Linear Algebra and its Applications

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1 Linear Algebra and its Applications 435 (20) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: On certain finite-dimensional algebras generated by two idempotents A. Böttcher a, I.M. Spitkovsky b, a Fakultät für Mathematik, TU Chemnitz, D-0907 Chemnitz, Germany b Department of Mathematics, The College of William and Mary, Williamsburg, VA 2387, USA ARTICLE INFO Article history: Received September 200 Accepted 5 March 20 Available online 20 April 20 SubmittedbyR.A.Brualdi AMS classification: Primary 5A30 Secondary 5A09 6R50 46H5 47L55 ABSTRACT This paper is concerned with algebras generated by two idempotents P and Q satisfying (PQ) m = (QP) m and (PQ) m (QP) m.the main result is the classification of all these algebras, implying that for each m 2 there exist exactly eight nonisomorphic copies. As an application, it is shown that if an element of such an algebra has a nondegenerate leading term, then it is group invertible, and a formula for the explicit computation of the group inverse is given. 20 Elsevier Inc. All rights reserved. Keywords: Finite-dimensional algebra Idempotent Skew and oblique projection Drazin inversion Group inversion. Introduction and main results Let B be an associative complex algebra and let P, Q B be two idempotents, that is, elements satisfying P 2 = P and Q 2 = Q. We assume that there is a natural number m such that (PQ) m = (QP) m. In that case the algebra A generated by P and Q isthesetofallelementsa of the form A = a P + b Q + a 2 PQ + b 2 QP + a 3 PQP + b 3 QPQ + +a 2m (PQ) m Q + b 2m (QP) m Q + c(pq) m () with coefficients from C. Corresponding author. addresses: aboettch@mathematik.tu-chemnitz.de (A. Böttcher), ilya@math.wm.edu, imspitkovsky@gmail.com (I.M. Spitkovsky) /$ - see front matter 20 Elsevier Inc. All rights reserved. doi:0.06/j.laa

2 824 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) The present paper has its source of motivation in Liu, Wu, and Yu s article [5], who studied group invertibility in matrix algebras generated by two idempotents P and Q under the assumption that the idempotents satisfy additional constraints such as (PQ) 2 = (QP) 2 or (PQ) 2 = 0. (Note that if (PQ) 2 = 0, then (PQ) 3 = (QP) 3 = 0.) In the case where (PQ) 2 = (QP) 2, representation ()takesthe form A = a P + b Q + a 2 PQ + b 2 QP + a 3 PQP + b 3 QPQ + c(pq) 2. Under the assumption that a 0 and b 0,Liuetal.[5] introduced X := P + ( Q + + a ) ( 2 PQ + + b 2 a b a b a b a b a b ( a 2 + b 2 + a ) 2b 2 b a 3 a b a b a 2 b PQP ( a 2 + b 2 + a ) 2b 2 a b 3 a b a b a b 2 QPQ + ( σ 2 2 a 2 + b 2 a 2b 2 b a 3 a 2b 2 a b 3 a b a b a 2 b a b 2 ) PQ ) (PQ) 2, where σ = a + b + a 2 + b 2 + a 3 + b 3 + c and /σ := 0forσ = 0, and they showed by a direct computation that this X satisfies A 2 X = A, X 2 A = X, AX = XA. We wanted to understand where they took this X from and what could happen if the condition (PQ) 2 = (QP) 2 is replaced by (PQ) m = (QP) m. We soon learned that one can easily find an X such that A 2 X = X and that this X is just the one of Liu, Wu, Yu in the case m = 2. The verification of the equalities X 2 A = X and AX = XA nevertheless remains nontrivial. The straightforward approach of [5] fails for general m due to the increasing computational complexity. However, after having some understanding of the structure of the algebras A, it becomes possible to derive the equalities X 2 A = X and AX = XA from the equality A 2 X = X. In the course of our investigation we also arrived at a classification of the algebras A.Sucha classification is clearly of interest by itself. Let P and Q be two idempotents satisfying (PQ) m = (QP) m. We assume that m 2; the case m = will be disposed of in Remark 4.2. Taking the smallest possible m, we may also assume that (PQ) m (QP) m.considerthelist = P, Q, PQ, QP, PQP, QPQ,..., (PQ) m,(qp) m,(pq) m P,(QP) m Q,(PQ) m. (2) The number of factors of a product in the list will be called the order of the product. For convenient referencing, we state the following simple observation as a lemma. Lemma.. If an element of order j in the list equals (PQ) m, then all elements in the list of order at least j + are also equal to (PQ) m. Proof. This is immediate from the fact that if (PQ) m is multiplied from the left or the right by P or Q, then the result is again (PQ) m. Suppose exactly k elements of the list are equal to (PQ) m, and put l = 0if(PQ) m = 0 and l = if (PQ) m 0. We refer to the triple (m, k,l)as the type of the pair (P, Q).Wealwayshavek 4. Indeed, if k 5, then either the last five elements of the list coincide, which gives (PQ) m = (QP) m,ora product of order less than 2m 2mustbeequalto(PQ) m and by Lemma., this implies again that (PQ) m = (QP) m = (PQ) m. Here are our main results.

