# QR decomposition of normally distributed matrices

Assume $$M$$ is an $$N \times k$$ Gaussian matrix, i.e., its entries are i.i.d. standard normal random variables, with $$N>>k$$. Take $$D=\text{diag}(\lambda_1, \dotsc ,\lambda_N)$$ for some fixed real scalars. I am interested in finding the p.d.f. of the $$N \times k$$ "unitary" matrix $$Q$$ from the QR decomposition of $$DM$$ (and possibly $$D^2M$$, etc.).

It is known that if $$k=N$$ and $$D=I_N$$, the identity matrix, then $$Q$$ is distributed with respect to the Haar measure on the Lie group of orthonormal matrices of order $$N$$. Can you provide any insight on the general case for $$k and/or general $$D$$?

I also tried to look for the simplest case, i.e., $$k=1$$. Then the QR decomposition coincides with a simple normalization. I have found this result for common variance, i.e., the case $$\lambda_1=\dotsc =\lambda_N$$. Can this be easily generalized for the general case with different $$\lambda_i$$?

I attempted in the simplest case to scale the matrix $$M$$ (which is for $$k=1$$ just an $$N$$ dimensional random vector). Indeed, then the above-mentioned result is applicable and one gets $$DM=DUR,$$ where $$UR$$ is the QR decomposition of $$M$$ and the p.d.f. of entries of $$U$$ is known from the above. Nonetheless, I haven't found any easy way to connect the p.d.f. of $$DU$$ with the one of $$Q$$. Thanks in advance.

• @whumber is the one to give credit for the special case. He/she also mentioned the p.d.f. in the comment for $N=2, k=1$, however I do not see the idea of how was this obtained and hence cannot try to generalize it for higher dimensions $N$. – michalOut Aug 4 '16 at 11:51
• What are the $\lambda_i$? Are they arbitrary values or have they somehow been derived from $M$ itself (such as eigenvalues of $MM^\prime$)? – whuber Aug 4 '16 at 14:23
• In general they are arbitrary. At the moment I am focusing on the case where $\lambda_1=1$ and $\lambda_i=\lambda^{i-1}$ for $i=2,3,\dotsc ,N$, i.e., they are geometrically decaying with a fixed parameter $\lambda \in (0,1)$. – michalOut Aug 4 '16 at 14:31
• @whumber , could you please clarify the simplest case with $N=2$ and $k=1$? – michalOut Aug 7 '16 at 9:39