aa r X i v : . [ m a t h . F A ] M a y When is ( A + B ) † = A † + B † ? K.C. Sivakumar
Department of MathematicsIndian Institute of Technology MadrasChennai 600 036, India.
Abstract
We address the question as to when it is true that ( A + B ) † = A † + B † , where † denotes the Moore-Penrose inverse. A similar questionis addressed for the group inverse. AMS Subject Classification (2010):
Keywords:
Moore-Penrose inverse, group inverse, inverse of the sum ofmatrices. 1
Introduction
The motivation for this short note is the work of [2], where the authors seek tosolve the equation a + b = a + b not only for reals or complex numbers, but alsofor matrices with real entries; in the first case there is no solution, while in thesecond and third cases, it is shown that there are infinitely many solutions.Specifically, they show that ( A + B ) − = A − + B − holds for real matrices oforder n , if and only if n is even and describe a method of constructing suchmatrices. Recall that a real vector space V is said to have a complex structureif there is a linear operator J on V such that J = − I . It is easy to observe thata finite dimensional real vector space admists a complex structure if and onlyif its dimension is even. In [3], it is shown that the identity above holds in afinite dimensional vector space V if and only if V admits a complex a structure,thereby obtaining the same conclusion as in [2], as a consequence. Here, weconsider a more general question of asking when the identity above extendsto generalized inverses. More precisely, we present sufficient conditions on(possibly) rectangular matrices A and B with complex entries such that theequation ( A + B ) † = A † + B † holds, where † stands for the Moore-Penroseinverse (see Remark 4.5). We also consider the case of the group inverse. The symbol C m × n denotes the set of all complex matrices of order m × n . For A ∈ C m × n , we use R ( A ) to denote its range space and N ( A ) to denote itsnull space. For any matrix X with complex entries, X ∗ denotes the conjugatetranspose. Let us recall that for A ∈ C m × n , the Moore-Penrose (generalized)inverse (or the pseudo inverse) of A , denoted by A † is the unique matrix X ∈ C n × m that satisfies the equations AXA = A, XAX = X, ( AX ) ∗ = AX and ( XA ) ∗ = XA . One of the many ways of showing the existence of theMoore-Penrose inverse is by using the full-rank factorization. A matrix A ∈ C m × n is said to have a full-rank factorization if there exist F ∈ C m × r and G ∈ C r × n such that rank ( F ) = rank ( G ) = rank ( A ) = r and A = F G . Itthen follows that A † = G ∗ ( GG ∗ ) − ( F ∗ F ) − F ∗ . In fact, in this case, one has F † = ( F ∗ F ) − F ∗ and G † = G ∗ ( GG ∗ ) − , so that F † is a left inverse of F , while G † is a right inverse of G . More generally, for any matrix X , one has theformulae: X † = ( X ∗ X ) † X ∗ = X ∗ ( XX ∗ ) † . The following properties will befrequently used: R ( A ) = R ( AA † ) and R ( A ∗ ) = A † A = R ( A † ). In particular,it follows that if x ∈ R ( A ), then AA † x = x , which may be extended to theidea that if R ( B ) ⊆ R ( A ) then AA † B = B . Similarly, A † A acts like identity2n R ( A ∗ ).For the reader who is encountering the Moore-Penrose inverse for the firsttime, here is a motivation: For the linear system Ax = b , given A ∈ C n × n and b ∈ C n with A nonsingular, one has x = A − b as the unique solution. Now,consider the system Ax = b , given A ∈ C m × n and b ∈ C m . Set x = A † b . If thesystem has a unique solution, then x is that unique solution; if it has infinitelymany solutions, then x is the solution that has the additional property thatit has the least (euclidean) norm, among all the solutions; if the system doesnot have a solution, but has a unique least squares solution, then x is thatsolution, and finally, if the