3 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Theorem.2. Given integers m 2, k {, 2, 3, 4}, andl {0, }, there exist an n and idempotent matrices P and Q in C n n such that the type of (P, Q) is (m, k,l). Theorem.3. For i =, 2, letp i and Q i be idempotents in an associative complex algebra B i satisfying (P i Q i ) m i = (Q i P i ) m i and (P i Q i ) m i (Q i P i ) m i for some m i 2 and denote by A i the algebra generated by P i and Q i. The algebras A and A 2 are isomorphic if and only if the pairs (P, Q ) and (P 2, Q 2 ) have the same type. We now turn to group inversion in A. LetA be of the form (). Since A is finite-dimensional, the element A is always Drazin invertible in A. This means that there exist a smallest natural number k and an X A such that A k+ X = A k, X 2 A = X, and AX = XA. The number k is referred to as the Drazin index of A. The element A is said to be group invertible if its Drazin index is, that is, if there exists an element X A such that A 2 X = A, X 2 A = X, and AX = XA. IfsuchanX exists, it is unique. We refer the reader to Drazin s original paper [9] for generalized invertibility in associative algebras and to the books [,5] for the topic in algebras of matrices. Note that our algebras are finite-dimensional and therefore have faithful representations as algebras of matrices. Squaring A we get A 2 = α P + β Q + α 2 PQ + β 2 QP + α 3 PQP + β 3 QPQ + +α 2m (PQ) m Q + β 2m (QP) m Q + γ(pq) m. Note that α = a 2 and β = b 2.Denotingbyσ the sum of the coefficients in (), σ = a + b + a 2 + b 2 + +a 2m + b 2m + c, we also have α + β + α 2 + β 2 + +α 2m + β 2m + γ = σ 2.Let X = x P + y Q + x 2 PQ + y 2 QP + x 3 PQP + y 3 QPQ + +x 2m (PQ) m Q + y 2m (QP) m Q + z(pq) m. (3) Comparing the coefficients of P, Q, PQ, QP,...,(PQ) m in the equation A 2 X = A, wegetthe4m equations P : α x = a, Q : β y = b, PQ : α x 2 + (α + α 2 )y = a 2, QP : β y 2 + (β + β 2 )x = b 2, PQP : α x 3 + (α + α 2 )y 2 + (α 2 + α 3 )x = a 3, QPQ : β y 3 + (β + β 2 )x 2 + (β 2 + β 3 )y = b 3, PQPQ : α x 4 + (α + α 2 )y 3 + (α 2 + α 3 )x 2 + (α 3 + α 4 )y = a 4, QPQP : β y 4 + (β + β 2 )x 3 + (β 2 + β 3 )y 2 + (β 3 + β 4 )x = b 4,... (PQ) m : σ 2 z + λ = γ, where λ is a linear combination of x, y, x 2, y 2,...,x 2m, y 2m.Ifa 0 and b 0, then the first 4m 2 of these equations can be solved successively and the solution is unique. The last equation is uniquely solvable for σ 0. If σ = 0, we put z = γ λ. Here is our result concerning group inversion.

4 826 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Theorem.4. If a 0 and b 0,thenXisthegroupinverseofA,thatis,wehaveA 2 X = A, X 2 A = X, AX = XA. We remark that invertibility in algebras generated by two idempotents has been the subject of intense studies for many decades, starting with work by Krein, Krasnoselski, Milman, Dixmier, Davis, Halmos, Pedersen, Giles, and Kummer, to name only a few of the pioneering figures. See the surveys [4,0,]. In the 990s, notable progress was made by Roch and Silbermann [7] and by Gohberg and Krupnik [2,3], who solved the problem of characterizing invertibility in Banach algebras generated by two idempotents (subject to no further constraints) via a symbol calculus. Moore Penrose invertibility in the C -algebra generated by two orthogonal projections was studied by one of the authors in [8]. In recent years, Drazin invertibility has received increasing attention. See, for example, [3,6 8,5,9]. For m = 2, Theorem.4 is due to Liu et al. [5]. Their group inverse X cited above is exactly the solution of the 7 equations for X arising from the equation A 2 X = X. Example.5. In the case where A = ap + bq (a 0, b 0) and (PQ) m = (QP) m,the4m equations yield the group inverse m [( j + X = + j ) ( j P(QP) j + + j + ) ] Q(PQ) j a b a b j=0 m 2 j=0 [( j + a + j + ) ( j + P(QP) j Q + b a ( + a + m m ) (PQ) m, b a b + j + ) ] Q(PQ) j P b with /(a+b) := 0ifa+b = 0. In the case where B is a Banach algebra and P and Q satisfy additional hypotheses, such as PQP = P, the first terms of this formula are already in [9]. Section 2 contains the proof of Theorems.2 and.3, while Theorem.4 will be proved in Section 3. Several additional topics are discussed in Section Existence and classification Proof of Theorem.2. If C and B are matrices in C k k, I is the k k identity matrix, and 0 denotes the zero matrix of order k, then P = I C, Q = 00 (4) 0 0 B I are idempotent matrices in C 2k 2k.Wehave (PQ) m = (CB)m (CB) m 2 C, (QP) m 0 0 =, 0 0 (BC) m 2 B (BC) m (PQ) m P = (CB)m (CB) m C, (QP) m 0 0 Q =, 0 0 (BC) m B (BC) m (PQ) m = (CB)m (CB) m C, (QP) m 0 0 =. 0 0 (BC) m B (BC) m

