Distribution of inverse of a random matrix I got stuck into a problem and couldn't find its
satisfactory answer anywhere.
My question is simple. Suppose I have a fat random matrix (i,e., $R$ has dimensions $k\times d$ where $k<d$) whose
elements are chosen from a i.i.d. standard normal distribution $N(0,1)$.
Suppose I find its pseudo-inverse, given by: $R^+ = (R' R)^{-1} R'$.

*

*Will this pseudo-inverse matrix still remain random ?

*If yes, will it contain elements distributed with normal distribution?

*If yes, what would be the mean and variance of this this normal distribution?

I am asking these questions because I have experimented with a lot of random matrices (with elements distributed with
$N(0,1)$. When in plot a histogram of pseudo inverse elements, it comes a normal distribution with mean $= 0$ and variance
$= 1/(\mbox{variance of }R \times d^2)$ ; where d are the columns in R.)
I have tried to find the PDF using Jacobian transform but i could not figure out how will it shape up the variance.
I would be thankful if you could guide me or clarify my problem.
 A: This is not a full answer, but firstly for 1, yes, of course. What's more, the elements of $R^+$ are independent of each other, because the elements of $R$ are independent of each other, and the pseudo-inverse preserves matrix rank.
For 3, $E[R^+_{ij}]=0$, due to symmetry, and $E[(R^+_{ij})^2]=\frac{1}{d(d-k-1)}$, as can be shown from an analysis in this paper. If you look into that paper, keep in mind that the singular inverse-Wishart distributed matrix would be $R^{+\top}R^+$, so the diagonal elements of the mean of that matrix, given in section 4.1, are simply $k$ times the variance of each element, i.e. $E[R^{+\top}R^+]=kE[(R^+_{ij})^2]I_d$. This is because the diagonal elements in the matrix product $R^{+\top}R^+$ each sum $k$ squared (independent) elements of $R^+$. Plugging in the derived equation from the paper, which in this post's notation would read $E[R^{+\top}R^+]=\frac{k}{d(d-k-1)}I_d$, gives us our answer. One could also get here via a similar path using the mean of the non-singular Wishart distributed random matrix, $E[R^+R^{+\top}]$.
As for 2, while each element of $R^+$ is generally not Gaussian (consider k=d=1), I would certainly believe they approach Gaussianity as the dimensions increase, but I don't know anything about that.
