topology If p = q and det(s + ), the pencl s + s called a regular pencl Pencls not satsfyng these condtons are referred to as sngular pencls A complet

Transcription

1 Orbt Closures of Sngular atrx Pencls D Hnrchsen Insttut fur Dynamsche Systeme Unverstat Bremen 8 Bremen Germany J O'Halloran Dept of athematcal Scences Portland State Unversty Portland, OR 9, USA Abstract Equvalence of matrx pencls (pars of pq matrces over C ) s gven by the GL p GL q -acton of smultaneous left and rght multplcaton The orbts under ths group acton were descrbed by Kronecker n 89 n terms of pencl nvarants column ndces, row ndces, and elementary dvsors In ths paper we descrbe the topologcal closures of these orbts, a problem motvated by our work n control theory For applcatons n control theory, the pencls of nterest have no row ndces Frst we gve necessary condtons for one pencl to le n the closure of another; then we show that, n the case of pencls wth no row ndces, these necessary condtons are sucent Introducton A p q matrx pencl over C s smply a par (; ) of p q matrces wth entres n C (often denoted s +) Two pencls s + ; = ;, are sad to be equvalent f, for some P GL p (C ) and Q GL q (C ), s + = sp Q? + P Q? () In other words, the equvalence classes are the orbts under the followng acton of G = GL p (C ) GL q (C ) on the set C (pq) of p q pencls G C (pq)! C (pq) () (P; Q) (s + ) = sp Q? + P Q? The orbts under ths acton were classed by Kronecker n 89 and a canoncal form appears n [] The classcaton s gven by a complete lst of nvarants for the acton () column ndces, row ndces, and elementary dvsors In terms of quvers (see []), a p q pencl denes a representaton of dmenson type (p; q) of the quver!! () Ths quver s classed as a \tame" quver Equvalence of pencls concdes wth somorphsm of representatons of the quver () Hence the orbts under the acton () may be vewed as somorphsm classes of representatons of ths quver The problem addressed n ths paper s a descrpton of the topologcal closure of each orbt Because an orbt under an algebrac group acton over an algebracally closed eld s a constructble set [, 8], orbt closure under the Zarsk topology concdes wth orbt closure under the standard The second author wshes to thank Unverstat Bremen for ts support durng the wrtng of ths paper second author was also supported by a grant from the atonal Scence Foundaton The

2 topology If p = q and det(s + ), the pencl s + s called a regular pencl Pencls not satsfyng these condtons are referred to as sngular pencls A complete descrpton of orbt closures of regular pencls was gven n [] The problem of descrbng closures of orbts under equvalence of sngular p q matrx pencls s central to the authors' recent work on sngular control systems In partcular, the analyss of hgh gan feedback depends on a soluton to ths orbt closure problem To apply feedback to a lnear control system descrbed by derental equatons _x(t) = Ax(t) + Bu(t); x(t) C n ; u(t) C m ; A C nn ; B C nm () (where x(t) represents the state and u(t) the control at tme t) we set u(t) = F x(t)+v(t), F C mn, obtanng the system _x(t) = (A + BF )x(t) + Bv(t) Consder a famly of systems whch arse from a gven system _x(t) = (A + BF )x(t) + Bu(t) ; C () _x(t) = Ax(t) + Bu(t) by applyng feedback If some entres of the n n matrx A + BF become arbtrarly large as tends to, we call ths a hgh gan feedback famly If members of the famly are replaced wth equvalent systems n such a way that a lmt exsts, we refer to ths as a lmt of the hgh gan feedback famly The equvalence transformatons we consder are left multplcaton of equaton (), change of x-coordnates, and change of u-coordnates Hence the transformed famly of systems s L R? _x(t) = L (A + BF )R? x(t) + L BW? u(t) () If ths famly has a lmt as!, the lmt of the rst coecent matrx L R? resultng n a system of equatons may be sngular, E _x(t) = A x(t) + B u(t) () A system of the form () s called a generalzed state space system In order to secure unque solvablty of (), we restrct our attenton to the systems for whch the n n pencl se? A s regular det(se? A) (8) To each generalzed state space system E _x(t) = Ax(t) + Bu(t) we assocate the n (n + m) pencl [se?a B] The feedback and equvalence transformatons of the system () may be realzed n terms of equvalence transformatons of the n (n + m) pencl [si? A B] In partcular, the transformed systems () correspond to L [si? A B] " #? R?W F R W Therefore, f the famly of systems () has a lmt as goes to, then that lmt s a pencl [se? A B ] n the closure of the G-orbt of the pencl [si? A B] In Secton we wll see that n (n + m) pencls of the form [se? A B] whch satsfy the regularty condton (8) have no row ndces Thus our descrpton of lmts under hgh gan feedback depends on a characterzaton of orbt closures of n (n + m) pencls wth no row ndces The paper s organzed as follows In Secton we dscuss the pencl equvalence acton () and ts nvarants In Secton we develop combnatoral tools to be used n provng the man theorem (9)

