2.3 Nilpotent endomorphisms

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1 s a block dagonal matrx, wth A Mat dm U (C) In fact, we can assume that B = B 1 B k, wth B an ordered bass of U, and that A = [f U ] B, where f U : U U s the restrcton of f to U Nlpotent endomorphsms ([1], p ) In ths secton we wll consder those lnear endomorphsms f End C (V ) whose only egenvalue s 0 Ths necessarly mples that χ f (λ) = λ n We wll see that for such endomorphsms there s a (ordered) bass B of V such that [f ] B s nearly dagonal Defnton 231 An endomorphsm f End C (V ) s called nlpotent f there exsts r N such that f r = 0 EndC (V ), so that f r (v) = 0 V, for every v V A matrx A Mat n (C) s called nlpotent f the endomorphsm T A End C (C n ) s nlpotent Lemma 232 Let f End C (V ) be a nlpotent endomorphsm Then, the only egenvalue of f s λ = 0 so that χ f (λ) = λ dm V Proof: Suppose that v V s an egenvector of f wth assocated egenvalue λ Therefore, we have v 0 and f (v) = λv Suppose that f r = 0 Then, 0 = f r (v) = f f (v) = f f (λv) = λ r v Thus, as v 0 we must have λ r = 0 (Proposton 125) mplyng that λ = 0 For a nlpotent endomorphsm f (resp matrx A Mat n (C)) we defne the exponent of f (resp of A), denoted η(f ) (resp η(a)), to be the smallest r N such that f r = 0 (resp A r = 0) Therefore, f η(f ) = r then there exsts v V such that f r 1 (v) 0 V For v V we defne the heght of v (wth respect to f ), denoted ht(v), to be the smallest nteger m such that f m (v) = 0 V, whle f m 1 (v) 0 V Hence, for every v V we have ht(v) η(f ) Defne H k = {v V ht(v) k}, the set of vectors that have heght no greater than k; ths s a subspace of V 41 Let f End C (V ) be a nlpotent endomorphsm Then, we obvously have H η(f ) = V, H 0 = {0 V } and a sequence of subspaces Let us denote so that we have {0 V } = H 0 H 1 H η(f ) 1 H η(f ) = V dm H = m, 0 = m 0 m 1 m η(f ) 1 m η(f ) = dm V We are gong to construct a bass of V : for ease of notaton we let η(f ) = k Assume that k 1, so that f s not the zero endomorphsm of V 1 Let G k be a complementary subspace of H k 1 so that H k = H k 1 G k, and let (z 1,, z p1 ) be an ordered bass of G k Then, snce z j H k \H k 1 we have that f k 1 (z j ) 0 V, for each j 40 Ths s a well-defned functon snce U s f -nvarant 41 Exercse: show ths 62

2 2 Consder the vectors f (z 1 ), f (z 2 ),, f (z p1 ) We have, for each j, f k 1 (f (z j )) = f k (z j ) = 0 V, snce z j H k, so that f (z j ) H k 1, for each j In addton, we can t have f (z j ) H k 2, else mplyng that z j H k 1 0 V = f k 2 (f (z j )) = f k 1 (z j ), Moreover, the set S 1 = {f (z 1 ), f (z 2 ),, f (z p1 )} H k 1 \ H k 2 s lnearly ndependent: ndeed, suppose that there s a lnear relaton c 1 f (z 1 ) + + c p1 f (z p1 ) = 0 V wth c 1,, c p1 C Then, snce f s a lnear morphsm we obtan so that c 1 z c p1 z p1 H 1 H k 1 f (c 1 z c p1 z p1 ) = 0 V, Hence, we have c 1 z c p1 z p1 H k 1 G k = {0 V }, so that c 1 z c p1 z p1 = 0 V Hence, because {z 1,, z p1 } s lnearly ndependent we must have c 1 = = c p1 = 0 C Thus, S 1 s lnearly ndependent 3 span C S 1 H k 2 = {0 V }: otherwse, we could fnd a lnear combnaton wth some c 0 Then, we would have c 1 f (z 1 ) + + c p1 f (z p1 ) H k 2, 0 V = f k 2 (c 1 f (z 1 ) + + c p1 f (z p1 )) = f k 1 (c 1 z c p1 z p1 ), so that c 1 z c p1 z p1 H k 1 G k = {0 V } whch gves all c j = 0, by lnear ndependence of the z j s But ths contradcts that some c s nonzero so that our ntal assumpton that span C S 1 H k 2 {0 V } s false Hence, we have span C S 1 + H k 2 = span C S 1 H k 2 H k 1 In partcular, we see that m k m k 1 m k 1 m k 2 4 Let G k 1 be a complementary subspace of H k 2 span C S 1 n H k 1, so that H k 1 = H k 2 span C S 1 G k 1, and let (z p1+1,, z p2 ) be an ordered bass of G k 1 5 Consder the subset S 2 = {f 2 (z 1 ),, f 2 (z p1 ), f (z p1+1),, f (z p2 )} Then, as n 2, 3, 4 above we have that S 2 H k 2 \ H k 3, S 2 s lnearly ndependent and span C S 2 H k 3 = {0 V } Therefore, we have so that m k 1 m k 2 m k 2 m k 3 span C S 2 + H k 3 = span C S 2 H k 3 H k 2, 63

