On complementary matrix algebras - Springer Link

Integral Equations and Operator Theory Vol. 13 (1990)

0378-620X/90/020165-I051.50+0.20/0 (c) 1990 Birkh~user Verlag, Basel

ON COMPLEMENTARY MATRIX ALGEBRAS Man-Duen Choi, Heydar Radjavi, and Peter Rosenthal*) It is proved that complementary algebras of linear operators on 12n do not necessarily have nontrivial complementary invariant subspaces. This settles a conjecture of Gohberg, Lancaster, and Rodman in the negative. A positive result is also proved under certain additional hypotheses. INTRODUCTION Gohberg, Lancaster, and Rodman [1, p. 342] presented the following conjecture. If (21 and (22 are algebras of linear operators on En that are complementary (i.e., (21 f'l(22 = {0} and (21+(2 2 is the algebra ~(~;n) of all operators on [;n)

then there exist nontrivial

complementary subspaces 711.1 and ~Jll.2 of 12n with ~Jll,i invariantunder (2i" We show this to be false in general. A positive result holds if the algebras are assumed to be orthogonally complementary. THE COUNTEREXAMPLES The following result shows the existence of a counter example for almost every n. The reader interested only in the simplest example (for n = 4) will find it is paragraph (i) of the proof. THEOREM 1. For each integer n >_4 (except possibly 7 and 11) there exist complementary algebras of operators on f n with no nontrivial complementary invariant subspaces.

PROOF. The result will be established in several steps. (i) We start with the case n = 4. Fix a nonzero vector e in 124 and let (21 be the algebra of all operators A with Ae = 0. Pick a basis {ei} for 1124 whose matrices relative to {ei } have the form

*) This research was supported by the Natural Sciences and Engineering Research Council of Canada.

166

Choi, Radjavi and Rosenthal

0 i]

0 0 Observe that

a

, a,l~,7, S e C .

"/

121 has only one nontrivial invariant subspace

~ e , and that 122 has only

2-dimensional ones. Thus the two algebras cannot have complementary invariant subspaces. Since Q1 and (22 have dimensions 12 and 4 , the proof will be completed if we show that (21+(22 = J~(C 4) . This follows from the fact that e is a cyclic vector for (22 ' i.e., (22 e = ~4. Given B in j ~ ( ~ 4 ) , let A 2 e (22 be such that A2e = Be. Then letting A 1 = B - A 2 , we get Ale=0.

Thus A l e

121'B=A l+A2,and

B e (21 +(22 for all B in j : ( E 4 ) .

r 2 (ii) Assume n = ~ m i with integers m i >__2. Fix a nonzero vector e and let 121 be i=l the algebra of all operators on s

with Ae = 0 . Then (21 is of dimension n 2 - n and has only

one nontrivial invariant subspace q:e of dimension 1. We proceed to construct a complementary algebra (22 of dimension n such that each proper invariant subspace of 122 is of codimension > 2 . Let ]3m=(B~

...~B:Be

2 J:(~:m)) C j : ( ~ m ) .

Then ]3 m is an algebra of dimension m 2 . Observe that it has many invariant subspaces, but the dimension of each invariant subspace must be one of the numbers 0, m, 2m ..... (m-1)m, m 2 . Now if

{ei) n n r i-1 is any given basis for ~: with n = ]~ m? and m. -> 2 we let -

j=l

23 Then dim ~ = ~ dim

m. J

=

J

J

23ml ~ ... ~ ~3mr

= ~ 2 = n. Every invariant subspace of ~B is of the form

~]]~ = ~1.1 9 ... 9 ~r[~r ,

'

Choi, Radjavi and Rosenthal

where ~ j

167

is an invariant subspace of ~ m j 9 Thus if ~

is proper, then at least one ~l].j has

codimension 2 mj _> 2. It follows that ~1. has codimension > 2. Since the dimensions of (21 and ~

add up to n 2 , it suffices, as in paragraph (i), to find an algebra (22 similar to ~

for

which e is a cyclic vector. Equivalently, we choose the basis {ei} such that e = 5". cie i with the sequence {ci} given below and then let 122 = ~ . For an integer m form the sequence S(m) = {dj } of m 2 terms as follows: 10 if j=l, m+2, 2m+3 . . . . . (m-1)m+m ; dJ=

otherwise.