5 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Type (m,, 0). Let B and C be the Jordan block of order 2m with zeros on the main diagonal, C = B =. 0 0 Then (CB) m C = (BC) m B = 0 and (CB) m = (BC) m 0, and hence (PQ) m = (QP) m = 0, (PQ) m P 0, (QP) m Q 0, and (PQ) m (QP) m.consequently,(p, Q) is of the type (m,, 0). Type (m, 2, 0). Defining C, B C m m by C =, B =, we have (CB) m = 0 and (BC) m 0, which gives (PQ) m = (QP) m = 0, (PQ) m P = 0, (QP) m Q 0, and (PQ) m (QP) m.thus,(p, Q) is of the type (m, 2, 0). Type (m, 3, 0). Let C = B be the Jordan block of order 2m 2 with zero on the main diagonal. In that case (CB) m = (BC) m = 0 and (CB) m 2 = (BC) m 2 0, yielding that (PQ) m = (QP) m = 0, (PQ) m P = (QP) m Q = 0, and (PQ) m (QP) m. It follows that (P, Q) has the type (m, 3, 0). Type (m, 4, 0). Let C, B C (m ) (m ) be the matrices C =, B =. 0 0 Then (CB) m 2 C = C(BC) m 2 = 0 and (BC) m 2 B 0. We therefore obtain that (PQ) m = (QP) m = (PQ) m P = (QP) m Q = (PQ) m = 0, and (QP) m 0, which shows that (P, Q) is of the type (m, 4, 0). Type (m, k, ). If the pair (P 0, Q 0 ) has the type (m, k, 0), then the pair (P, Q) given by P = I 0, Q = I 0 0 P 0 0 Q 0 is of the type (m, k, ). To prepare the proof of Theorem.3, we consider again the list (2). Suppose the type of (P, Q) is (m, k,l). We construct a new list 0 as follows. If the type is (m, k, 0),wedeletethek elements which

6 828 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) equal (PQ) m = 0, and if the type is (m, k, ), we keep (PQ) m in the list but delete the other k elements which equal (PQ) m. In the first case, 0 has the length 4m k, and in the second, 0 is of length 4m k. Lemma 2.. The elements of 0 are linearly independent. Proof. Suppose there exists a nontrivial linear combination of elements of 0 that is equal to zero and let j be the smallest order of the monomials actually involved in this linear combination (that is, having a nonzero coefficient). Assume first that j < 2m 2. There may be at most two such monomials. Choose one of them, and denote it by M. Without loss of generality let this monomial start with P and denote its coefficient by u j. Multiply the linear combination by P from the left and by P/Q from the right if j is odd/even. This operation does not change M. Now derive two equalities from the given one. For the first, multiply from the right by such a product which completes M to (PQ) m, and for the second, multiply from the left by Q and from the right by a product which completes M to (QP) m. In both cases, all other terms in the linear combination become (PQ) m, and their coefficients do not change. What follows is that u j (PQ) m = u j (QP) m.consequently,u j = 0. This is a contradiction, which implies that in fact j is at least 2m 2. Thus, we are left with the five-term equality u(pq) m + v(qp) m + x(pq) m P + y(qp) m Q + w(pq) m = 0. (5) Multiplication of this equality by (PQ) m shows that u + v + x + y + w = 0 provided that (PQ) m 0. Multiplying (5) fromtheleftbyp and from the right by Q, weobtain u(pq) m + (v + x + y + w)(pq) m = 0. (6) If (PQ) m = (PQ) m,thenu = 0 (since then (PQ) m is not in 0 ). If (PQ) m (PQ) m = 0, then (6) takestheformu(pq) m = 0, so that again u = 0. Finally, if (PQ) m (PQ) m and (PQ) m 0, then rewriting (6)as u((pq) m (PQ) m ) = 0 yields the same conclusion u = 0. So,wehaveprovedthatu = 0in(5). Analogously, v = 0. It remains to consider x(pq) m P + y(qp) m Q + w(pq) m = 