3 In Secton we gve necessary condtons for a p q pencl to be n the closure of the orbt of a gven p q pencl In Secton we restrct our attenton to n (n + m) pencls wth no row ndces and show that, n ths settng, these necessary condtons are sucent Specal cases of ths result appear n [] and n [8] (for regular pencls and for pencls assocated wth controllable state space systems respectvely) The Classcaton of atrx Pencls In ths secton we present Kronecker's classcaton of matrx pencls under the equvalence acton () of G = GL p (C ) GL q (C ) on the set of p q pencls C (pq) We denote the G-orbt of a pencl s + by O(s + ) The column ndces, row ndces, and elementary dvsors are a complete set of nvarants for ths acton The column ndces of a p q pencl are dened as follows (see []) Choose a mnmal degree soluton X (s) (C [s]) q to the equaton (s + )X(s) = For each, choose a mnmal degree soluton X (s) (C [s]) q to the equaton (s + )X(s) = whch s not contaned n the C [s]-span of X (s); ; X? (s) Ths process ends after a nte number c of steps (c q) We dene = deg X c?+ (s); = ; ; c, so that c The row ndces r of a pencl s + are dened smlarly as mnmal degrees of solutons of X(s)(s + ) = The elementary dvsors of a pencl s + are a set of homogeneous polynomals n C [s; t] assocated wth the pencl Let n = mn(p; q) and, for each j, j n, let D n?j+ be the greatest common dvsor of the set of j j mnors of s + t We use the conventon that the term of the greatest common dvsor wth the hghest power of s has coecent Let h = n+? rank(s +); then D j = for j < h The nvarant factors are the homogeneous polynomals ( j < h f j = h j n D j D j+ (D n+ = ) () Let fe ; ; e k g be the rreducble factors of D h Splttng the nvarant factors f j nto rreducble factors, we have f j = As a consequence of () and (), we get ky D j = f j f j+ f n = e r j h j n () ky P ǹ =j e r ` ; h j n () The factors fe r j g hjn;k are the elementary dvsors of the pencl s + Because Q D h s a homogeneous polynomal, ts factorzaton nto rreducble components s of the form t r k = (s? t) r Hence elementary dvsors are ether of the form t r (nnte elementary dvsors) or of the form (s? t) r, C (nte elementary dvsors) For each rreducble factor e of D h, we have r ;h r ;h+ r ;n (see [, vol ]) Theorem (Kronecker []) Two pq pencls are equvalent f and only f they have the same row ndces, column ndces, and elementary dvsors In applyng pencl theory to the theory of lnear systems of derental equatons, we make the followng observaton Lemma If the system E _x(t) = Ax(t)+Bu(t) satses the solvablty condton det(se?a), then the pencl [se? A B] has no row ndces