3 6 Let G k 2 be a complementary subspace of span C S 2 H k 3 n H k 2, so that and (z p2+1,, z p3 ) be an ordered bass of G k 2 H k 2 = H k 3 span C S 2 G k 2, 7 Consder the subset S 3 = {f 3 (z 1 ),, f 3 (z p1 ), f 2 (z p1+1),, f 2 (z p2 ), f (z p2+1),, f (z p3 )} Agan, t can be shown that S 3 H k 3 \ H k 4, S 3 s lnearly ndependent and span C S 3 H k 4 = {0 V } We obtan m k 2 m k 3 m k 3 m k 4 8 Proceed n ths fashon to obtan a bass of V We denote the vectors we have obtaned n a table (231) z 1, z p1, f (z 1 ), f (z p1 ), z p1+1, z p2, f k 1 (z 1 ), f k 1 (z p1 ), f k 2 (z p1+1), f k 2 (z p2 ), z pk 1 +1, z pk, where the vectors n the th row have heght k + 1, so that vectors n the last row have heght 1 Also, note that each column determnes an f -nvarant subspace of V, namely the span of the vectors n the column Lemma 233 Let W denote the span of the th column of vectors n the table above Set p 0 = 1 Then, dm W = k j, f p j + 1 p j+1 Proof: Suppose that p j + 1 p j+1 Then, we have Suppose that there exsts a lnear relaton W = span C {z, f (z ),, f k j 1 (z )} c 0 z + c 1 f (z ) + + c k j 1 f k j 1 (z ) = 0 V Then, applyng f k j 1 to both sdes of ths equaton gves c 0 f k j 1 (z ) + c 1 f k j (z ) + + c k j 1 f 2k 2j 2 (z ) = 0 V Now, as z has heght k j (ths follows because the vector at the top of the th column s n the (k j) th row, therefore as heght (k j)) the prevous equaton gves c 0 f k j 1 (z ) + 0 V V = 0 V, so that c 0 = 0, snce f k j 1 (z ) 0 V Thus, we are left wth a lnear relaton c 1 f (z ) + + c k j 1 f k j 1 (z ) = 0 V, and applyng f j k 2 to ths equaton wll gve c 1 = 0, snce f (z ) has heght k j 1 Proceedng n ths manner we fnd that c 0 = c 1 = c j k 1 = 0 and the result follows Thus, the nformaton recorded n (231) and Lemma 233 proves the followng Theorem 234 Let f End C (V ) be a nlpotent endomorphsm wth exponent η(f ) = k Then, there exsts ntegers d 1,, d k Z 0 so that kd 1 + (k 1)d d k 1 + 1d k = dm V, 64

4 and f -nvarant subspaces wth dm C W (j) W (k) 1,, W (k), W (k 1) 1,, W (k 1) d 2,, W (1) 1,, W (1) d k V, = j, such that d 1 V = W (k) 1 W (k) d 1 W (k 1) 1 W (k 1) d 2 W (1) 1 W (1) d k Moreover, there s an ordered bass B (j) [f (j) W ] (j) B = of W (j) such that We call such matrces 0-Jordan blocks Hence, we can wrte the matrx of f relatve to B =,j B(j) as a block dagonal matrx for whch all of the blocks are 0-Jordan blocks and are of nonncreasng sze as we move from left to rght Moreover, the geometrc multplcty of 0 as an egenvalue of f s equal to the number of blocks of the matrx [f ] B and ths number equals the sum d 1 + d d k = dm E 0 Proof: Everythng except for the fnal statement follows from the constructon of the bass B made pror to the Theorem The last statement s shown as follows: we have that E 0 = H 1, so that the 0-egenspace of f conssts of the set of all heght 1 vectors n V 42 Moreover, the constructon of the bass B shows that a bass of H 1 s gven by the bottom row of the table (231) and that ths bass has the sze specfed Corollary 235 Let A Mat n (C) be a nlpotent matrx Then, A s smlar to a block dagonal matrx for whch all of the blocks are 0-Jordan blocks Proof: Consder the endomorphsm T A End C (C n ) and apply Theorem 234 Then, we have a bass B such that [T A ] B takes the desred form Now, use Corollary 177 and [T A ] S (n) = A to deduce the result Defnton 236 Let n N A partton of n s a decomposton of n nto a sum of postve ntegers If we have a partton of n then we denote ths partton n = n n l, wth n 1,, n l N, n 1 n 2 n l, 1 r1 2 r2 n rn l l, where we are assumng that 1 appears r 1 tmes n the partton of n, 2 appears r 2 tmes etc For example, consder the partton of = , then we denote ths partton 42 Check ths