0, (7) where x + y + w = 0unless(PQ) m = 0. After multiplication by P from the left, (7) becomes x(pq) m P + (y + w)(pq) m = 0. Reasoning analogous to the one we used when considering (6) shows that x = 0. Similarly, y = 0, so that (7) boilsdowntow(pq) m = 0. But then w = 0for(PQ) m = 0 (since (PQ) m is not in 0 ) and w = 0for(PQ) m 0. This completes the proof. Lemma 2.2. For i =, 2, letp i and Q i be idempotents in an associative complex algebra B i and suppose (P i, Q i ) is of the type (m i, k i,l i ).DenotebyA i the algebra generated by P i and Q i.if(m, k,l ) = (m 2, k 2,l 2 ),thena and A 2 are isomorphic. Proof. We claim that the map : A A 2 defined by (a P + b Q + a 2 P Q + b 2 Q P + +c(p Q ) m) = a P 2 + b Q 2 + a 2 P 2 Q 2 + b 2 Q 2 P 2 + +c(p 2 Q 2 ) m (8)

7 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) is a well-defined algebra isomorphism. Indeed, by the construction of the lists () 0 and (2) 0 associated with A and A 2, respectively, we may assume that the linear combinations on the left and right of (8) are formed by elements of () 0 and (2) 0 only. As (P, Q ) and (P 2, Q 2 ) have the same types, a specific product of P and Q appears on the left if and only if the same product with P and Q replaced by P 2 and Q 2 is present on the right. Now suppose we have two linear combinations A and B of elements of the list () 0 and A B = u P + v Q + u 2 P Q + v 2 Q P + +c(p Q ) m = 0. The products are all from the list () 0. Since these are linearly independent by virtue of Lemma 2., it follows that all coefficients are zero. This implies that (A) = (B). Thus, is well-defined. It is clear that is linear, multiplicative, and surjective. To show that is injective, suppose (8) iszero. The products P 2, Q 2, P 2 Q 2, Q 2 P 2,...,(P 2 Q 2 ) m areallfromthelist (2) 0. Lemma 2. tells us that the coefficients must all be zero and hence a P + b Q + a 2 P Q + b 2 Q P + +c(p Q ) m = 0, as desired. Lemma 2.2 yields the if portion of Theorem.3. By Lemma 2., the dimension of the algebra generated by P and Q is equal to the length of the list 0.Thus,thedimensionis4m k ifl = 0 and 4m k if l =. Since every finite-dimensional associative algebra over C of dimension M has a faithful representation as a subalgebra of C (M+) (M+),itfollowsthatthen in Theorem.2 can actually be chosen to be 4m k + l. We are left with the only if part of Theorem.3. It suffices to prove that if A and A 2 have the same dimension but (P, Q ) and (P 2, Q 2 ) are of different types, then A and A 2 are not isomorphic. Thus,therearetwocaseswehavetostudy:thecaseofthetypes(m, k, 0) and (m, k +, ) with k 3 and the case of the types (m +, 4, 0) and (m,, ). We denote by L n the linear hull of all products in the list whose order is at least n. Lemma 2.3. Let (P, Q) have the type (m, k, 0). IfP A is an idempotent, then one of the following is true: (a) P = 0; (b) P = P + LwithL L 2 ; (c) P = Q + LwithL L 2 ; (d) P = P + Q (PQ + QP) + (PQP + QPQ) (PQPQ + QPQP) +. Proof. Let P = xp + yq + L with L L 2.ThenP 2 = x 2 P + y 2 Q + L with L L 2. Since P and Q are always on the list 0,itfollowsthatforP to be an idempotent it is necessary that x 2 = x and y 2 = y. For (x, y) = (, 0) and (x, y) = (0, ) we arrive at (b) and (c), respectively. Suppose (x, y) = (0, 0). We claim that then P = 0. To prove this, assume P 0. We have P = x 2 P + y 2 Q + x 3 PQP + y 3 QPQ + and may assume that the coefficient of a product is zero if this product is not in the list 0.Because P 0, there is a smallest n 2 such that (x n, y n ) (0, 0).Thus,P = x n U + y n V + L where L L n+ and U, V are products of the order n.since(x n, y n ) (0, 0), one of the products U and V is a member of the list 0.Weget P 2 = x 2 n U2 + x n y n (UV + VU) + y 2 n V 2 + L with L L n+.asu 2, UV + VU, V 2 are also in L n+, we see that P = P 2 L n+, which implies that x n U + y n V L n+. What results is that a member of the list 0 is a linear combination of the remaining members of the list, and this contradicts Lemma 2.. Thus,if(x, y) = (0, 0), wehave case (a).

