usually, full rank means rank(A)=min{# cols, # rows}; so, it is possible to consider nonsquare full rank matrices. this allows to address full rank factorization without mentioning which kind of full rank each factor has (full row rank or full column rank).

WHICH generalized inverse are you considering? Moore-Penrose? Drazin? von Neumann? There are hundreds!

which partial order are you considering in ", A_i \neq A_j ,"? Lowner? Frobenius? Rank partial order?

@Peter You are correct, I correct myself and call them full row rank matrices In the case of having: A_rc, Rank(A)=min(r,c). 

I am quoting Wikipedia here: " The pseudoinverse of any matrix A exists and is unique". This is not a definitive reference, but you can definitely dig more. This statement implies that the pseudoinverse operator applied twice successively generates a unique matrix. In other words, equal pseudoinverses implies equal original matrices.

https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse#Existence_and_uniqueness

Peter, you said:"But also nobody says that the pseudoinverse of the pseudoinverse is the same matrix as you started with".  This is consistent with what I mentioned in my previous response, whereby I missed to be explicit about the following. Bullet number 4 of the basic properties of the pseudoinverse states that the pseudoinverse of the pseudoinverse is the identity map.  As such, the equivalence classes of "pseudo-equivalent" matrices are singletons. 

https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse#Basic_properties

Peter, your point is well taken, the main motivation of this though was "pseudo-inverse of a matrix is not unique" and in my case, it only has to satisfy first Penrose condition of BgcrBrcBgcr=Irr ,r

usually, full rank means rank(A)=min{# cols, # rows}; so, it is possible to consider nonsquare full rank matrices. this allows to address full rank factorization without mentioning which kind of full rank each factor has (full row rank or full column rank).

WHICH generalized inverse are you considering? Moore-Penrose? Drazin? von Neumann? There are hundreds!

which partial order are you considering in ", A_i \neq A_j ,"? Lowner? Frobenius? Rank partial order?

Dear Reza, I understand your question, it is under the subject of partial orders and generalized inverses. For example, we have minus order, star order, which occur in general linear Models. You want the mathematical terminology of the type of partial order you mentioned above.  Ask google using: partial orders types and generalized inverses, you will get many articles,  look for your type.

For example, in this article, the author used the minus partial order to define a new one called based partial order.

http://www.sciencedirect.com/science/article/pii/002437959190096F

These ones, called left and right partial orders defined by the prjoectors A- A and AA-.

http://www.sciencedirect.com/science/article/pii/002437959290206P

Conclusion: there are many "new" partial orders defined by the known "famous" partial orders related to generalized inverses. some of them related to the rank and others related to the projectors or to the Moore-Penrose inverse, Drazin inverse or group inverse, because these concepts are unique and so allow to define new concepts. 

Note: for the full rank it is as what said Pedro, so I will not repeat.

Thank you dear Prof. Wiwat for up-voting my answer. I noticed that the researcher has changed the question. So, I want to say for those who read my answer, it was about the first version of the question, not this new one. Also, I wish from dear researchers on RG, do not change the question, because the answers will be incompatible with the new version of the question. I propose to post the new version as a new question . Please :) 

Prof. Hanifa. I am agreed with your answer.

Let A†  be the Moore-Penrose inverse of A, and A#  the group inverse of A, we have the following facts:

1. (A†)† = A.

2. (A#)# = A, if A# exists.

If the generalized inverse is means Moore-Penrose inverse or group inverse:  every i ≠ j, {1 < i, j < n}, Ai ≠ Aj, then Bi ≠ Bj.