On the Moore-Penrose and the Drazin inverse of two projections on Hilbert space

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1 On the Moore-Penrose and the Drazin inverse of two projections on Hilbert space Sonja Radosavljević and Dragan SDjordjević March 13, 2012 Abstract For two given orthogonal, generalized or hypergeneralized projections P and Q on Hilbert space H, we gave their matrix representation We also gave canonical forms of the Moore-Penrose and the Drazin inverses of their product, difference and sum In addition, it is showed when these operators are EP and some simple correlations between mentioned operators are established 1 Introduction Motivation for writing this paper came from publicatons of Deng and Wei (5, 6) and Baksalary and Trenkler, (1, 2) Namely, Deng and Wei studied Drazin invertibility for product, difference and sum of idempotents and Baksalary and Trenkler used matrix representation of the Moore-Penrose inverse of product, difference and sum of orthogonal projections Our main goal is to give canonical form of the Moore-Penrose and the Drazin inverse for product, difference and sum of two orthogonal, generalized or hypergeneralized projections on an arbitrary Hilbert space Using the canonical forms, we can examine when the Moore-Penrose and the Drazin inverese exist Also, we can describe the relation between inverses (if any), estimate the Drazin index and establish necessary and sufficient conditions under which these operators are EP Although some of the results are the same or similar to results in mentioned papers, our results are different sice for the starting operators we used generalized and hypergeneralized pojectons and not only orthogonal projections There is also a partial difference in the sense of examined properties Throughout the paper, H will stand for Hilbert space and L(H) will stand for set of all bounded linear operators on space H The symbols R(A), N (A) and A will denote range, null space and adjoint operator of operator A L(H) Operator P L(H) is idempotent if P P 2 and it is an orthogonal projection if P P 2 P Generalized and hypergeneralized projections were inroduced in 7 by Groβ and Trenkler 1

2 Definition 11 Operator G L(H) is (a) a generalized projection if G 2 G, (b) a hypergeneralized projection if G 2 G Set of all generalized projecton on H is denoted by GP(H) and set of all hypergeneralized projecton is denoted by HGP(H) Here A is the Moore-Penrose inverse of A L(H) ie the unique solution to the equations AA A A, A AA A, (AA ) AA, (A A) A A Notice that A exists if and only if R(A) is closed Then AA is orthogonal projection onto R(A) parallel to N (A ), and A A is orthogonal projection onto R(A ) parallel to N (A) Consequently, I AA is orthogonal projection onto N (A ) and I A A is orthogonal projection onto N (A) An essential property of any P L(H) is that P is an orthogonal projection if and only if it is expressible as AA, for some A L(H) For A L(H), an element B L(H) is the Drazin inverse of A, if the following hold BAB B, BA AB, A n+1 B A n, for some non-negative integer n The smallest such n is called the Drazin index of A By A D we denote Drazin inverse of A and by ind(a) we denote Drazin index of A If such n does not exist, ind(a) and operator A is generalized Drazin invertible Its invers is denoted by A d Operator A is invertible if and only if ind(a) 0 If ind(a) 1, A is group invertible and A D is group inverse, usually denoted by A # Notice that if the Drazin inverse exists, it is unique Operator A L(H) is Drazin invertible if and only if asc(a) < and dsc(a) <, where asc(a) is the minimal integer such that N (A n+1 ) N (A n ) and dsc(a) is the minimal integer such that R(A n+1 ) R(A n ) In this case, ind(a) asc(a) dsc(a) n Recall that if R(A n ) is closed for some integer n, then asc(a) dsc(a) < Operator A L(H) is EP if AA A A, or, in the other words, if A A D A # There are many characterization of EP operators In this paper, we use results from Djordjević and Koliha, (4) In what follows, A will stand for I A and P A will stand for AA 2 Auxiliary results Let P, Q L(H) be orthogonal projectons and R(P ) L Since H R(P ) R(P ) L, we have the following representaton of projections P, P, Q, Q 2