system is not consistent and has infinitely manyleast squares solution, then x is the unique least squares solution with theleast norm.Another generalized inverse, this time for square matrices, is recalled next.Let A ∈ C n × n . If there exists X ∈ C n × n such that AXA = A, XAX = X and AX = XA , then such an X must be unique and is referred to as the groupinverse of A . It is denoted by A . A necessary and sufficient condition forthe existence of the group inverse is the condition that rank ( A ) = rank ( A ),which is of course, the same as R ( A ) = R ( A ), which in turn, is equivalent tothe condition N ( A ) = N ( A ). The nomenclature for the group inverse comesfrom the fact that the set consisting of A and its positive powers, A and itspositive powers, forms a group under matrix multiplication, where AA is theidentity element and A is the inverse of A . It is useful to note that the groupinverse of A , if it exists, is a polynomial in A . Once again, a formula for thegroup inverse may be given in terms of a full-rank factorization: if A = F G isa full-rank factorization, then A exists if and only if GF is invertibe. In thatcase, one also has A = F ( GF ) − G. Analogous to the Moore-Penrose inverse,one has: R ( A ) = R ( AA ) = R ( A ). If R ( B ) ⊆ R ( A ) then AA B = B .It may be emphasized that while the Moore-Penrose inverse exists for allmatrices, the group inverse of a given matrix need not exist. For instance, nonilpotent matrix possesses the group inverse. Of course, if A is square andnonsingular, then one has A † = A = A − .The following result will also be useful. Theorem 2.1. [1] For C ∈ C n × n , let X be a matrix satisfying Xp = 0 ⇐⇒ Cp = 0 and Xp = q ⇐⇒ Cq = p f or all p, q ∈ R ( C ) . Then X = C . First, we collect some prelminary properties.
Theorem 3.1.
Let
A, B ∈ C n × n . Suppose that AB ∗ + BB ∗ = 0 and B ∗ A + B ∗ B = 0 . We then have: ( a ) AB † + BB † = 0 and B † A + B † B = 0 . ( b ) N ( A ) ⊆ N ( B ) and N ( A ∗ ) ⊆ N ( B ∗ ) . ( c ) BA † + BB † = 0 and A † B + B † B = 0 . ( d ) BA † A = AA † B = B and BA † B = − B. ( e ) A † BB † = − B, B † AA † = B † and A † BA † + B † BA † = 0 . ( f ) BA † and A † B are hermitian.Proof. ( a ) We have 0 = AB ∗ + BB ∗ = ( A + B ) B ∗ and so, AB † + BB † = ( A + B ) B † = ( A + B ) B ∗ ( BB ∗ ) † = 0 . Also, 0 = B ∗ A + B ∗ B = B ∗ ( A + B ), which yields B † ( A + B ) = ( B ∗ B ) † B ∗ ( A + B ) = 0 . ( b ) From the equality B ∗ A + B ∗ B = 0 , it follows that if x ∈ N ( A ) , then B ∗ Bx = 0, which implies that Bx = 0, due to the well known condition N ( B ∗ B ) = N ( B ). Thus N ( A ) ⊆ N ( B ). Taking conjugate transposes of AB ∗ + BB ∗ = 0 , we get BA ∗ + BB ∗ = 0 . Now, if x ∈ N ( A ∗ ), then BB ∗ x = 0,which yields B ∗ x = 0, showing that N ( A ∗ ) ⊆ N ( B ∗ ).( c ) Taking the transposes of AB ∗ + BB ∗ = 0, one obtains 0 = BA ∗ + BB ∗ = B ( A ∗ + B ∗ ). Arguing as earlier, we have 0 = B ( A † + B † ) , yielding the firstidentity. The second identity follows similarly.( d ) One has ( A † A ) ∗ B ∗ = A † AB ∗ = B ∗ , where the last equality is due to thefact that R ( B ∗ ) ⊆ R ( A ∗ ) (which in turn, is due to N ( A ) ⊆ N ( B )). Upontaking transposes, one obtains B = BA † A . Next, since R ( B ) ⊆ R ( A ), one has4 A † B = B . From the second identity of ( c ), upon premultiplying by B , onehas BB † B + BA † B = 0, i.e BA † B = − B .( e ) Using the second identity of ( c ), we have A † B = − B † B . Thus, one has A † BB † = − B † BB † = − B † , proving the first part. Also, AA † ( B † ) ∗ = ( B † ) ∗ ,since R (( B † ) ∗ ) = R ( B ) (which is contained in R ( A )). Upon taking transposes,we get B † AA † = B † . Upon post multiplying the second identity of ( c ) by A † ,we obtain the third part.( f ) By ( c ), we have BA † = − BB † , proving that BA † is hermitian. The secondpart is similar. Theorem 3.2.