4 Proof Suppose the pencl [se? A B] has a row ndex Then there s a polynomal vector X(s) (C [s]) n, X(s) =, such that X(s)[sE? A B] = and so det(se? A) The solvablty condton det(se? A) s not nvarant under pencl equvalence However, at the end of ths secton we show that every n (n + m) pencl wth no row ndces s equvalent to one satsfyng the solvablty condton We see from Lemma that, for applcatons to systems theory, we can restrct our attenton to n (n + m) pencls wth no row ndces From the canoncal form whch appears n [], t s easy to see that an n (n + m) pencl wth no row ndces has rank n For such a pencl, ths canoncal form may be descrbed as follows Let s + be the n (n + m) pencl wth column P ndces = ( ; ; c ), elementary dvsors fe r j c g jn;k, and no row ndces Let j j= and r = n? j j For each column ndex > let C be the ( + ) pencl C = s s s ( +) For each nte elementary dvsor (s? t) r j, let J(s? t; r j ) be the r j r j \Jordan block" pencl s? J(s? t; r j ) = () s? r j r j and for each nnte elementary dvsor t r j, let J(t; r j ) be the r j r j pencl J(t; r j ) = s s r j r j L For any nte collecton fb gì= of p q pencls B of the form (), (), (), let B ì= be the \block dagonal" pencl B B () () B` P` p P` q () For ease of notaton, we consder C for = as a pencl wth zero rows and column so that, for example, c C = C C C j?pj?p j +c where j = and = for > j From [], we know that, up to reorderng of blocks, a pencl s + wth no row ndces s equvalent ()

5 to exactly one pencl of the form " S R where the sngular part S (gven by the column ndces = ( ; ; c )) L s the j j (j j +c) block () and the r r regular part R (gven by the elementary dvsors) s ;j J(e ; r j ) Then for n (n + m) pencls wth no row ndces, the canoncal form s C # (8) c C ;j J(e ; r j ) = C j J(e ; r ; ) J(ek; rk;n) (9) where a block J(e ; r j ) s absent f r j = It s easy to verfy from the canoncal form that, for n (n + m) pencls s + wth no row ndces the number of column ndces c s m () rank (s + ) = n () mx X r = r j = n? = deg D () ;j ( Dj (R) j r D j (s + ) = () r + j n In terms of representatons of the quver (), the blocks n the canoncal form dene a drect sum decomposton of the P representaton as P follows The canoncal form (9) denes a representaton m of dmenson type ( + m + ;j r ;j; n), e t denes two lnear transformatons from C P m +m+ P ;j r ;j to C n The doman decomposes as C P m +m+ and the codoman decomposes as P C n = ;j r ;j = m C m! C ;j! C r ;j where C s the zero vector space The blocks of the canoncal form (9) dene lnear transformatons from C + to C, = ; ; m, and from C r ;j to C r ;j for all pars (; j) ote that, for each zero column ndex, there s a lnear transformaton from C to, e a \ block" To see that each block s, n fact, ndecomposable, we employ the ndecomposablty crteron gven n [] a representaton s ndecomposable f and only f every semsmple element of ts stablzer s a scalar multple of the dentty element I I of G = GL p (C ) GL q (C ) Ths condton can be vered by drect computaton Thus the canoncal form (9) s a decomposton of the quver representaton nto ndecomposable components For applcatons n systems theory, we assocate to each generalzed state space system E _x = Ax + Bu the pencl [se? A B] That s to say, \system pencls" are n (n + m) pencls whose last m columns are constant We call a system pencl [se? A B] an nput pencl f det(se? A) A C r ;j A