Let {cj} be the sequence of n terms formed by first taking S ( m l ) , then S(m2) ..... and finally S(m r) . It is then easily verified that e = Y~cie i is cyclic for 122 . (iii) Next we treat the case n = 6 . In this case the algebra (21 is formed as in the above example: it is the algebra of all operators A with Ae = 0 , where e is a nonzero vector. Fora basis {ei} of ~;6 define (22 by the matrices 0

~1

~2

0

0

0

0

~3

%

0

0

0

0

~5

~6

0

0

0

0

0

0

0

o~I

c~2

0

0

0

0

~

~4

0

0

0

0

o~5

C~61

, o~ie

~

.

We verify that (22 has no invariant subspace of dimension 5, and thus any proper invariant subspace of (22 has codimension at least 2, and cannot complement C e , the only nontrivial invariant subspace of (21 " Since [21 and (22 have dimensions adding up to 36, we must only choose the basis {0,1,0,0,0,1

{ei} so that e = ~ cie i is cyclic for (22 " A suitable sequence

{ci} is

} .

(iv) Now assume n is of the form 6 + ]~ mi2 with m i > 2 . The construction is

168

Choi, Radjavi and Rosenthal

similar to the examples given above: just take 121 as in (ii) and let 122 be the direct sum of the corresponding algebras given in (ii) and (iii). (The vector e should be the direct sum of the corresponding vectors in those examples.) (v) The case n = 5 has to be treated separately: let

~2 ~3 ~4

ro

o

o

o

~1

0

0

0

1

0

0

1

0

0

0

0

c~1

~2

~5

~6

(214 0

a3

~4

~7

~8

0

0

0

a1

a4

1

0

0

0

0

0

0

0

53

a4

0

0

1

0

0

: 0~i,~je (~}, U =

and (22 = U*121U + IEI. It is easy to verify that each proper invariant subspace of 121 (and thus each proper invariant subspace of (22) is of dimension 1 or 3. Therefore 121 and 122 do not have complementary invariant subspaces. Since dim 121 = 12 and dim 122 = 13, the equation 121 + 122 = ~ ( C 5 )

is equivalent to 121 fq 122 = {0} which is valid by straight-forward

entry-wise verification. (vi) We can now complete the proof. Observe that once examples are found for four consecutive integers starting with n O , then there is an example for each n _> n O . For in the formulas given in (ii) and (iv) we can add another term m 2 with m = 2 . Let n O = 12 and observe that 1 2 = 2 2 + 2 2 + 2 2 , 1 3 = 3 2 + 2 2 , 1 4 = 6 + 2 2 + 2 2

and 1 5 = 6 + 3 2

. Suitable

decompositions for 8, 9 and 10 are also found easily. Thus, in view of (iii) and (vi) we have established counter examples for every n _>4 except n = 7 and n = 11. REMARK.

[]

Professor Gohberg has asked the following question an affirmative

answer to which would establish a weaker form of the conjecture. If 12 is an algebra of matrices which has a complementary algebra, must there exist

some

complementary algebra ~ such that

(2 and ~ have nontrivial complementary invariant subspaces? However, even this question has a negative answer. To see this consider the algebra 122 in paragraph (i) of the proof of Theorem 1, i.e., the algebra on ~4 defined by {A~A:Ae

~(IE2)}.

Choi, Radjavi and Rosenthal

169

We know that (22 has a complementary algebra, namely (21 in (i). We also know that every nontrivial invariant subspace of (22 is 2-dimensional. Assume, if possible, that (22 had a complementary algebra ~ with a nontrivial invariant subspace ~ Then ~

complementing one of (22"

has dimension 2. Since ~ is 12-dimensional, it necessarily contains all operators on

1114 leaving ~

invariant. Thus ~ contains the identity, and since so does (22 ' we arrive at a

contradiction. SOME AFFIRMATIVE RESULTS. We start with a lemma from [1]. LEMMA 1. If (21 and (22 are complementary linear spaces of operators on C n with ~ i aninvariantsubspaceof (2i,theneither

~1.1 A ~I112= {0} or ~rl.1 + ~ . 2 = ~ n .

The following result is an easy consequence of this lemma. THEOREM 2. Under the additional hypothesis that one of the complementary algebras (21 and (22 is triangularizable, the conjecture is true. PROOF. Let (21 be the triangularizable algebra. Since (22 is a proper subalgebra of ~ ( E n ) , Burnside's Theorem (see [1] or [2]) implies that (22 has a nontrivial invariant subspace 911.2 . Since (21 has an invariant subspace of every dimension between 0 and n , we can choose one, say ~Yn.1 , of dimension n - d i m ~ 2 9 Then ~q'l and ~/t.2 are complementary by Lemma 1.