3 L(H) with respect to the decomposition of space P1 0 IL 0 L P P 0 I L A B Q B D L IL A B Q B I D L L L L, (1), (2), (3) L with A L(L) and D L( ) being Hermitian and non-negative Next two theorems are known for matrices on C n, see 2, (4) Theorem 21 Let Q L(H) be represented as in (3) holds Then the following (a) A A 2 + BB, or, equivalently, AA BB, (b) B AB + BD, or, equivalently, B B A + DB, (c) D D 2 + B B, or, equivalently, DD B B Proof Since Q Q 2, we obtain A B A B A B D B 2 + BB AB + BD A B D B A + DB B B + D 2 B D implying that A A 2 + BB, B AB + BD and D D 2 + B B Theorem 22 Let Q L(H) be represented as in (3) Then (a) R(B) R(A), (b) R(B) R(A), (c) R(B ) R(D), (d) R(B ) R(D), (e) A B BD, (f) A B BD, (g) A is a contraction, (h) D is a contraction, (i) A BD B I L A A 3

4 Proof (a) Since A A 2 + BB, we have R(A) R(A 2 + BB ) R(AA + BB ) To prove that R(AA + BB ) R(A) + R(B), observe operator matrix A B M For any x R(MM ), there exisit y H such that x MM y M(M y) and x R(M) On the other hand, for x R(M), there is y H and x My Besides, MM x MM My My x and MM MM (MM ) P R(MM ) implying x R(MM ) Hence, R(M) R(MM ) and and we have R(A) + R(B) R(M) R(MM ) R(AA + BB ) R(A) R(A) + R(B) implying R(B) R(A) (b) Since A I A, from Theorem 21 (a), we get A A 2 + BB The rest of the proof is analogous to the point (a) of this theorem (c), (d) Similarly (e) Since B AB+BD, we have A B A (AB+BD) A AB+A BD and using the facts that A A P R(A ) and R(B) R(A ), we get A AB B and A B B + A BD, or, equivalently B A BD Postmultiplying this equation by D and using item (d) of this Theorem, in its equivalent form BD D B, we obtain (e) (f) Analogously to the previous proof (g) Since A A, from Theorem 21 (a), we have that I L AA I L (A BB ) A + BB, and the right hand side is nonnegative as a sum of two nonnegative operators implying that A is a contraction (h) This part of the proof is dual to the part (g) (i) From Theorem 21 (a), item (f) of this Theorem and the fact that hermitian operator A commutes with its MP-inverse, it follows that BD B A BB A AA A (I A)A A A A A A A A A by taking into account that A A A A Now we get establishing the condition A BD B I A A, 4

5 Following the results of Groβ and Trenkler for matrices, we will formulate a few theorems for generalized and hypergeneralized projections on arbitrary Hilbert space We start with the result which is very similar to Theorem (1) in 7 Theorem 23 Let G L(H) be a generalized projection Then G is a closed range operator and G 3 is an orthogonal projection on R(G) Moreover, H has decomposition H R(G) N (G) and G has the following matrix representaton G1 0 R(G) R(G) G N (G) N (G) where restriction G 1 G R(G) is unitary on R(G) Proof If G is a generalized projection, then G 4 (G 2 ) 2 (G ) 2 (G 2 ) (G ) G From GG G G 4 G follows that G is a partial isometry implying that G 3 GG P R(G), G 3 G G P N (G) Thus, G 3 is an orthogonal projection onto R(G) N (G) R(G ) Consequently, R(G) is a closed subset in H as a range of an orthogonal projection on a Hilbert space From Lemma (12) in 4 we get the following decomposition of the space H R(G ) N (G) R(G) N (G) Now, G has the following matrix representation in accordance with this decomposition G1 0 R(G) R(G) G, N (G) N (G) where G 2 1 G 1, G 4 1 G 1 and G 1 G 1 G 1G 1 G 3 1 I R(G) Theorem 24 Let G, H GP(H) and H R(G) N (G) Then G and H has the following representation with respect to decomposition of the space G1 0 R(G) R(G) G, N (G) N (G) H1 H H 2 R(G) R(G), H 3 H 4 N (G) N (G) where H 1 H H 2 H 3, H 2 H 3 H 1 + H 4 H 3, H 3 H 1 H 2 + H 2 H 4, H 4 H 3 H 2 + H 2 4, 5