Let
A, B ∈ C n × n be related in such a way that AB ∗ + BB ∗ = 0 and B ∗ A + B ∗ B = 0 . We then have: ( A + B ) † = A † + B † . Proof.
Set X = A † + B † . We show that X = ( A + B ) † by verifying the fourequations for the Moore-Penrose inverse. We have( A + B ) X = AA † + AB † + BA † + BB † = AA † + BA † , by using the fact that AB † + BB † = 0. By ( f ) of Theorem 3.1, BA † ishermitian and so ( A + B ) X is hermitian. Also, X ( A + B ) X = ( A † + B † )( AA † + BA † )= A † AA † + A † BA † + B † AA † + B † BA † = A † + B † AA † = A † + B † , where we have made use of the second and third parts of ( e ) of Theorem 3.1.Further, ( A + B ) X ( A + B ) = ( AA † + BA † )( A + B )= AA † A + AA † B + BA † A + BA † B = A + BA † A = A + B, where we have used all the formulae in ( d ) of Theorem 3.1. Finally, one has X ( A + B ) = A † A + A † B + B † A + B † B = A † A + A † B, where the second part of ( a ) of Theorem 3.1 was used. Again, by ( f ) ofTheorem 3.1, since A † B is hermitian, it follows that X ( A + B ) is hermitian,completing the proof. 5 Group inverse analogue
First, we collect some prelminary properties.
Theorem 4.1.
Let
A, B ∈ C n × n . Suppose that B exists and that one has thefollowing relationships between A and B : AB + BB = 0 and B A + B B = 0 . We then have: ( a ) N ( A ) ⊆ N ( B ) and N ( A ∗ ) ⊆ N ( B ∗ ) . ( b ) AB = BA = − B . ( c ) R ( A + B ) ⊆ N ( B ) and R ( A ∗ + B ∗ ) ⊆ N ( B ∗ ) . ( d ) R ( B ∗ ) ⊆ N ( A ∗ + B ∗ ) and R ( B ) ⊆ N ( A + B ) . ( e ) ( A + B ) , A and ( AB ) exist.Proof. ( a ) Let Ax = 0. Then from the second condition, one has0 = B Ax + B Bx = B Bx, which upon premultiplying by B , gives Bx = 0. Hence N ( A ) ⊆ N ( B ). Sim-ilarly, by taking the conjugate transposes of the first condition and premulti-plying by B ∗ , one has the implication A ∗ x = 0 = ⇒ B ∗ x = 0, showing that thesecond inclusion holds.( b ) We have AB = AB B = − BB B = − B , using the first identity. Employing the second identity, one has BA = B B A = − B B B = − B . ( c ) The second identity is the same as B ( A + B ) = 0 and this shows that R ( A + B ) ⊆ N ( B ) = N ( B ). The other inclusion is similarly proved, upontaking the conjugate transposes of the first identity and using the fact that( B ) ∗ = ( B ∗ ) .( d ) Consequence of ( c ).( e ) First, observe that ( A + B ) = A + B + AB + BA = A − B . We showthat N (( A + B ) ) = N ( A + B ). Let ( A + B ) x = 0 so that ( A − B ) x = 0.Upon premultiplying by B , one then obtains0 = B A x − B B x = − B BAx − Bx = − Ax − Bx, i.e. x ∈ N ( A + B ).Let A x = 0. Then Ax ∈ N ( A ) ⊆ N ( B ) and so BAx = 0 so that B x = 06which in turn implies that Bx = 0). Thus ( A + B ) x = ( A − B ) x = 0 and so( A + B ) x = 0. Thus, Ax = 0, proving that N ( A ) ⊆ N ( A ), so that A exists.By ( c ), R ( AB ) = R ( B ) = R ( B ) and N ( AB ) = N ( B ) = N ( B ). Since B exists, the subspaces R ( B ) and N ( B ) are complementary and so are R ( AB )and N ( AB ), proving the existence of the group inverse of AB . Remark 4.2.