6 Proposton A system pencl [se? A B] s equvalent to an nput pencl f and only f t has no row ndces Proof The necessty of the condton follows drectly from Kronecker's Theorem and Lemma To prove the converse we move the constant columns of the sngular part S to the rght sde of the matrx (9) Ths shows that every n (n + m) pencl wth no row ndces s equvalent to a pencl of the form J(s; ) J(s; j ) J(e ; r ; ) J(e k ; r k;n ) ` `j where ` s the column vector of length wth zeros n all but the last entry and n the last entry Snce () s an nput pencl the proposton s proved When there s no regular part n () (e r ;j = for all ; j), the canoncal form () corresponds to the Brunovsky canoncal form for controllable state space systems (see []) Combnatorcs of Young Dagrams Combnatoral problems are often approached by vewng them n terms of Young dagrams Gven a nte non-ncreasng sequence ( ) of nonnegatve ntegers, the assocated Young dagram conssts of rows of boxes wth boxes n the th row For example, the Young dagram assocated wth the lst (; ; ; ) s row row row row Let = f; ; ; g be the set of natural numbers and = [ fg For ease of exposton, we dentfy the set of Young dagrams (and the set of decreasng nte sequences n ) wth the followng subset D of D = f = ( ) + for all and = for only a nte number of g () We wll refer to elements of D as dagrams In the natural dentcaton wth dagrams, the set of column ndces of n (n + m) pencls wth no row ndces corresponds to the set D(m; n) = f D = for > m and mx () () ng () Denton For any dagram = ( ) n D, the dagram = ( ) dual to s gven by = cardfj j g; (8)

7 Wth respect to Young dagrams, the nteger s the number of boxes n the th column of the Young dagram of For example, the Young dagram dual to () s row = column of () row = column of () row = column of () row = column of () row = column of () (9) When dual dagrams are vewed n terms of ther Young dagrams, then t s clear that and X = X () ( ) = () In the followng we wll be consderng the dagrams dual to the lsts of column and row ndces of matrx pencls Usng (8) and (), t s easy to verfy that the set of dagrams dual to those n D(m; n) are D(m; n) = f D m and X ng () In [], Deconcn, Esenbud, and Proces dscuss a partal orderng on D whch generalzes the domnance orderng on parttons of ntegers and s dened as follows f kx ote that, n case P = P, we have kx for all k () f and only f () But ths s not true n general For any subset D D we say that ; D, < are adjacent n D f there s no D such that < < The proof of the followng proposton appears n [] Proposton ([]) Let ; D, < The dagrams and are adjacent n D f and only f one of the followng three condtons holds ) There s a natural number k such that = for all < k; k = ; k = ; and j = j = for all j > k ) There s a natural number k k+ = k+ + such that = for all = k; k + ; k = k? ; and ) There are natural numbers k < l such that = for all = k; l; k = k? ; l = l + ; and = k for all, k l The three adjacency condtons are llustrated n Fgure Corollary (a) Let ; D(m; n) ; < Then and are adjacent n D(m; n) f and only f they satsfy one of the condtons ), ), or ) of Proposton

8 Add a box Subtract a box Add a box Add a box Subtract a box ) ) ) Fgure Obtanng adjacent dagrams by addng and subtractng boxes (b) Suppose ; D(m; n); P = P Then < are adjacent n D(m; n) f and only f ; satsfy one of the condtons ) or ) of Proposton (c) Suppose ; D(m; n); P < P Then < are adjacent n D(m; n) f and only f + = and k = k for k Proof (a) Because of (), adjacency n D and n D(m; n) are equvalent for ; D(m; n) (b) By (a) and () < are adjacent n D(m; n) f and only f < are adjacent n D Hence (b) follows from Proposton P P (c) Snce <, < are adjacent f and only f condton () of Proposton holds wth ; nstead of ; Dualzng ths condton and applyng () we obtan (c) Orbt Closure ecessary Condtons n the General Case In ths secton we consder the acton () of G = GL p (C ) GL q (C ) on the set C (pq) of p q pencls s + Kronecker's Theorem () provdes a complete set of nvarants for ths acton column ndces ( ), row ndces ( ), and nvarant factors D j =D j+ The followng theorem gves necessary condtons, n terms of these nvarants, for a pencl s + to le n the closure of O(s + ) In Secton we show that these condtons are sucent n the case of pencls wth no row ndces We conjecture that these condtons are sucent n the general case For ease of exposton, we denote the nvarants of the pencl s + by ( ) = ( ) c, ( ) = ( ) r, D j and the nvarants of the pencl s + by ( ) = ( ) c, ( ) = ( ) r, D j ote that = for > q and = for > p Theorem If s + s n the closure of O(s + ), then ) P k? P k ) P k? P k ) D j dvdes D j k(c? c) for all k = ; ; q k(r? r) for all k = ; ; p for all j = h; ; n (h = n +? rank (s + )) To prove ths theorem we consder closed G-nvarant subsets C C (pq) dened n terms of the nvarants sif s + C, t follows that s + C for all s + O(s + ) Relatonshps between the nvarants of s + and those of s + then follow We begn by denng addtonal lsts of pencl nvarants ( k ) and ( T k ) whch may be computed from the column ndces (respectvely row ndces), see [] Let c be the number of column ndces and r the number of row ndces of the pencl s + 8

