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Recently 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).
If i find its pseudo inverse, given by: $R^+ = (R' R)^{-1} R'$.

  1. Will the this pseudo inverse matrix will still remain random ?
  2. If yes, will it contain elements distributed with normal distribution?
  3. 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/(variance of $R$ $\times d^2$) ; where d are the columns in R.)

I have tried to find 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. Thanks,

share|improve this question
    
I cleaned up your math by putting it into LaTeX. I think though that your pseudoinverse formula is wrong. –  Mark Meckes Dec 13 '12 at 14:46
4  
1. Yes, of course. 2. No. Think about the case k=d: the inverse of a single normal random variable is not normal. This paper looks like a good place to find the kind of information you want: ugr.es/~ramongs/articulos%20en%20pdf/cimat1.pdf –  Mark Meckes Dec 13 '12 at 14:49
    
Thanks Mark. But can you tell a condition under which inverse of a single normal random variable becomes normal. When i plot histogram of the inverse of \mathbf{R} with k and d very large, i get a nearly normal distribution. –  Salman Dec 15 '12 at 8:01

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