6 Furthermore, H 2 0 if and only if H 3 0 Proof Let H R(G) N (G) Then representation of G follows from Theorem (1) in 7 and let H has representaton H1 H H 2 H 3 H 4 Then, from H 2 H1 2 + H 2 H 3 H 1 H 2 + H 2 H 4 H 3 H 1 + H 4 H 3 H 3 H 2 + H4 2 H 1 H3 H2 H4 H, conclusion follows directly If H 2 0, then H 3 H 1 H 2 + H 2 H 4 0 and H 3 0 Analogously, H 3 0 implies H 2 0 Theorem 25 Let G L(H) be a hypergeneralized projection Then G is a closed range operator and H has decomposition H R(G) N (G) Also, G has the following matrix representaton with the respect to decomposition of the space G1 0 R(G) R(G) G, N (G) N (G) where restriction G 1 G R(G) satisfies G 3 1 I R(G) Proof If G is a hypergeneralized projecton, then G and G commute and G is EP Using Lemma (12) in 4, we get the following decomposition of the space H R(G) N (G) and G has the required representation 3 The Moore-Penrose and the Drazin inverse of two orthogonal projections We start this secton with theorem which gives matrix representation of the Moore-Penrose inverse of product, difference and sum of orthogonal projections Theorem 31 Let orthogonal projections P, Q L(H) be represented as in (1) and (2) Then the Moore-Penrose inverse of P Q, P Q and P + Q exists and the following holds (a) (P Q) AA 0 B A 0 L L R(P Q) R(A) and 6

7 (b) (P Q) A A BD B A DD L L and (c) (P + Q) R(P Q) R(A) R(D) 1 2 (I + A A ) BD D B 2D DD L L and R(P + Q) L R(D) Proof (a) Using representatons (1) and (3) for orthogonal projections P, Q L(H), the well known Harte-Mbekhta formula (P Q) (P Q) (P Q(P Q) ) and Theorem 21(a), we obtain (P Q) A 0 B 0 From P Q(P Q) P R(P Q), we obtain (P Q)(P Q) A 2 + BB 0 A B AA 0 B A 0 AA 0 B A 0 AA 0 or, in the other words, R(P Q) R(A) (b) Similarly to part (a), we can calculate the Moore-Penrose inverse of P Q as follows, (P Q) (P Q) ((P Q)(P Q) ) A B A 2 + BB AB + BD B D B A + DB B B + D 2 A B A 0 B D A A BD B A DD For the range of P Q we have P R(P Q) (P Q)(P Q) A A A + BD B ABD + BDD B A A + DD B B BD + DDD A A 0 D, 7

8 implying R(P Q) R(A) R(D) (c) The Moore-Penrose inverse of P + Q has the following representation with the respect to decomposition of the space (P + Q) X1 X 2 L L X 3 X 4 In order to calculate (P + Q), we will use Moore-Penrose equations From the first Moore-Penrose equation, (P + Q)(P + Q) (P + Q) P + Q, we have ((I + A)X 1 + BX 3 )(I + A) + ((I + A)X 2 + BX 4 )B I + A, ((I + A)X 1 + BX 3 )B + ((I + A)X 2 + BX 4 )D B, (B X 1 + DX 3 )(I + A) + (B X 2 + DX 4 )B B, (B X 1 + DX 3 )B + (B X 2 + DX 4 )D D The second Moore-Penrose equation, (P + Q) (P + Q)(P + Q) (P + Q), implies (X 1 (I + A) + X 2 B )X 1 + (X 1 B + X 2 D)X 3 X 1, (X 1 (I + A) + X 2 B )X 2 + (X 1 B + X 2 D)X 4 X 2, (X 3 (I + A) + X 4 B )X 1 + (X 3 B + X 4 D)X 3 X 3, (X 3 (I + A) + X 4 B )X 2 + (X 3 B + X 4 D)X 4 X 4, while the third and fourth Moore-Penrose equations, ((P + Q)(P + Q) ) (P + Q)(P + Q) and ((P + Q) (P + Q)) (P + Q) (P + Q), give X 3 X 2 Further calculations show that (I + A)X 1 + BX 2 I L, (I + A)X 2 + BX 4 0, B X 1 + DX 2 0, B X 2 + DX 4 DD According to Theorem 22 (b), (c), from B X 1 + DX 2 0 we get D B X 1 + X 2 0, or equivalently, X 2 D B X 1 From (I +A)X 1 +BX2 I L and Theorem 22 (i), we get (2I A A )X 1 I L ie X 1 (2I A A ) (I+A A ) Theorem 21 (c) and B X 2 +DX 4 DD imply B BD + DX 4 DD Finally, we have X 2 BD, X 3 D B, X 4 2D DD and (P + Q) 1 2 (I + A A ) BD D B 2D DD 8