9 Denton Gven a pencl s +, the assocated rank nvarants ( k ) and ( T k ) are dened by k = k (s + ) = ranka k ; T k = T k (s + ) = rank Ak ~ ; k where A k = C (k+)p(k+)q ; Ak ~ = Lemma The rank nvarants k and T k and row ndces of s + by k = T k = k+ X k+ X C (k+)p(k+)q of a p q pencl s + are related to the column + (k + )(q? c) ; k () + (k + )(p? r) ; k () where ( ) and ( ) are the dual dagrams to the column and row ndces respectvely Proof Let X(s) (C [s]) q be gven by X(s) = x +sx + +s k x k ; x C q Then (s+)x(s) = f and only f x k x k? A k X(s) = x = It follows that (k + )q? k s the maxmal number of lnearly ndependent solutons of (s + )X(s) = whch have degree less than or equal to k Snce a lnearly ndependent set of solutons of degree gves rse to a lnearly ndependent set of solutons of degree + (multply by s), we have (k + )q? k = k+ X cardfj j < g = k+ X (c? ) = (k + )c? The equaton () follows An analogous argument establshes equaton () X k+ Lemma Let C k;j = fs + k (s + ) jg and let Ck;j T = fs + T k (s + ) jg The sets C k;j and Ck;j T are Zarsk closed G-nvarant subsets of C (pq) Proof We saw n Lemma that k and T k can be calculated from the G-nvarants ( ) and ( ), hence they are G-nvarants It follows that the sets C k;j and Ck;j T are G-nvarant for each par (k; j) For any matrx A, rank A j f and only f all hh mnors of A are zero for all h > j Each hh mnor of A k (s +) s determned by a choce of h rows and h columns I = (r ; ; r h ; c ; ; c h ) Let m I k (s + ) be the h h mnor of A k(s + ) determned by the choce I Let C I k = fs + m I k(s + ) = g 9