Note that since AB = BA and since the group inverse of amatrix is a polynomial in that matrix, the mutual commutativity relationshipsbetween A, B, A and B are applicable (note that the two conditions of theresult above already imply AB = B A ). This fact will be used frequently inour proofs. Applying Theorem 4.1 and Theorem 2.1, we prove the next result.
Theorem 4.3.
Let
A, B satisfy the conditions of Theorem 4.1. Then A BA = B . Proof.
Set X = A BA . We must show that X = B .First, let Bp = 0. Then Xp = A BA p = A A Bp = 0. Conversely,let Xp = 0 so that A BA p = 0. Then BA p ∈ N ( A ) ⊆ N ( B ), and BA p ∈ R ( B ). Since B exists, this means that BA p = 0. So, A Bp = 0so that Bp ∈ N ( A ) ⊆ N ( B ) as well as Bp ∈ R ( B ). So, Bp = 0. We haveshown that Xp = 0 ⇐⇒ Bp = 0.Next, let Bq = p, given p, q ∈ R ( B ). Then Xp = A BA p = A A Bp.
Now, p ∈ R ( B ) ⊆ N ( A + B ) and so Bp = − Ap . Thus Xp = − A A Ap = − A p = − A Bq.
Again, q ∈ R ( B ) ⊆ N ( A + B ) and so Bq = − Aq . Thus, Xp = A Aq = q ,since q ∈ R ( B ) ⊆ R ( A ).Finally, let Xp = q so that q = A BA p = A A Bp.
Now, A ( A ) = AA and since R ( B ) ⊆ R ( A ), upon premultiplying theequation above by A , we then have A q = Bp . Then BA q = B p andso ABAq = − ABp . This means that p + Aq ∈ N ( AB ) = N ( B ) and so, Bp + BAq = 0. Premultiplying by B and using the fact that p, q ∈ R ( B ) as7ell as B BA = AB B , one obtains p + Aq = 0. Premultiplying by B oneobtains B Aq = − B p , and so B Bq = B p. Premultiplying by B and usingthe fact that p ∈ R ( B ), we get Bq = p . Thus, one has Xp = q ⇐⇒ Bq = p f or all p, q ∈ R ( B ) . By Theorem 2.1, the conclusion follows.We are now in a position to prove the group inverse analogue of Theorem3.2.
Theorem 4.4.
Let
A, B ∈ C n × n be related in such a way that AB + BB = 0 and B A + B B = 0 . We then have: ( A + B ) = A + B . Proof.
Consider( A + B )( A + B ) = AA + BB + AB + BA = AA + BA , where we have used the fact that BB + AB = 0 . This means that one has( A + B )( A + B )( A + B ) = A + AA B + BA A + BA B. Now, premultiplying (( b ) of Theorem 4.1 viz.,) AB = − B , by B one obtains B = B B = − B AB . Also, BA A = A AB = B and so the sum of thelast two terms in the expression above equals zero. We have shown that( A + B )( A + B )( A + B ) = A + B. Next, from the first expression as above, one has( A + B )( A + B )( A + B ) = ( A + B )( AA + BA )= A AA + A BA + B AA + B BA = A + A BA , where we have made use of the fact that B AA + B BA = 0, since B A + B B = 0 . By Theorem 4.3, A BA = B and so the expression abovesimplifies to A + B . Finally,( A + B )( A + B ) = A A + A B + B A + B B = A A + A B, where we have made use of the identity B A + B B = 0 . As was alreadymentioned, since B and A commute, it also follows that( A + B )( A + B ) = ( A + B )( A + B ) . This completes the proof. 8 emark 4.5.