9 Like in part (b), P R(P +Q) (P + Q)(P + Q) 1 2 (I + A)(I + A A ) BD B (I + A)BD + 2BD BDD 1 2 B (I + A A ) DD B B BD + 2DD DDD IL 0 D, which proves that R(P + Q) L R(D) To prve the existence of the Moore-Penrose inverse of P Q, P Q and P + Q, it is sufficient to prove that these operaors have closed range Since Q is an orthogonal projection, R(Q) is closed subset of H Also, A B R(Q) Q(H) B D L R(A) + R(B) R(B ) + R(D) R(A) + R(D), because items (a), (c) of Theorem 22 state that R(B) R(A) and R(B ) R(D) This implies that R(A) and R(D) are closed subsets of L and respectively If R(A) is closed, then for every sequence (x n ) L, x n x and Ax n y imply x L and Ax y Now, (I A)x n x y and x y L, (I A)x x y which proves that R(I A) is closed Consequently, R(P Q), R(I A) and R(I + A) are closed which completes the proof Similar to Theorem 31 in 6, we have the following result Theorem 32 Let orthogonal projections P, Q L(H) be represented as in (1) and (3) Then the Drazin inverses of P Q, P Q and P + Q exist, P Q and P + Q are EP operators and the following holds A (a) (P Q) D D (A D ) 2 B and ind(p Q) ind(a) + 1, (b) (P Q) D (P Q) and ind(p Q) 1, (c) (P + Q) D (P + Q) and ind(p + Q) 1 Proof (a) Theorem 31 proves that R(P Q) is closed subset of H Thus, the Drazin inverse for this operators exists According to representations (1) and (3) of projections P, Q, their product P Q and the Drazin inverse (P Q) D can be written in the following way P Q A B, (P Q) D X1 X 2 L X 3 X 4 L Equations that describe Drazin inverse are (P Q) D P Q(P Q) D X1 AX 1 X 1 AX 2 X 3 AX 1 X 3 AX 2 X1 X 2 X 3 X 4 (P Q) D, 9

10 (P Q) D X1 A X P Q 1 B AX1 AX 2 P Q(P Q) D, X 3 A X 3 B (P Q) n+1 (P Q) D A n+1 X 1 A n+1 X 2 A n A n 1 B (P Q) n Thus, from the first equation we have X 1 AX 1 X 1, X 1 AX 2 X 2, X 3 AX 1 X 3, X 3 AX 2 X 4, from the second equation X 1 A AX 1, AX 2 X 1 B, X 3 A 0, X 3 B 0, and the third equation implies A n+1 X 1 A n, A n+1 X 2 A n 1 B It is easy to conclude that X 1 A D, X 3 0, X 4 0 Equations X 1 AX 2 X 2 and AX 2 X 1 B give X1 2 B X 2 Finally, (P Q) D A D (A D ) 2 B To estimate the Drazin index of P Q, suppose that ind(a) n Then (P Q) n+2 (P Q) D A n+2 A n+1 B A D (A D ) 2 B A n+1 A n+1 A D B A n+1 A n B (P Q) n+1 implying that ind(p Q) ind(a) + 1 (b) Since (P Q)(P Q) (P Q) (P Q) and R(P Q) R(A) R(D) is closed, P Q is EP operator as normal operator with closed range and (P Q) (P Q) D Besides, (P Q) 2 (P Q) D (P Q)(P Q) (P Q) P Q and ind(p Q) 1 (c) Similarly to (b), P + Q is EP operator and (P + Q) (P + Q) D, ind(p + Q) 1 Theorem 33 Let orthogonal projections P, Q L(H) be represented as in (1) and (3) Then the following holds (a) If P Q QP or P QP P Q, then (P + Q) D IL 1 2 A 0, (P Q) D A 0 0 D 10