10 Then we have \ C k;j = Ck I where the ntersecton s taken over all I = (r ; ; r h ; c ; ; c h ) such that j < h n Because m I k (s + ) s a polynomal n the entres of the pencl s +, the set CI k s a Zarsk closed subset of C (pq) Hence C k;j s closed A smlar argument establshes that Ck;j T s closed Lemma Let f be a non-zero homogeneous polynomal of degree d n C [s; t] m d, the set S m f s Zarsk-closed n C m+ = f(a ; ; a m ) C m+ f dvdes mx = a s t m? g Then, for any Proof The set S m f s a lnear subspace of C m+ Lemma For each non-zero homogeneous polynomal f C [s; t] and each j, j n, the set C f;j = fs + f dvdes D j (s + )g s a Zarsk closed subset of C (pq) Proof Each j j mnor of s + t s gven by a choce of j rows and j columns I = (r ; ; r j ; c ; ; c j ) For each such choce I, let Dj I (s + ) be the j j mnor of s + t determned by I and dene I C (pq)! C j+ s +! coe D I j (s + ) The set Cf;j I = fs + f dvdes DI j (s + )g s the nverse mage of Sj f under the morphsm I, where S j f s the set dened n Lemma It follows from Lemma that the set CI f;j s Zarsk closed We have \ C f;j = Cf;j I where the ntersecton runs over all choces I of j rows and j columns Hence the set C f;j s Zarsk closed Proof of Theorem From Lemma t follows that C k;j s a closed G-nvarant subset of C (pq) Therefore, f s + C k;j, then s + C k;j for all s + n the orbt closure O(s + ) So f s + O(s + ), then k (s +) k (s +) for each k, k = ; ; n Smlarly, T k (s + ) T k (s + ) Hence the nequaltes ) and ) follow from Lemma Usng Lemma, a smlar argument establshes that D j (s + ) dvdes D j (s + ) for all j = h; ; n Orbt Closure Sucent Condtons for Pencls wth no Row Indces Throughout ths secton we consder only n (n + m) pencls wth no row ndces (e pencls whch are equvalent to an nput pencl) In Secton we observed that such a pencl has exactly m column ndces () and has rank n () Restrctng our attenton to these pencls, the statement of Theorem s consderably smpled; condton ) vanshes and condton ) becomes the partal orderng () on dagrams In ths secton we show that the necessary condtons of Theorem are n fact sucent for pencls wth no row ndces

11 Theorem Let s + and s + be n (n + m) pencls wth no row ndces Then s + s n the closure of O(s + ) f and only f the followng condtons hold ) ( ) ( ) ) D j (s + ) dvdes D j (s + ) for all j = ; ; n Let s + and s + be n(n+m) pencls wth no row ndces We wrte s + s + f the nvarants of these pencls satsfy condtons ) and ) of Theorem and s + < s + f s + s + and O(s +) = O(s +) We say that s + < s + are adjacent f there s no pencl s ~ + ~ wth no row ndces such that s + < s ~ + ~ < s + If s + < s +, then there s a nte chan of n (n + m) pencls wth no row ndces s + = s + < s + < < s q + q = s + such that s + s adjacent to s By transtvty of orbt closure, t suces to show that s + O(s + ) for adjacent pencls s + < s + Hence our next step s to descrbe adjacent pencls In the sequel, we use the followng constructons Gven an r r regular pencl R n canoncal form and an rreducble polynomal f, f = t or f = s? t, let R f be the (r + ) (r + ) pencl obtaned from R as follows If f dvdes D (R) (e f = e for some ), replace J(e ; r ; ) by J(e ; r ; + ) If f s not a factor of D (R), let R f = R J(f; ) Then t follows from () that D j (R f ) = ( fd (R) j = D j (R) j r + () where, by denton, D r+ (R) = For each rreducble factor e of D (R) let h be the ndex whch satses r ;j = r ; for j h and r ;h + < r ;h and let R e be the (r? ) (r? ) pencl obtaned from R by replacng J(e ; r ;h ) wth J(e ; r ;h? ) From () we have D j (R e ) = ( Dj (R)=e j h D j (R) h + j ; j r? (8) Proposton Let s + and s + be n(n+m) pencls wth no row ndces Suppose these pencls are n canoncal form and have regular parts R and R respectvely If s + < s + are adjacent then one of the followng three condtons holds ) ( ) = ( ) and R < R ) ( ) < ( ) are adjacent n D and R = R ) P n < P n, ( ) < ( ) are adjacent, and R = R f for some rreducble f Frst we consder the case P n = P n If ( ) = ( ), then clearly R < R and we Proof have condton ) Suppose ( ) < ( ) If R = R, then we have R < R and hence s + = C ( ) R < C ( ) R < C ( ) R = s + whch contradcts the assumpton that s + and s + are adjacent Thus R = R If ( ) and ( ) are not adjacent, then, by Corollary, there s a dagram (~ ) n D(m; n) such that ( ) < (~ ) < ( ) Then P n ~ = P n and s + = C ( ) R < C (~ ) R < C ( ) R = s +