8 830 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Let finally (x, y) = (, ). Then P = P + Q + x 2 PQ + y 2 QP + L =: P + Q + B (9) with L L 3.WehaveP 2 = (P + Q) 2 + (P + Q)B + B(P + Q) + B 2.Clearly,B 2 L 3 and (P + Q)B = x 2 P + y 2 QP + L, B(P + Q) = x 2 PQ + y 2 QP + L with L, L L 3. It follows that P 2 = P + Q + PQ + QP + 2x 2 PQ + 2y 2 QP + L (0) with L L 3. Comparison of (9) and (0) shows that x 2 = ifpq is in 0 and y 2 = ifqp belongs to 0.IncaseoneoftheproductsPQ, QP is not in 0,wehaveL 3 ={0} and hence P = P + Q PQ or P = P + Q QP, which is as in (d). Otherwise we get P = P + Q PQ QP + L with L L 3.If L 3 ={0}, the proof is complete. So assume L 3 contains PQP or QPQ. Wethenhave P = P + Q PQ QP + x 3 PQP + y 3 QPQ + N =: P + Q PQ QP + C () where N L 4, with the convention that x 3 = 0ifPQP is not in 0 and y 3 = 0ifQPQ is not in 0.The square P 2 equals (P + Q PQ QP) 2 + (P + Q PQ QP)C + C(P + Q PQ QP) + C 2. Since C 2 L 4 and (P + Q PQ QP)C = x 3 PQP + y 3 QPQ + N, C(P + Q PQ QP) = x 3 PQP + y 3 QPQ + N, with N, N L 4,itfollowsthat P 2 = P + Q PQ QP PQP QPQ + 2x 3 PQP + 2y 3 QPQ + N (2) with N L 4. Now compare () and (2). If QPQ is not in 0, then the terms with QPQ may be assumed to be absent in (2). In that case PQP is on the list 0 and hence x 3 =. Analogously, if PQP is not in 0,theny 3 =. Finally, in case both PQP and QPQ are on the list 0,wegetx 3 = y 3 =. Continuing in this way we obtain x 4 = y 4 =, x 5 = y 5 =, x 6 = y 6 =,... as long as the products associated with the coefficients belong to 0. This proves that P must be as in (d). Lemma 2.4. If (P, Q ) has the type (m, k, 0) with k 3 and (P 2, Q 2 ) is of the type (m, k +, ) then the corresponding algebras A and A 2 are not isomorphic. Proof. Assume : A 2 A is an isomorphism and put P = (P 2 ), Q = (Q 2 ).ThenP and Q are idempotents which generate A, and the type of the pair (P, Q) is (m, k +, ).WedenoteP, Q, A by P, Q, A. By Lemma 2.3 and by symmetry, we may assume that we have one of the following three cases: (C) P = P + x 2 PQ + y 2 QP +, Q = P + a 2 PQ + b 2 QP +, (C2) P = P + x 2 PQ + y 2 QP +, Q = Q + a 2 PQ + b 2 QP +, (C3) P = P + x 2 PQ + y 2 QP +, Q = P + Q (PQ + QP) +.

9 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Case (C). In this case we have PQ = P + a 2 PQ + y 2 QP + L, (PQ) 2 = P + a 2 PQ + y 2 QP + L and so on, until (PQ) m = P + a 2 PQ + y 2 QP + L with L, L, L L 3. Analogously, (QP) m = P + x 2 PQ + b 2 QP + L with L L 3.As(P, Q) is of the type (m, k +, ), the equality (PQ) m = (QP) m holds and hence (a 2 x 2 )PQ + (y 2 b 2 )QP L 3. If PQ belongs to the list 0, this implies that a 2 = x 2, and in case QP is in 0,wegety 2 = b 2.Ineither case, P Q = (x 3 a 3 )PQP + (y 3 b 3 )QPQ + L 3. It follows that (P Q) 2 L 5, (P Q) 3 L 7,..., (P Q) m L 2m+. By Lemma., L 2m+ ={0}.Thus, (P Q) m = 0. (3) Now consider the lists and 0 associated with P and Q. The element (P Q) m is a nontrivial linear combination of the first 2m products on the list.as 0 has length 4m k 4m 4 2m,the first 2m products of belong all to 0 and are therefore linearly independent. This shows that (3)is impossible. Case (C2). The product PQ belongs to the linear hull of {PQ} L 3. This implies that (PQ) 2 is in the linear hull of {(PQ) 2 } L 5, that (PQ) 3 lies in the linear hull of {(PQ) 3 } L 7, and so on. Eventually, (PQ) m is an element of the linear hull of {(PQ) m } L 2m+ ={0}, that is, (PQ) m = 0. But this cannot happen if the type is (m, k +, ). Case (C3). It can be checked straightforwardly that UQ = U for every U from the list. Consequently, PQ = P. This implies that P and PQ are linearly dependent. But as (P, Q) is of the type (m, k +, ), thefirst4m k productsofthelist must be in 0 and thus linearly independent. Because 4m k 7 k 4, the elements P, Q, PQ, QP are linearly independent. This contradiction shows that case (C3) is impossible as well. Lemma 2.5. If (P, Q ) has the type (m +, 4, 0) and (P 2, Q 2 ) is of the type (m,, ) then the corresponding algebras A and A 2 are not isomorphic. Proof. This can be proved in the same way as Lemma 2.4.Thistime(P, Q) has