Where do the sufficient conditions for Theorem 3.2 and Theorem4.4 come from? Recall that for
A, B ∈ C m × n , one says that A ≤ ∗ B (whichis referred to as the “star partial order”) if AA ∗ = BA ∗ and A ∗ A = A ∗ B .Analogously, the notation A ≤ B (which is referred to as the “sharp partialorder”) signifies the fact that AA = BA and A A = A B (assuming thatthe group inverse A exists). Pioneering contributions were made on matrixpartial orders by Mitra [5]. There the author shows that if A ≤ ∗ B , then onehas the identity ( B − A ) † = B † − A † , while ( B − A ) = B − A holds if A ≤ B . It is now clear that the conditions of Theorem 3.2 are equivalent to therequirement that − B ≤ ∗ A (which therefore implies the identity in the title ofthis note), whereas the condition − B ≤ A holds if and only if the hypothesis ofTheorem 4.4 hold (which in turn, leads to the group inverse identity). However,the objective of this note is to divest the problem at hand from the notion ofmatrix partial orders, and also to present an independent and a self-containedtreatment. It would be interesting to derive some characterizations for the twoidentities, studied in this note, to hold. In what follows, we present a class of matrices that satisfy the conditionsof Theorem 3.2 and Theorem 4.4. First, we consider Theorem 3.2.
Example 4.6.
We give a recursive procedure to compute matrices that satisfythe identity AB ∗ + BB ∗ = 0 and B ∗ A + B ∗ B = 0 . Let a, b ∈ C be such that ab ∗ + bb ∗ = 0 . Let A, B ∈ C × be defined by A = (cid:18) a α α α (cid:19) and B = (cid:18) b β β β (cid:19) , where α i , β i , i = 1 , , are to be determined. One may verify that AB ∗ + BB ∗ =0 translates into the following equations: ( α + β ) β ∗ = 0( a + b ) β ∗ + ( α + β ) β ∗ = 0( α + β ) b ∗ + ( α + β ) β ∗ = 0( α + β ) β ∗ + ( α + β ) β ∗ = 0 . Many choices are available and an easy option leads to the pair of matrices A = (cid:18) a b − b α (cid:19) and B = (cid:18) b − bb − α (cid:19) , where α is arbitrarily chosen. Then A + B = (cid:18) a + b
00 0 (cid:19) . In this case, onemay verify that both AB ∗ + BB ∗ = 0 and B ∗ A + B ∗ B = 0 . aving constructed the basis step, one may now proceed to construct matriceswith one extra row and column, satisfying the required identities, given a pairof matrices of lower order. More specifically, let A, B ∈ C m × n be such that AB ∗ + BB ∗ = 0 and B ∗ A + B ∗ B = 0 . Let u ∈ C m be chosen such that ( A + B ) ∗ u = 0 and v ∈ C n be selected so that ( A + B ) v = 0 . Set M = (cid:18) A u v ∗ α (cid:19) and N = (cid:18) B − u − v ∗ − α (cid:19) . Then
M, N ∈ C ( m +1) × ( n +1) and one has M + N = (cid:18) A + B
00 0 (cid:19) . One mayverify that
M N ∗ + N N ∗ = 0 and N ∗ M + N ∗ N = 0 . There are more generalchoices for the matrices
M, N and we have given just one easy method ofdetermining them.Here is a numerical example: Let a = 1 and b = 0 so that A = (cid:18) (cid:19) and B = (cid:18) − (cid:19) . Then A † = A − = I and B † = B so that A † + B † = (cid:18) (cid:19) . Also A + B = (cid:18) (cid:19) and so ( A + B ) † = A + B = A † + B † .Now, define u = (0 , T and v = (0 , T so that ( A + B ) u = 0 = ( A + B ) ∗ v =0 . Define M = α and N = − − − − α . Then one may verify that M † + N † = ( M + N ) † . Next, we construct matrices that satisfy the conditions of Theorem 4.4.
Example 4.7.
