Hi, 
I have a problem that bugs me quite a while. 

Let R1 and R2 be two (m x n) random matrices with m>n. 
Each entry of R1 and R2 is independent and identically chosen from 
normal distribution (0, c), where c is a constant. 


My question is:

(1) What are the statistical properties of the product of R1 
and pseudo-inverse(R2)? denoted as R1*pseudo-inverse(R2). Here 
pseudo-inverse(R2) is the pseudoinverse of matrix R2 such that 
pseudo-inverse(R2)*R2 = I. 


(2) How different the matrix product X*R1*pseudo-inverse(R2) is from
X? Here X can be viewed as either a constant matrix or a random matrix
(no prior knowledge about X). If constant matrix is easier to analyze,
then let us discuss constant X first. You can also define the word
"different" by yourself, for example, distribution is different,
physically position is different, or ...

If both questions are too difficult to answer, I prefer to see more
discussions about question 2.

Thanks a lot 
Roy

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Clarification of Question by elgoog_elgoog-ga on 05 Apr 2005 12:11 PDT

Here, the matrix rank rank(R1) = rank(R2) = n. 
In this case, pseudoinverse of R2 = (R2^T R2)^(-1) R2^T.

Ive had a few thoughts...

1 pseudo inverses arent unique... but that shouldnt matter for the distribution...

So we assume this first... that we can partition R2 this way

| R2_1 |
  _

| R2_2 |

Where R2_1 is invertible.  If its not this way its not a big deal...
the form of the pseudo inverse will just change

There is a pseudo inverse in the form

| R2_1^(-1) | 0 |  

So ultimately when we do the product of the pseudo inverse of R2 and R1

we get R2_1^(-1) * R1_1

now in the simple case of m=2 n=1...

you will basically have a normal over a normal... which is a cauchy...

i cant think of the other ones... but they may be cauchy as well... 

do you have any thoughts...

for part B i dont think you can say much except that it will likely
have infinite mean since cauchy has such and E(XY) = E(X)E(Y) will be
inf (even if E(x) = 0 unless X is identically 0)

any questions... williamvolterman@gmail.com

Thank you very much for the reply.
Given R1_1 * R2_1^(-1) and the simple case of m=2, n=1, the result is
cauchy. I agree with that. The problem is I am dealing with large
matrices. m>>1, and n>>1. What will the result be?

Ultimately you will need to know the distribution of the inverse of a
matrix of normals...

for a two by two you have individual elements like this

a / (ad-bc) = 1 / (d-bc/a) no clue what the distribution of this is...

Then the product of the matrices has elements

A a / (ad-bc) - B c / (ad-bc) ... honestly i have no clue what this
could be. but like i said... it probably has an infinite mean