the type (m,, ) and P, Q are of the form as in the cases (C) to (C3) in the proof of Lemma 2.4.Inthecase(C) we get as in that proof that (P Q) m = 0. The length of 0 is 4m 2m, which implies that all products of order at most 2m belong to 0 and that therefore (P Q) m cannot be zero. In the case (C2) we obtain as in the proof of Lemma 2.4 that (PQ) m = 0, which is impossible for the type (m,, ).Inthe case (C3), wehaveagainpq = P, and since the list 0 is of length 4m 7, the elements P and PQ are linearly independent. Thus, we arrive again at a contradiction. Proof of Theorem.3. Lemma 2.2 proves the if part. From Lemmas 2.4 and 2.5 we infer that the algebras are not isomorphic in the critical cases where they have the same dimension but are generated by pairs of different types. In all other cases the algebras generated by pairs of different types

10 832 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) have different dimensions and hence cannot be isomorphic. This completes the proof of the only if portion. 3. Group inversion Proof of Theorem.4. Suppose first that σ 0. Then the 4m equations of the introduction provide us with an X A such that A 2 X = A. Analogously we get a Y A such that YA 2 = A. Consequently, A is group invertible in A (see [9,4,6]). Let G A be the group inverse. We show that the equation A 2 X = 0 has only the solution X = 0inA. This implies that the equation A 2 X = A is uniquely solvable in A, and since it is satisfied by both the X of Theorem.4 and G, we arrive at the desired conclusion that X = G. Thus, let X be of the form (3) and suppose A 2 X = 0. We may assume that X is a linear combination of terms in 0.TheproductA 2 X may also be written as a linear combination of the elements in 0, and as these are linearly independent, all coefficients must be zero. The coefficients of P and Q are α x and β y.sinceα = a 2 0 and β = b 2 0, we conclude that x = y = 0. The coefficient of PQ is α x 2 + (α + α 2 )y = α x 2.IfPQ 0,wegetα x 2 = 0 and thus x 2 = 0. In case PQ is not in 0,thetermx 2 PQ is missing in (3) and hence the coefficient of PQ in A 2 X is automatically zero. Analogously, the coefficient of QP in A 2 X is β y 2 + (β + β 2 )x = β y 2, which gives y 2 = 0ifQP 0 and shows that y 2 QP is not present in (3) and that the coefficient of QP in A 2 X is automatically zero if QP / 0. Continuing in this way we arrive at the conclusion that each of the coefficients x, y, x 2, y 2,...,x 2m, y 2m is either zero or missing in (3). The coefficient of (PQ) m in A 2 X is σ 2 z + λ = σ 2 z with σ 0. If (PQ) m 0,wegetz = 0, while if (PQ) m = 0, the term z(pq) m in (3) is missing. In summary, we have shown that X = 0. This completes the proof in the case where σ 0. Now suppose σ = 0. In this case we denote the elements () and (3) bya 0 and X 0.PutA = A 0 + (PQ) m. Clearly, the sum of the coefficients of A is. From what was already proved we know that A has a group inverse X of the form (3). The proof of the theorem will be complete once we have shown that X 0 := X (PQ) m satisfies A 2 0 X 0 = A 0, X0 2 A 0 = X 0, and A 0 X 0 = X 0 A 0. The product of an element of the form () or(3) and (PQ) m is equal to the product of the sum of the coefficients of this element and (PQ) m. The sums of the coefficients of A 0, X 0, A, X are 0, 0,,, respectively. Since A 2 (PQ)m = (PQ) m,itfollowsthat A 2 X 0 = A 2 (X (PQ) m ) = A (PQ) m = A 0, and as A 0 (PQ) m = (PQ) m A 0 = 0 and hence A 2 = (A 0 + (PQ) m ) 2 = A A 0(PQ) m + (PQ) m A 0 + (PQ) 2m = A (PQ)m, we obtain that A 2 0 X 0 + (PQ) m X 0 = A 0, which gives A 2 0 X 0 = A 0. Analogously, because X 2(PQ)m = (PQ) m,wehave X 2 A 0 = X 2 (A (PQ) m ) = X (PQ) m = X 0, and taking into account that X 2 = X2 0 + (PQ)m,wegetX 2 0 A 0 + (PQ) m A 0 = X 0 and thus X 2 0 A 0 = X 0. Finally, the equality A X = X A reads (A 0 + (PQ) m )(X 0 + (PQ) m ) = (X 0 + (PQ) m )(A 0 + (PQ) m ), which cancels down to A 0 X 0 = X 0 A Additional remarks Remark 4.. Theorem.4 is no longer true without the assumption that a 0 and b 0. To see this, let