12 contradctng the assumpton that s + and s + are adjacent Therefore ( ) and ( ) are adjacent and condton ) holds P P m m Fnally, we consder the case < Then r = deg D > deg D = r If ( ) and ( ) are not adjacent, then by Proposton we have ( ) ( ) < (~ ) ( ) where mx mx = and (~ ) = ( ; ; k ; ) wth k = and k+ = and ether ( ) = ( ) or (~ ) = ( ) From () we see that ( ) and (~ ) are n D(m; n) Let e be an rreducble factor of D =D and let s ~ + ~ = C(~ ) R e By (8) and () we have ~D j = ( Dj =e j h D j h + j ; j r? (9) and ~ Dj = for j r, hence ~ Dj dvdes D j for j = ; ; n Because D j dvdes D j for all j and r < r we have by (), (9) and () nx `=j r ` nx `=j r `; j = ; ; n and D j = for j r () oreover, snce e dvdes D =D we have ( nx `= r `) + nx `= r ` () ow assume P ǹ =k r ` = P ǹ=k r ` for some k h Choosng j = k + n () we conclude that r r k r k = = r P P ǹ and hence r = ` ǹ= r ` whch contradcts () Therefore nx nx ( `=j r `) + `=j r ` ; j h so that D j dvdes ~ Dj = D j =e for j h It then follows from (9) and () that D j dvdes ~D j for j = ; ; n We have s + = C ( ) R C ( ) R < C (~ ) R e C ( ) R = s + where C ( ) R = C ( ) R or C (~ ) R e = C ( ) R Ths contradcts the assumpton that s + and s + are adjacent Therefore ( ) and ( ) are adjacent and deg D = + deg D Let f = D =D and let s ~ + ~ = C( ) R f It follows from () that D j dvdes Dj ~ for all j and ~D j dvdes D j for all j Therefore s + = C ( ) R C ( ) R f < C ( ) R = s + Because s + and s + are adjacent, we conclude that R = R f Proposton reduces the proof of Theorem to three cases, the rst of whch was dealt wth n [] The other two cases wll be addressed n the followng lemmata In the followng we use the notaton ntroduced n Secton ; n partcular, f C s any matrx wth ` rows then C C denotes the pencl [`; C] where ` s the zero vector n C `

13 Lemma C? C j+ s n the closure of O(C C j ) for j Proof Transform C by movng column to the end and row to the bottom; then C C j s equvalent to C? s If =, add column to column to obtan the pencl C j+ C j () ultply the rst column by and take the lmt as! to establsh the concluson If > we proceed as follows Add column to column + to obtan C? s s C j = C? s C j+ Subtractng row + from row and addng column to column + results n C? s () () Repeat the last two steps; e gven C? s C j+ C j+ (s n the (k; + k + ) place) () then subtract row + k from row k and add column k + to column + k + The result of ths process s that s moves to the (k + ; + k + ) place, k? Applyng ths process repeatedly, we eventually have s n the (? ; ) place (because j there s an (? ; ) place) Then one more applcaton of the process results n C C j C? C j+ ()

14 In the followng, let I k denote the k k dentty matrx Then we have lm! " # I? I j+ C? = C j+ " # C? C j+ Therefore C? C j+ s n the closure of O(C C j ) for j I?? I j+ I m () Lemma C? J(f; r) s n the closure of O(C J(f; r? )) for ; r, f = s? t or f = t Proof We begn wth the case f = s? t If =, add column of C J(f; r? ) to column (n case r = omt ths step), then multply column by?, and then add column to column to obtan s J(s? t; r) ultplyng the rst column by and takng the lmt as!, we see that C J(f; r) O(C J(f; r? )) If >, we proceed as follows Transform C as n the rst step () of Lemma and add column to column + to obtan C? s s J(s? t; r? ) (8)