[]

THEOREM 3. For n = 2 and n = 3 the conjecture is true. PROOF. The case n = 2 follows from Theorem 1 since every proper subalgebra is triangularizable. Now let (21 and (22 be complementary proper algebras on ~ 3 . We claim that ~i has a nonzero invariant subspace ~]'i with dim ~1.1 + dim ~'n.2 = 3 , so that Lemma 1 completes the proof. Otherwise, all the nontrivial invariant subspaces of (21 have the same constant dsimension as those of (22 . By considering (21" and (22* if necessary, we can assume that every nontrivial invariant subspace of either algebra is 2-dimensional. Since not both (21 and (22 can have dimension greater than 4 , we let dim (21 < 4 . Since (21 has a

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Choi, Radjavi and Rosenthal

2-dimensional minimal invariant subspace its dimension is exactly 4 by Burnside's theorem. We leave it to the reader to verify that every 4-dimensional algebra on ~3 with a minimal invariant subspace of dimension 2 is similar to ~ ( C 2) (9 0 . Thus {21 must have a one-dimensional invariant subspace, which is a contradiction.

[]

Our next theorem is about orthogonally complementary algebras. The inner product used in ~ ( C n) is the natural one: for operators A and B let (A,B) = tr(AB*). For a subset ,8 of ,~(IE n) , the set {A e ~ ( ~ n ) : (A,S) = 0 'V' S e ,8} is a subspace of ~ ( E n ) ; we denote it by ,8 -L . We first prove a lemma. LEMMA2. Let

(2 beapropersubalgebraof {0}=~],0C

7n,1C...C

~(l~ n) andlet

~Ek=Cn

be a maximal chain of invariant subspaces for Q . Denote by Pj the orthogonal projection onto ~ j 0 ~[~j-1 ' J -- 1..... k . Assume that

P1 (2 P1 ~ {0} . Then exactly one of the following

alternatives holds. (i) There is a subalgebra (20 of Q0=PI(20

(2 such that

and P 1 Q o P I = P I ( 2

PI"

(ii) There is a j ~ 1 such that for all A in [2, tr(P 1 A P1) = tr(PjA Pj). ln particular, if Q_L is also an algebra, then (i) holds. PROOF. We shall use the following known result about the diagonal blocks of (2 (whose proof is contained, e.g., in [3]). There exists a partition. S 1 U S 2 U ... O S m of the set of integers { 1,2 ..... k} such that for each i there is a subalgebra (/i of (2 with Pj Q i P j = P j Q P j

or {0}

according as j belongs to S i or not. Furthermore, for each i , S i has the following property: Let s be a fixed member of S i . Then for each t in S i there is an invertible linear a'ansformation

Choi, Radjavi and Rosenthal

171

Rt: Pt cn ._, Ps~n such that Pt APt = Rt-l(Ps A Ps)Rt for every A in Q. To prove the lemma, let S 1 be the set that contains 1, ff it contains any other integer j , then (ii) follows immediately. If S 1 = { 1} , then the subalgebra 121 of the preceding paragraph satisfies [(1-P I) 121]n = [(t-PI)~21(1-P1)] n = {0}, so that the algebra {20 = {21n satisfies Q0 = P1120" To verify the equation P1 {20 P1 = P112 P1 just observe that P1 {2 P1 contains Pl (because {211~'[1.1 equals ,~(~1.1) by Bumside's theorem) and thus P1 {20 P1 = P1 (2 P1 just observe that P1 0 P1 contains P1 (because {2I]~ 1 equals /'.(~ll)

by Bumside's

theorem) and thus P1 121 P1 C_ (P1 (2 P1) n = P1 (]l n P1 9 This shows that (i) holds. Now if (ii) holds then, for A in {2, tr[A(P1-Pj)*] = tr(API) - tr(PjA) = tr(PlAP1) - tr(PjAPj) = 0. Thus P1 - P2 is in 12.a_. This implies that (2 & is not an algebra, because otherwise (P1 - pj)2 =Pl + P j e 12" and thus tr(P1AP1) =-tr(PjAPj) or tr(PlAP1) = 0 for all A in 12, which is a contradiction. [] THEOREM 4. If an orthogonal decomposition

(2 and

r

12_t. are proper subalgebras of ,~(~n), then there is

= ~1.1 @ ... ~ ~ k

relative to which (2 and (]a_ are upper and lower block-triangular respectively. Furhtermore, there is a subset S of the set of integers { 1..... k} such that

172

Choi, Radjavi and Rosenthal

k (2= {(Tij)i,j= 1 :Tij=0

whenever i > j

or i = j ~

S}.

