(
E, f | E ) and α G ∈ Aut( f | G , V ) then α V = α E + α G ∈ Aut( f, V ). Hence α E z + α G w = α V ( z + w ) ∈ h ˆ z i c , and we obtain X = h z i c E ⊕ h w i c G ⊆ h ˆ z i c . (4.9)Since X is characteristic it is obvious that h ˆ z i c ⊆ X . Therefore X = h ˆ z i c ,and X is of the form (4.7). From (2.22) follows (cid:16) h z i c E (cid:17) H | E = h f µ ρ +1 u ρ i ⊕ h f µ τ +1 u τ i . The proof of Lemma 4.1 shows that h w i c G = ⊕ ≤ i ≤ m ; i = ρ,τ h f µ i u i i . X H = (cid:16) h z i c E (cid:17) H | E ⊕ (cid:10) w (cid:11) c G = ⊕ i =1 ,...,m h f µ ′ i u i i , (4.10)with µ ′ i , i = 1 , . . . , m , defined by (4.5). The subspace X H is hyperinvariant.Hence Theorem 2.1 implies ~µ ′ = ( µ ′ , . . . , µ ′ m ) ∈ L ( ~t ).Now consider a subspace X = D X mi =1 f µ i u i E c assuming that the inequalities (4.8) hold and ~µ ′ = ~µ + ( ~e ρ + ~e τ ) ∈ L ( ~t ).Define z = f µ ρ u ρ + f µ τ u τ , w = X ≤ i ≤ m ; i = ρ,τ f µ i u i , and ˆ z = z + w. Then X = h ˆ z i c . Since X is characteristic it follows from Lemma 3.2 that π E ˆ z = z ∈ X ∩ E , π G ˆ z = w ∈ X ∩ G . Set Y = h z i c E and W = h w i c G . ThenLemma 3.4(iii) implies Y = h z i c ∩ E . The inequalities (4.6) ensure that W is hyperinvariant in G and that W = ⊕ i = ρ,τ h f µ i u i i . Let us show first that X = ( Y + W ) c . Since X is characteristic we have ( Y + W ) c ⊆ X . Conversely, z + w ∈ Y + W yields X = h z + w i c ⊆ ( Y + W ) c . It follows from (4.8) that h z i c E = Y is not hyperinvariant in E . Hence Y H | E = h f µ ρ +1 u ρ , f µ τ +1 u τ i = h f µ ′ ρ u ρ , f µ ′ τ u τ i . Then ~µ ′ ∈ L ( ~t ) implies that Y H | E + W is hyperinvariant in V . We applyTheorem 3.5 choosing Y s = Y H | E and conclude that X is not hyperinvariant. Example 4.3.
We consider a map f with elementary divisors λ, λ , λ , λ and V = ⊕ i =1 h u i i such that ~t = (cid:0) e( u i ) (cid:1) = (1 , , , ≤ µ < , ≤ µ < µ < µ , − µ < − µ , has the unique solution ( µ , µ ) = (0 , ~µ = (0 , , µ , µ ) and ~µ ′ = (1 , , µ , µ ) . Then ~µ ′ ∈ L ( ~t ) if and only if ( µ , µ ) = ( j, j ), j = 2 , . . . ,
6. Define g j = u + f u + f j u + f j u and X j = h g j i c , j = 2 , . . . ,
6, and X = h u + f u i c .Then e( u + f j − u + f j − u ) = e( u ) , j = 5 , , implies X = X = X . We obtain 4 different characteristic not hyper-invariant subspaces, namely X, X , X , X . There are n H = 30 hyperin-variant subspaces in V . 21 .1 Concluding remarks We have restricted our study to the case of two unrepeated elementary di-visors. Using methods of our paper one can prove more general results. Forexample one can extend Theorem 1.3(i) as follows.
Theorem 4.4.
Let | K | = 2 . Suppose λ t ρ , . . . , λ t ρk are unrepeated elementarydivisors of f such that t ρ < · · · < t ρ k . Let U = ( u , . . . , u m ) ∈ U be agenerator tuple with e( u ρ j ) = t ρ j , j = 1 , . . . , k . Suppose ( µ ρ , . . . , µ ρ k ) ∈ N k and set z = f µ ρ u ρ + · · · + f µ ρk u ρ k . If µ ρ j < t ρ j , j = 1 , . . . , k, (4.11) and µ ρ < µ ρ < · · · < µ ρ k and < t ρ − µ ρ < · · · < t ρ k − µ ρ k , (4.12) then X = h z i c is a characteristic and not hyperinvariant subspace of V . In Example 4.5 below we illustrate Theorem 4.4. We also construct anon-hyperinvariant subspace that is not the characteristic hull of a singleelement. Such subspaces will be studied in a subsequent paper.
Example 4.5.
Let | K | = 2. Suppose f has elementary divisors λ, λ , λ such that V = h u i ⊕ h u i ⊕ h u i and( t , t , t ) = (cid:0) e( u ) , e( u ) , e( u ) (cid:1) = (1 , , . Then ( µ , µ , µ ) = (0 , ,
2) is the only solution of the set of inequalities0 ≤ µ < , ≤ µ < , ≤ µ < ≤ µ < µ < µ , < − µ < − µ < − µ . Set z = u + f u + f u . (4.13)Then X = h z i c ∈ Chinv(
V, f ) \ Hinv(
V, f ). The fact that X is not hyper-invariant can be verified as follows. We have e( z ) = 3 and the indicatorsequence of z is H ( z ) = (0 , , , ∞ , ∞ ). Define Y = { x ∈ V | H ( x ) = H ( z ) } .Then Y = { z + v + v ; v ∈ f h u i ; v ∈ f h u i} . Hence h z i c = h Y i = h z, f u , f u i . (4.14)Then π z = f u / ∈ h Y i . Therefore h Y i is not hyperinvariant in V .22he following subspace W is also in Chinv( V, f ) \ Hinv(
V, f ). We shallsee that it is not the characteristic hull of a single element. Let W = h z , z i c and z = u + f u , z = f u . Then h z , f u i c = h z i and h z i c = Im f ∩ Ker f = Im f = h f u , f u i .Therefore W = h z i ⊕ h f u i . We have π z = u / ∈ W . Hence W is nothyperinvariant. Suppose W = h w i c for some w ∈ W . Then w = x + y with x ∈ h z i , y ∈ h f u i . Because of z ∈ W and h( z ) = 0 it is necessarythat h( w ) = 0. Hence h( x ) = 0. Therefore h x i = h z i , and x = γz forsome γ ∈ Aut( f, V ). Hence we can assume w = z + y . If h( y ) ≥ w ) = e( z ) and h w i c = h z i c ( W . If h( y ) = 2 then y = β ( f u ) for some β ∈ Aut( f, V ), and therefore h w i c = h z + f u i c = h z i c , where z is given by (4.13). We have seen before that π z = f u / ∈ h z i c .Hence h w i c = h z + z i c ( W . Therefore h w i c = W for all w ∈ W . (cid:3) Acknowledgement:
We are grateful to a referee for helpful suggestions andcomments.