5 For a nlpotent endomorphsm f End C (V ) we defne ts nlpotent class to be the set of all nlpotent endomorphsms g of V for whch there s some ordered bass C V wth where B s the bass descrbed n Theorem 234 [f ] B = [g] C, We defne the partton assocated to the nlpotent class of f, denoted π(a), to be the partton 1 d k 2 d k 1 k d1 obtaned n Theorem 234 We wll also call ths partton the partton assocated to f For a matrx A Mat n (C) we defne ts nlpotent class (or smlarty class) to be the nlpotent class of the endomorphsm T A We defne the partton assocated to A to be the partton assocated to T A Theorem 237 (Classfcaton of nlpotent endomorphsms) Let f, g End C (V ) be nlpotent endomorphsms of V Then, f and g le n the same nlpotent class f and only f the parttons assocated to f and g concde Corollary 238 Let A, B Mat n (C) be nlpotent matrces Then, f and g are smlar f and only f the parttons assocated to A and B concde Proof: We smply note that f T A and T B are n the same nlpotent class then there are bases B, C C n such that [T A ] B = [T B ] C Hence, f P 1 = P S (n) B, P 2 = P S (n) C then we must have so that P 1 1 AP 1 = P 1 2 BP 2, P 2 P 1 1 AP 1 P 1 2 = B Now, snce P 2 P 1 1 = (P 1 P 1 2 ) 1 we have that A and B are smlar precsely when T A and T B are n the same nlpotent class The result follows 231 Determnng parttons assocated to nlpotent endomorphsms Gven a nlpotent endomorphsm f End C (V ) (or nlpotent matrx A Mat n (C)) how can we determne the partton assocated to f (resp A)? Once we have chosen an ordered bass B of V we can consder the nlpotent matrx [f ] B Then, the problem of determnng the partton assocated to f reduces to determnng the partton assocated to [f ] B As such, we need only determne the partton assocated to a nlpotent matrx A Mat n (C) 1 Determne the exponent of A, η(a), by consderng the products A 2, A 3, etc The frst r such that A r = 0 s the exponent of A 2 We can determne the subspaces H snce H = {x C n ht(x) } = ker T A In partcular, we have that dm H s the number of non-pvot columns of A 3 d 1 = dm H η(a) dm H η(a) 1 4 d 2 = dm H η(a) 1 dm H η(a) 2 d 1 5 d 3 = dm H η(a) 2 dm H η(a) 3 d 2 6 Thus, we can see that d = dm H η(a) ( 1) dm H η(a) d 1, for 1 η(a) Hence, the partton assocated to A s π(a) : 1 d η(a) 2 d η(a) 1 η(a) d1 66

6 Example 239 Consder the endomorphsm f : C 5 C 5 ; x 1 x 2 x 2 x 3 x 4 0 x 4 0 x 5 0 Then, wth respect to the standard bass S (5) we have that A def = [f ] S (5) = You can check that A 2 = 0 so that η(a) = 2 Then, - d 1 = dm H 2 dm H 1 = 5 3 = 2, snce H 1 = ker T A has dmenson 3 (there are 3 non-pvot columns of A) - d 2 = dm H 1 dm H 0 d 1 = = 1, snce H 0 = {0} Hence, the partton assocated to A s π(a) : = 5; there are three 0-Jordan blocks - two of sze 2 and one of sze 1 You can check that the followng matrx B s nlpotent B = and that the partton assocated to B s - We have B 2 = 0 so that η(b) = 2 π(b) : = 5 - d 1 = dm H 2 dm H 1 = 5 4 = 1, snce H 1 = ker T B has dmenson 4 (there are 4 non-pvot columns of B) - d 2 = dm H 1 dm H 0 d 1 = = 3, snce H 0 = {0} Thus, A and B are not smlar, by Corollary 238 However, snce the matrx C =, has assocated partton π(c) : 1 3 2, then we see that B s smlar to C, by Corollary 238 Moreover, there are four 0-Jordan blocks of B (and C) - one of sze 2 and three of sze 1 67

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