Let a, b ∈ C be such that ab + bb = 0 . Here x = x , if x = 0 and x = 0 , if x = 0 . Let A, B ∈ C × be defined by A = (cid:18) a α α α (cid:19) and B = (cid:18) b β β β (cid:19) , where α i , β i , i = 1 , , are to be determined. In orderfor AB + BB = 0 to be satisfied, the said group inverse must exist.Let us start with the case when b = 0 , so that a + b = 0 (and so a = 0 ). Since B is singular, one has β = β β b . We have the following full rank factorizationfor B : B = F G, where F = (cid:18) β (cid:19) and G = (cid:0) b a β (cid:1) . lso, since B exists, one must have GF = 0 and so b + β = 0 . Thus, B = 1( b + β ) (cid:18) b β β β (cid:19) . For the requirement AB + BB = 0 to be satisfied, one has: ( α + β ) β = 0 β β b ( α + β ) = 0( α + β ) b + ( α + β ) β = 0( α + β ) β + β β b ( α + β ) = 0 . Here is a choice that leads to a nonzero A + B : α = β = 0 so that β = 0 and choose α = − b − α . Then B = (cid:18) b b (cid:19) , A = (cid:18) a α − b − α (cid:19) and A + B = (cid:18) a + b α + b − α − b (cid:19) . One may verify that B A + B B = 0 also holds.Next, let us consider the possibility when b = 0 , while a is arbitrary andnonzero. One has B = (cid:18) β β β (cid:19) . The possibility that β = β = 0 and β = 0 is ruled out, since B would be nilpotent and so the group inverse does not exist.One needs to take into account two cases. Case ( i ) : β = 0 . Then β = 0 (otherwise B would be nonsingular). Thus, B = F G, where F = (cid:18) β (cid:19) and G = (cid:0) β (cid:1) . One must have GF = β = 0 and so B = 1 β (cid:18) β β (cid:19) . For the requirement AB + BB = 0 to be satisfied, the following must hold: α β = 0 α β = 0( α + β ) β = 0( α + β ) β = 0 . Now, since β = 0 , one has α = 0 and α + β = 0 as well. Thus, one obtainsthe trivial situation, since A + B = 0 . Case ( ii ) : β = 0 . Then β = 0 . Arguing as above, one has B = 1 β (cid:18) β β (cid:19) . Again, for AB + BB = 0 to hold, one must have the following: β + ( α + β ) β = 0 α β + ( α + β ) β = 0 . Imposing the condition B ( A + B ) = 0 , in addition, one obtains: α β = α β = 0( α + β ) β = ( α + β ) β = 0 . So, α = 0 and α + β = 0 . Assuming that β = − a , we may choose β = − α β a + β . By taking α = − β , one may verify that all the six equationsabove hold. Thus, one has A = (cid:18) a − β − β (cid:19) , B = (cid:18) β β (cid:19) so that one has anontrivial expression A + B = (cid:18) a + b
00 0 (cid:19) . Unlike the case of the Moore-Penrose inverse, there does not appear to bea recursive process to construct matrices satisfying the group inverse identity.Let us present a numerical illustration of the procedure above. Let a = 1 and b = 0 . Define A = (cid:18) − − (cid:19) , B = (cid:18) (cid:19) so that one has A + B = (cid:18) (cid:19) . Then ( A + B ) = A + B . Also, A + B = A − + B = (cid:18) − − (cid:19) + (cid:18) (cid:19) = (cid:18) (cid:19) . We point to some directions for further study. We have considered the questionof when a generalized inverse of a sum of two matrices equals the sum oftheir generalized inverses. As mentioned earlier, the problem of determiningnecessary and sufficient conditions for the two identities to hold, remains open.Next, one might be interested in asking the same question for sums involvingthree or more matrices. Apparently, an answer to that can turn out to bequite complex, considering the effort involved for the case of two matrices, aspresented here. The second quest may be towards proving similar formulaefor other clasical, as well new classes of generalized inverses, like the Drazininverse, the core inverse or the Drazin-Moore-Penrose inverse. A third directionis to address the problem of determining when the studied identities hold, forelements in a ring. It is noteworthy that all the proofs presented here are linearalgebraic and are free of multilinear notions like the rank or the determinant.Hence, with a little modification, they may be extended to the case of infinitedimensional spaces. 12 eferences [1] A. Ben-Israel and T.N.E. Greville,
Generalized Inverses: Theory and Ap-plications , , Springer-Verlag, New York, 2003.[2] E. Boman and F. Uhlig, When is a + b = a + b anyway? , College Math. J., (2002) 296-300.[3] J.P. D’Angelo, When is the sum of inverses the inverse of the sum? ,https://faculty.math.illinois.edu/ jpda/jpd-gardner-2012-web.pdf[4] C.D. Meyer, Jr.,
The role of the group generalized inverse in the theory offinite Markov chains , SIAM Rev. (1975) 443-464.[5] S.K. Mitra, On group inverses and the sharp order , Linear Algebra Appl.,92