11 (b) If P QP P, then (P + Q) D 1 2 I L 0 (c) If P QP Q, then (P + Q) D IL 1 2 A 0, (P Q) D 0 D, (P Q) D A 0 (d) If P QP 0, then (P + Q) D IL 0 P + Q, (P Q) D Proof Let or (a) If (P + Q) D X1 X 2 X 3 X 4 P Q P QP A B A 0 L A 0 B 0 A B IL 0 0 D L QP P Q, P Q P Q then B B 0, I L + A is invertible and (I L + A) 1 I L 1 2A and according to Theorem 21 (c), D D 2 Thus, we can write Q A 0, P +Q IL + A 0, (P +Q) n (IL + A) n 0 Verifying the equation (P + Q) 2 (P + Q) D (IL + A) 2 X 1 (I L + A) 2 X 2 DX 3 DX 4 IL + A 0 P + Q we get X 2 X 3 0, DX 4 D The other two equations, (P + Q) D (P + Q)(P + Q) D (P + Q) D and (P + Q) D (P + Q) (P + Q)(P + Q) D, give X 4 DX 4 X 4, X 4 D DX 4 ie X 4 D Thus, (P + Q) D IL 1 2 A 0 11

12 Formula (P Q) D A 0 0 D follows form Theorem 32 (b) and the fact that A A 2 implies A D A 2 A (b) If P QP P, then A I L and Theorem 21 implies B B 0 Then, IL 0 Q and from part (a) of this Theorem we conclude 1 (P + Q) D 2 I L 0, (P Q) D 0 D (c) From P QP Q we get B B D 0 and A A 2 Now, I L + A is invertible and (P + Q) D (P + Q) 1 (IL + A) 1 0 IL 1 2 A 0 and (P Q) D A 0 (d) If P QP 0, then A 0 and since R(B) R(A), we conclude B B 0 In this case, IL 0 Q, P + Q implying (P + Q) D P + Q IL 0, (P Q) D IL 0 P Q 0 D Theorem 34 Let orthogonal projections P, Q L(H) be represented as in (1) and (3) Then (P Q) D (QP ) (P Q) (QP ) Moreover, if P Q QP, then P Q is EP operator and (P Q) D (P Q), ind(p Q) 1 Proof Corollary 52 in 8 states that (P Q) is idempotent for every orthogonal projections P and Q Thus we can write (P Q) I 0 K 0 12

13 and P Q (P Q) (I + K K) 1 (I + K K) 1 K Denote by A (I + K K) 1 and B (I + K K) 1 K AK Then A B P Q and according to Theorem 33 (a), (P Q) D A D (A D ) 2 B I + K K (I + K K) 2 (I + K K) 1 K I + K K (I + K K)K I + K K 0 I K I 0 I K K 0 (QP ) (P Q) (QP ), I K where we used ((P Q) ) ((P Q) ) (QP ) If P and Q commute, then P Q is normal operator with closed range which means it is EP operator (P Q) (P Q) D 4 The Moore-Penrose and the Drazin inverse of generalized and hypergeneralized projections Some of the results obtained in the previous section we can extend to generalized and hypergeneralized projections Theorem 41 Let G, H L(H) be two generalized or hypergeneralized projections Then the Moore-Penrose inverse of GH exists and has the following matrix representation (GH) (G1 H 1 ) D 1 0 (G 1 H 2 ) D 1 0 where D G 1 H 1 (G 1 H 1 ) + G 1 H 2 (G 1 H 2 ) > 0 is invertible Proof From Theorems 23, 24 and 25, we see that R(G) R(G 1 ) is closed and pair of generalized or hypergeneralized projections has matrix form G1 0 H1 H G, H 2 H 3 H 4, 13

