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First let me describe the setting of the problem.

I have a random matrix $A\in \mathbb{R}^{m\times n},\ (m<n)$ with $a_{ij}\sim \mathcal{N}(0,I)$ i.i.d. Let there be a given set of $K (K<m)$ columns of this matrix which are linearly independent; call the set $S$. Let $T\subset S$ and $P_T$ be the orthogonal projection operator on $T$, i.e. $P_T=A_SA_S^\dagger A_S^T$ and let $P_T^\perp=I-P_T$. Here $A_S$ is the submatrix of $A$ having columns from the set $S$.

Then, my question is,

If $\psi$ is a column of $A$ outside the set $S$, are the eigenvectors of $P_T^\perp$ stochastically independent of $\psi$?

Intuitively I think the answer is yes, since, by definition, the entries of the eigenvectors of $P_T^\perp$ are functions of the entries of the columns in $S$, and so they are independent of $\psi$. But since stochastic independence is such a subtle issue, I am not confident enough to rely upon my observation.

So It is my request that someone kindly see if my logic is sound, or if there is some mistake then any hint leading to the correct direction is most welcome. Thanks in advance.

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  • $\begingroup$ think of it this way: only generate the $nK$ random numbers that go into $S$, and find the set of eigenvectors $E$ of the projector; then generate $n$ more independent random numbers that form $\psi$; the independence of $\psi$ and $E$ is then assured by construction, isn't it? $\endgroup$
    – Carlo Beenakker
    Jun 10, 2015 at 9:38
  • $\begingroup$ Yes, @CarloBeenakker, the constructive method guarantees the independence. So, the logic that I thought of is not wrong, right? Actually I had the doubt in my mind because the projection operator can be written as $P_T=(M\mid N)\Lambda (M\mid N)^{-1}$ where $M, N$ are some bases for $A_{T},A_{T}^\perp$. But I think those bases are also formed from the $nK$ elements of $S$. Please see if what I am thinking is correct. Anyway, thanks for your comment! $\endgroup$
    – Samrat Mukhopadhyay
    Jun 10, 2015 at 13:21

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