15 Add tmes row to row and subtract tmes column from column + C? +? s + s? J(s? t; r) J(s? t; r? ) (9) where s the (? ) matrx wth (; ) entry equal to and other entres ow apply the followng algorthm repeatedly for k? = C? +? s + We are gven a pencl n whch all rows of the upper rght (? ) r block are zero except for the kth row Subtract approprate multples of the last r rows from row k so that only constant terms reman n the last r entres of row k All other rows of ths block wll reman zero and the upper left (? ) block may acqure constant entres n ts rst column Then subtract approprate multples of column k + from the last r columns so that row k of the upper rght (? ) r block s zero After? applcatons of the algorthm, we obtan C? + J(s? t; r) () where s a constant (? ) matrx whose last? columns are zero ultply the matrx () on the left by I? I r and on the rght by I?? I r I m and take the lmt as! ; we obtan (C? + ) J(s? t; r) Because C? + has rank?, t has no row ndces and thus one column ndex It s easy to see that D (C? + ) = ; hence the sngle column ndex s? Therefore C? + s equvalent to C? Ths establshes that C? J(f; r) s n the closure of O(C J(f; r? )) ext we consder the case f = t If r =, smply multply the last row of C by and the last column by? and take the lmt as! to establsh that C? J(t; ) O(C ) If r >, we proceed as follows Frst assume that r > Addng column? k of C J(t; r? ) to column + k +, k?, we obtan C s s J(t; r? ) () or we obtan () wthout the C? part f = Subtractng row + k from row? k, k?, results n C s J(t; r? ) = C? s J(t; r) ()

16 (wth s n the (; ) place) If r, apply analogous transformatons to obtan C J(t; r? ) = C s s J(t; r? ) C s J(t; r? ) () whch s the same as the pencl () ultplyng the pencl () on the left by? I? I r and on the rght by I I r? and takng the lmt as!, we establsh that C? J(t; r) s n the closure of O(C J(t; r? )) Proof of Theorem The fact that condtons ) and ) are necessary follows from Theorem As noted n the remarks followng Theorem, we need only show that s + s n the closure of O(s + ) for adjacent pencls s + < s + Wthout loss of generalty, we may assume that s + and s + are n the canoncal form (9) By Proposton, adjacent pencls fall nto three cases In case ) the sngular part of s + s the same as the sngular part of s + Orbt closure for regular pencls was establshed n [, Theorem ] and so the concluson follows In case ) of Proposton the column ndces ( ) > ( ) satsfy ether condton ) or condton ) of Proposton, see Corollary (b); n ether case, there are natural numbers k < l such that k = k +, l = l?, and =, = k; l Snce s + and s + have the same blocks except for the kth and lth blocks, t suces to show that C l? C k + s n the closure of O(C l C k ); ths was establshed n Lemma Case ) of Proposton corresponds to statement (c) of Corollary In ths case =? and =, Hence s + and s + have the same blocks except for the rst block n the sngular part and the rst \f" block n the regular part Thus t suces to show that C? J(f; r) s n the closure of O(C J(f; r? )) for r, whch s what we establshed n Lemma Acknowledgment The authors would lke to thank H Kraft and E Oeljeklaus for ther advce concernng some questons n the proof of Theorem References [] P Brunovsky A classcaton of lnear controllable systems Kybernetka, {8, 9 [] C DeConcn, D Esenbud, and Proces Young dagrams and determnantal varetes Invent ath, 9{, 98 [] ER Gantmacher The Theory of atrces, volume and Chelsea, ew York, 99 [] D Hnrchsen and J O'Halloran A complete characterzaton of orbt closures of controllable sngular sy stems under restrcted system equvalence SIA Journal on Control and Optmzaton, 8{, 99 [] JE Humphreys Lnear Algebrac Groups Sprnger-Verlag, Berln-Hedelberg-ew York, 9 [] H Kraft and C Redtmann Geometry of representatons of quvers In P Webb, edtor, Representatons of Algebras, Proc of the Durham Symposum 98, London ath Soc Lecture otes, pages 9{ Cambrdge Unv Press, 98

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