14 Then GH G1 H 1 G 1 H 2 and analogously to the proof of Theorem 31 (a), we obtain mentioned matrix form Since R(GH) R(G 1 ) is closed, the Moore-Penrose (GH) exists Theorem 42 Let G, H L(H) be two generalized or hypergeneralized projections Then the Drazin inverse of GH exists and has the following matrix representation (GH) D (G1 H 1 ) D ((G 1 H 1 ) D ) 2 G 1 H 2 Proof Similarly to the proof of Theorem 32 (a) and using Theorem 25 Theorem 43 Let G, H L(H) be two generalized projections (a) If GH HG, then GH is EP and (GH) (GH) D (GH) (GH) 2 (GH) 1, (GH) (G1 H 1 ) 1 0 (b) If GH HG 0, then G + H is EP and (G + H) (G + H) D (GH) (G + H) 2 (G + H) 1, (G + H) G H 1 4 (c) If GH HG H, then G H is EP and (G H) (G H) D (GH) (G H) 2 (G H) 1, (G H) (G1 H 1 ) 1 0 Proof (a) If G, H L(H) are two commuting generalized projections, then from (GH) H G H 2 G 2 (HG) 2 (GH) 2 we conclude that GH is also generalized projection, and therefore EP operator Checking the Moore-Penrose equations for (GH) 2, we see that they hold From the uniqueness of the Moore-Penrose inverse follows (GH) 2 (GH) and (GH) (GH) D (GH) 2 14

15 From GH(GH) P R(GH), using matrix form of GH, we get G 1 H 1 (G 1 H 1 ) I, or equivalently, (G 1 H 1 ) (G 1 H 1 ) 1 Finally, (GH) (GH) D (GH) (GH) 2 (GH) 1 (b) If GH HG 0, then (G + H) 2 G 2 + H 2 G + H (G + H) and G + H is a generalized projection The rest of the proof is similar to part (a) (c) If GH HG H, then (G H) 2 G 2 H 2 G H (G H) and the rest of the proof is similar to part (a) Matrix representations are easily obtained by using canonical forms of G and H given in Theorem 24 Theorem 44 Let G, H L(H) be two hypergeneralized projections (a) If GH HG, then GH is EP and (GH) (GH) D (GH) 2 (GH) 1, (GH) (G1 H 1 ) 1 0 (b) If GH HG 0, then G + H is EP and (G + H) (G + H) D (G + H) 2 (G + H) 1, (G + H) G H4 1 (c) If GH HG H, then G H is EP and (G H) (G H) D (G H) 2 (G H) 1, (G H) (G1 H 1 ) 1 0 Proof (a) GH is EP operator and (GH) 4 GH, so it is a hypergeneralized projection Since (GH) 2 (GH), operator GH commutes with its Moore- Penrose inverse and (GH) (GH) D From GH(GH) I 0, follows (GH) (GH) 1 Thus, (GH) (GH) D (GH) 2 (GH) 1 (b) If GH HG 0, then (G + H) 2 (G + H) and H 1 H 2 H 3 0 implies G1 0 G + H, (G + H) G H 4 0 H 4 15

16 From (G + H)(G + H) P R(G+H) P R(G) + P R(H) and (G + H)(G + H) G1 G H 4 H 4 I 0 0 I we conclude that G 1 G 1 1, H 4 H 1 4 and (G + H) (G + H) 1 (c) Similarly to (b) References 1 O M Baskalary and G Trenkler, Column space equalities for orthogonal projectors, Applied Mathematics and Computations, 212 (2009) O M Baskalary and G Trenkler, Revisitaton of the product of two orthogonal projectors, Linear Algebra Appl 430 (2009), D S Djordjević and V Rakočević, Lectures on generalized inverses, Faculty of Sciences and Mathematics, University of Niš, D S Djordjević and J Koliha, Characterizing hermitian, normal and EP operators, Filomat 211 (2007) C Deng, The Drazin inverses of sum and difference of idempotents, Linear Algebra Appl 430 (2009) C Deng and Y Wei, Characterizations and representations of the Drazin inverse involving idempotents, Linear Algebra Appl 431 (2009) J Groβ and G Trenkler, Generalized and hypergeneralized projectors, Linear Algebra and its Applications 264 (1997) J J Koliha and V Rakocević, Range projections and the Moore- Penrose inverse in rings with involution Address Faculty of Sciences and Mathematics, University of Niš, PO Box 224, Niš, Serbia acircleperfect@gmailcom dragan@pmfniacrs 16