11 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) P = I I, Q = I 0 00 C 0 and put A = Q + PQ QP. Clearly, (PQ) 2 = (I + C)2 0, A = I + C I, A 2 = (I + C)2 I C 0 C 2 Choosing C so that I+C is the Jordan block of order 2 with zeros on the main diagonal, we get (PQ) 2 = 0 and thus (PQ) 3 = (QP) 3 = 0, while rank A = 3 and rank A 2 = 2, which implies that A is not group invertible. Remark 4.2. For commuting idempotents the subject of this paper is trivial. Indeed, write A = ap + bq + cpq A as A = aπ + bπ 2 + (a + b + c)π 3 := a(p PQ) + b(q PQ) + (a + b + c)pq. We have π i π j = δ ij π ij and hence A is isomorphic to the algebra of all diagonal matrices in C n n where n is the number of different nonzero elements on the list π,π 2,π 3. We remark that Theorem.3 is not true in the case m =, that is, for commuting idempotents. To see this, let P = 0, P 2 = 0, Q = Q 2 = Then A and A 2 are both isomorphic to the algebra of all 2-by-2 diagonal matrices. However, (P, Q ) has the type (, 2, ), while (P 2, Q 2 ) is of the type (,, 0). Letting α = /α if α 0 and 0 = 0, we see that X = a π + b π 2 + (a + b + c) π 3 is the group inverse of A. The situation is almost equally simple for algebras generated by an arbitrary finite number of commuting idempotents; see, for instance, Section 0 of [4]. Remark 4.3. Our paper [3] contains a criterion for Drazin invertibility in the W -algebra generated by two orthogonal projections on Hilbert space. No additional conditions are required there. In that paper it is also clarified when the Drazin inverse is the group inverse. See also Section 9 of [4]. We remark that if P and Q are orthogonal projections on Hilbert space, then (PQ) m = (QP) m if and only if PQ = QP. Remark 4.4. Let B be a Banach algebra and P, Q B be idempotents. Instead of requiring that (PQ) m = (QP) m (which implies that (PQ) m = (PQ) m+ and hence that the spectrum of PQ is a subset of {0, }), we now demand that the spectral radius of PQ is strictly less than. Then (PQ) j goes to zero exponentially fast as j. Thus, deleting in the expression for X in Example.5 the last term and replacing the sums by j=0, we obtain absolutely converging series for a well-defined element X B, [( j + X = + j a b j=0 [( j + a j=0 ) ( j P(QP) j + + j + ) ] Q(PQ) j a b + j + ) ( j + P(QP) j Q + b a + j + ) ] Q(PQ) j P. b One can verify straightforwardly that A 2 X = A, AX = XA, and X 2 A = X for A = ap +bq.consequently, X is the group inverse of A.

12 834 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) Remark 4.5. Suppose B is an associative algebra and P, Q B are idempotents, subject to no additional conditions. For m, we denote by J m the ideal in A generated by (PQ) m (QP) m, that is, the set of all finite sums of the form i A i ((PQ) m (QP) m )B i with A i, B i A {}, where a = a := a. Since P[(PQ) m (QP) m ]=(PQ) m (QP) m [(PQ) m (QP) m ]P, Q[(PQ) m (QP) m ]=(PQ) m (QP) m [(PQ) m (QP) m ]Q, we actually have J m = { k[(pq) m (QP) m ]+[(PQ) m (QP) m ]A : k Z, B A}. (4) It is easily seen that J J 2 J 3 and that J m = J m+ implies that J m = J m+k for all k. It is also clear that J m ={0} if and only if (PQ) m = (QP) m. Consider the quotient algebra A = A/J m and put P = P + J m, Q = Q + J m.obviously,(pq) m = (QP) m.ifj m J m,wealso have (PQ) m (QP) m.leta A be a finite sum of the form A = a P + b Q + a 2 PQ + b 2 QP + with a 0 and b 0 and determine the coefficients of the formal series X = x P + y Q + x 2 PQ + y 2 QP + by comparing the coefficients of A 2 X and A.IfJ m J m, we can apply Theorem.4 to A to conclude that the element X = X m + J m given by X m = x P + y Q + x 2 PQ + y 2 QP + +x 2m (PQ) m Q + y 2m (QP) m Q + x 2m (PQ) m is the group inverse of A = a P + b Q + a 2 PQ + b 2 QP +.Equivalently, A 2 X m A J m, X 2 m A X m J m, AX m X m A J m. Consequently, if A is group invertible in A, that is, if there is a G A such that A 2 G = A, G 2 A = G, AG = GA, theng X J m.thus,wehavethegroupinverseatleastmoduloj m. If B is finite-dimensional, and in particular if B = C n n, there is a smallest natural number m such that J J 2 J m = J m+ = J m+2 =. From what was said in the preceding paragraph, we get the group