(Thus we get a similar description for (2 -L : (2~_ = {(Tij)i,j= k 1 : Tij = 0 whenever i < j or i = j e S } . In particular, the conjecture is true in this case.) PROOF. It suffices to find a nontrivial subspace ~1, of •n invariant under (2 and co-invariantunder (2•

(i.e., ~q].a_ invariantunder Q.a_) with thefollowingproperty: in the

representation of 12 and ( 2 - relative to ]]], 9 ~j~.a_ :

(2=

0

(22

and

(2

= q~

~2

one of the algebras (21 and ~ 1 is ~(~T1,) and the other {0}. For then it is easily verified that ~ 2 is the orthogonal complement of (21 in J~(~[I, a') , so that the proof can be completed by induction. Let ~3, be a minimal nontrivial invariant subspacr of (2. (Such a subspace exists because (2 is a proper subalgebra). Note that the restriction algebra (21~FI, is either ~(~FL) or {0} , the latter being possible only if TL has dimension one. First assume (~t~-L= ,G(~J1,). Let P denote the orthogonal projection on ~1,. We shall show that ~

is co-invariant for (2_L, or PB(1-P) = 0 for all B in ( 2 " .

Observe that

(1-P)XP e C~" for any operator X . Thus for B in ( 2 1 , the two operators B-(1-P)B*P and (1-P)B(1-P)B*P belong to (2"J", and so does their difference PB(1-P)B*P. Now any operator of the form PYP in Q,L, and so does their difference PB(1-P)B*P. Now any operator of the form PYP in (~'Jis zero, because (2tlJ-l, contains all the operators on ~'L. Hence PB(1-P) = 0 . To show that = ~ ~I1,1 = ~

is the desired subspace, we use Lemma 2. Since (i) of the lemma holds for [2 (with and P1 = P), we deduce that P(2P is orthogonal to P ( 2 ~ P . Since (2[~-L= J~(~l.),

we conclude that p(2_Lp = {0} . Thus

(21 = J~(~l,) and ~ 1 = {0} .

Choi, Radjavi and Rosenthal

173

Next assume {2t~]q,= {0}. Then XP r {2" for any operator X . This implies that every nonzero co-invariant subspace of {2_L contains ~'1,. Let ~

be a minimal such subspace;

is proper because {2.a_ is proper. Denote the orthogonal projection on ~ that Q12•

by Q and note

~ {0} since P e {2_L. Thus Q{2_LQ[~], = ~ ( ~ )

by minimality. The rest of the proof is the "dual" of the one given in the preceding paragraph. We show that (1-Q)AQ = 0 for all A in (2 as follows. Observing that QX(1-Q) e (2 for arbitrary operator X , we obtain QA*(1-Q).A 9 12, QA*(1-Q)A(1-Q) 9 12 , and thus QA*(1-Q)AQ 9 {2. This implies that (1-Q)AQ = 0 . To complete the proof we must show that QAQ = {0}. We can do this by applying Lemma 2 to the algebra ((2") * = ((2*) -t-. [] COROLLARY . / f

(2 and (2 -t- are subalgebras of

d~(~ n) and I e {2, then

there is an orthogonal decomposition of ~n relative to which {2 consists of all upper blocktriangular operators and

(2-L consists of all strictly lower block-triangular ones.

COROLLARY

Let

12 and

~ be subalgebras of

positive-definite operator K such that K-I{2K = ~ "

[]

,~(IE n) . If there is a

,then (2 and ~ are complementary

algebras and have complementary invariant subspaces. PROOF. The hypotheses imply that the algebras K{2K and K ~ K are orthogonally complementary and the theorem is applicable to them.

O

REFERENCES 1.

I. Gohberg, P. Lancaster, and L. Rodman, Invariant Subspaces of Matrices with Applications, John Wiley & Sons, New York, 1986.

2.

N. Jacobson, Lectures in Abstract Algebra II: Linear Algebra, D. Van Nostrand, Princeton, 1953.

174

3~

Choi, Radjavi and Rosenthal

J.F. Watters, Block Triangularization of algebras of matrices, Linear Alg. & Appl. 32 (1980), 3-7.

Man-Duen Choi & Peter Rosenthal Department of Mathematics University of Toronto Toronto, Canada M5S 1A1

Submitted: Revised:

May 24, 1989 July 23, 1989

Heydar Radjavi Dept. of Math., Stats. & Comp. Sci. Dalhousie University Halifax, Canada B3H 3J5