See

About the group inverse and Moore-Penrose inverse of a product

 Gouveia, M.C., Puystjens, R., 1991, Linear Algebra and Its Applications

150 (C), pp. 361-369

There, the authors address to the case

(PAQ)^\dag and (PAQ)^#

in the case P,Q are invertible (well, a little bit more general, in fact), assuming A^\dag or A^\# exist.

Moore Penrose inverse satisfies 4 cfonditions of generalized inverse which is not group inverse

https://books.google.com.tr/books?id=o3-97W8vCdIC&redir_esc=y

please check this book

 Note: All generalized inverses of a product of AB such that one of the factors is invertible, ( for example, here  A  is invertible ) are of the form: B- A-1 where B- is a generalized inverse of B.

Hince:

1) (AB)+ = B+A-1, where B+ is the Moore-Penrose inverse of B under conditions ( because still the 3 and 4 conditions of Moore Penrose, i.e. orthogonality of projectors)

2) If rank C= rank C2 , then the same for B, because, as A is invertible, then  rankC=rank B. Consequently,  the group inverse B# of B exists and also C#  and C#=B#A-1 

For the general case, where A and B are not invertible, the generalized inverses of the product doesn't obey to the reverse order law (AB)- =B- A- as in  the normal case, only under conditions. 

Also there is an expression for the (AB)- constitute of terms of  A- and B- and projectors related to A and B  and their generalized inverses.

You can see my article; Some algebraic properties of the generalized inverses of the product of two matrices. Also you can find many knowledge on my thesis:  SUR LES PROPRIETES 

ALGEBRIQUES DES Gk- INVERSES DES MATRICES

Sorry, B- A-1 not A-1 B-. it obey to the reverse order. I corrected my answer above.

Dear  Prof. Dr.Hanifa Zekraoui,

Thank you so much for your kindly answers and advice. 

I will check the last formula (B- A-1) again, the first formula (A-1B-) is incorrect. 

Thank you Sir for your valuable Ph.D. thesis.

Regards,

Wiwat Wanicharpichat

I was checking the orthogonality of the projectors (3 and 4 conditions of Moore Penrose ),  there are cases where it is correct: (AB)+ =B+ A-1:

A-1 =A* or B has full row rank.  Or we can directly check Greville's conditions for the reverse order.Then when we want to calculate the Moore -Penrose inverse of a square matrix using Jordan form, it is incorrect to say that ( P-1 BP)+ =P-1 B+ P, only if  P-1=P*. That why, it is preferable to use SVD or the expression B+=B*(BB*)+ .  (BB* )+ is easy to calculate, because it is symmetric. 

(AB)† = B†A† if and only if both A†ABB* and A*ABB† are Hermitian.

(AB)† = B†A†  if and only if  A†ABB*A*ABB† = BB*A*A.

... .

There are a lot of equivalent conditions for  (AB)†= B†A†,   See for example:

T. N. E. Greville, Note on the Generalized Inverse of a Matrix Product,  SIAM Review, Vol. 8, No. 4 (Oct., 1966), pp. 518-521.

In fact, if  A is an invertible matrix then  A-1 = A† = A#.  We have  

(AB)†= B†A-1, under some conditions.

In general    (AB)†  ≠  B† A⁻¹.

 Proof:  To show that  (AB)† = B†A⁻¹.  It suffices to show that  B†A⁻¹  satisfies the four Penrose conditions. 

Clearly,

(1) AB(B†A⁻¹)AB  =  ABB†B  =  AB;

(2) B†A⁻¹(AB)B†A⁻¹  =  B†BB†A⁻¹  =  B†A⁻¹;

(3) [AB(B†A⁻¹)]T = (B†A⁻¹)T(AB)T  =  (A⁻¹)T(B†)TBTAT  = (A⁻¹)T(BB†)TAT  =(AT)⁻¹BB†AT  ≠  AB(B†A⁻¹);

(4) [(B†A⁻¹)AB]T  =  (AB)T(B†A⁻¹)T  =  BTAT(A⁻¹)T(B†)T  =  BT(B†)T  ≠ (B†A⁻¹)AB.

We see that the conditions (3) and (4) of Penrose conditions does not holds.

See

About the group inverse and Moore-Penrose inverse of a product

 Gouveia, M.C., Puystjens, R., 1991, Linear Algebra and Its Applications

150 (C), pp. 361-369

There, the authors address to the case

(PAQ)^\dag and (PAQ)^#

in the case P,Q are invertible (well, a little bit more general, in fact), assuming A^\dag or A^\# exist.

Dear Prof. Dr. Pedro Patriclo,

The conditions in that paper is  P’PA = A = AQQ’, that is P-1 = PT.  It is a strong conditions.

Regards,

Wiwat Wanicharpichat

Not quite; just that there exist P', Q' s.t. P'PA=A=AQQ'. P,Q and may not even be invertible!

If P,Q are non-singular, they are trivially satisfied.

The conditions P'PA=A=AQQ' is means  PTPA = A = AQQT, that is, P is left inner and Q is right inner, does not mean  P-1PA = A = AQQ-1.

In my last answer above, there was a misprint of -1, I corrected it: A-1 =A*, P-1=P*.

Conclusion of the answers:

1) As Prof. Wiwat excluded the case A-1 =At,   i.e. (A*), then (AB)+=B+A-1 only if B is of full row rank which is equivalent to Greville's conditions. Otherwise,  You calculate (AB)+ by using SVD or the expression cited  in my answer above.

2) For the group inverse there is no problem, under the condition of existence:

rank C=rank C2,  we have (AB)#=B# A-1 

Dear Pedro, for the group inverse, yes. just  P has to be  left invertible and Q right invertible, and of course rank A= rank A2 (the condition of existence). But for the Moore-Penrose, we need orthogonality or which is equivalent for the 3rd and the 4 th conditions. So if we take P just left invertible, that means it has a full column rank, we write PAQ=P(AQ) , then, we need  (AQ) to has full row rank to have (PAQ)+ =Q'A+ P', where P' and Q' are left and right inverses of P and Q respectively.

Excuse me, I just checked, for the group inverse, No. because there is no commutativity, i.e. the 3rd equation in the definition of the group inverse.

I only handle the Moore-Penrose Inverse.

The problem can be easily  reduced to the computation of the Inverse of  the singular Matrix B. The inverse of B is a Moore-Penrose Inverse, which is either a left or a right Inverse. The problem is that we cannot apply straightforward  the B+=(BTB)-1BT rule to compute the inverse(just for left) of B. In fact, the column/row-rank of B is not full and consequently  the above rule cannot be applied. What we should do is:

1. Compute the column-rank of B=r

2. Factorize B as product of two matrices C(n,r) and D(r,n)(See Gram-Schmidt Algorithm)

3. Apply the above Inverse rule to compute  A+, C+, D+

4.Apply the Product Inverse Rule  to compute (ACD)+ given  A+, C+, D+.

NB: The biggest problem is the fact that the column/row-rank of B is not full.

@Wiwat: Where, in the paper, can you conclude

"The conditions P'PA=A=AQQ' is means P^T PA = A = AQ Q^T," ?

I am completely lost with your reasoning!

@Pedro: The conditions P'PA=A=AQQ' is means P is left inner (it has left inverse) and Q is right inner (it has right inverse). I am sorry about that. In general  P'PA=A=AQQ' is not means P^T PA = A = AQQ^T, but if P, Q are orthogonal then the post is correct.

[1 0; 0 0]=P'=P=A (matlab notation)

P'PA=A and yet P is not left invertible.

@Pedro: Thank you.