inverse of matrices A A modulo J m.forexample,ifb = C 2 2 and P =, Q = 00, 00 then (PQ) m = P Q = (QP) m for all m and J = J 2 = J 3 = is the ideal of matrices of rank at most. More generally, if P and Q are n n matrices and (PQ) m (QP) m has rank r, then (4) implies that J m is a subset of the matrices of rank at most r.incasea is infinite-dimensional and (PQ) m (QP) m for all m, we can obtain at least in principle the group inverse of A A modulo m= J m. Remark 4.6. Suppose B is a Banach algebra with identity element I.LetP and Q be two idempotents in B and let A stand for the smallest closed subalgebra of B which contains I, P, Q. ForA B, we

13 A. Böttcher, I.M. Spitkovsky / Linear Algebra and its Applications 435 (20) denote by σ B (A) the spectrum of A in B. The following result was established by Roch and Silbermann [7], and Gohberg and Krupnik [2,3]. A full proof is also in [2]. Put T = (P Q) 2. For each point λ σ B (T) the map F λ :{I, P, Q} C 2 2 given by F λ (I) = 0, F λ (P) = 0, F λ (Q) = λ w(λ), 0 00 w(λ) λ where w(λ) := λ( λ) denotes any number the square of which equals λ( λ), extends to a continuous algebra homomorphism of A to C 2 2, and for each point λ σ B (P + 2Q) {0,, 2, 3} the map G λ :{I, P, Q} C given by G λ (I) = and G 0 (P) = 0, G 0 (Q) = 0, G (P) =, G (Q) = 0, G 2 (P) = 0, G 2 (Q) =, G 3 (P) =, G 3 (Q) = extends to a continuous algebra homomorphism of A to C.AnelementA A is invertible in B if and only if det F λ (A) 0 for all λ belonging to σ B (T) \{0, } and G λ (A) 0 for all λ σ B (P + 2Q) {0,, 2, 3}. Now suppose (PQ) m = (QP) m.since F λ [(PQ) m ]= ( λ)m ( λ) m w(λ), 0 0 F λ [(QP) m ( λ) m 0 ]=, ( λ) m w(λ) 0 we necessarily have σ B (T) {0, }. This reveals once more that the condition (PQ) m = (QP) m is highly restrictive. Anyway, because σ B (T) \{0, } is empty, invertibility in B of elements A A is completely determined by the at most four one-dimensional representations G λ. Let A be of the form (). Then G 0 (A) = 0, G (A) = a, G 2 (A) = b, G 3 (A) = σ (= the sum of all coefficients). Thus, if a, b,σ are nonzero, then A is invertible in B if and only if 0 is not in the spectrum σ B (P + 2Q), that is, if and only if P + 2Q is invertible in B, which is in turn equivalent to the invertibility in B of simply P + Q. Choosing B = A, we conclude that A is invertible in A if and only if P + Q is invertible in A. Analogously, considering di + A instead of A, wegetg 0 (di + A) = d, G (di+a) = d+a, G 2 (di+a) = d+b, G 3 (di+a) = d+σ.consequently,ifd / {0, a, b, σ }, then di + A is always invertible. References [] A. Ben-Israel, T.N.E. Greville, Generalized Inverses, second ed., Springer, New York, [2] A. Böttcher, I. Gohberg, Yu. Karlovich, N. Krupnik, S. Roch, B. Silbermann, I. Spitkovsky, Banach algebras generated by N idempotents and applications, Oper. Theory Adv. Appl. 90 (996) [3] A. Böttcher, I. Spitkovsky, Drazin inversion in the von Neumann algebra generated by two orthogonal projections, J. Math. Anal. Appl. 358 (2009) [4] A. Böttcher, I. Spitkovsky, A gentle guide to the basics of two projections theory, Linear Algebra Appl. 432 (200) [5] S.L. Campbell, C.D. Meyer, Generalized Inverses of Linear Transformations, Pitman, London, 979 (and SIAM, Philadelphia, 2009). [6] D.S. Cvetković-Ilić, Chunyuan Deng, Some results on the Drazin invertibility and idempotents, J. Math. Anal. Appl. 359 (2009) [7] Chun Yuan Deng, The Drazin inverses of products and differences of orthogonal projections, J. Math. Anal. Appl. 335 (2007) [8] Chun Yuan Deng, The Drazin inverses of sum and difference of idempotents, Linear Algebra Appl. 430 (2009) [9] M.P. Drazin, Pseudo-inverses in associative rings and semigroups, Amer. Math. Monthly 65 (958) [0] A. Galántai, Projectors and Projection Methods, Kluwer Academic Publishers, Boston, [] A. Galántai, Subspaces, angles and pairs of orthogonal projections, Linear and Multilinear Algebra 56 (2008) [2] I. Gohberg, N. Krupnik, Extension theorems for invertibility symbols in Banach algebras, Integral Equations Operator Theory 5 (992)

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