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Suppose that $A$ is an $n\times n$ diagonal matrix with positive diagonal elements and $\Pi$ is a random $k\times n$ matrix that could be (a) i.i.d. Gaussian, or (b) $k$ rows of a random orthogonal matrix

The matrix I am considering is $X = \mathbb{E}\{\Pi^T (\Pi A^{-1}\Pi^T )^{-1} \Pi\}$. I know that $\mathbb{E}\{\Pi A^{-1}\Pi^T\} = \frac{\|A^{-1}\|_1}{n} I$ but $\Pi$ is correlated with what gets multiplied outside the inverse. So I have no good idea how to analyse the eigenvalues of $X$. Numerical experiments suggests that large eigenvalues of $X$ are about the same. Any explanations for this?

Thanks.

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  • $\begingroup$ A projection of $\mathbb R^n$ onto a $k$-dimensional subspace is not a $k \times n$ matrix, it's $n \times n$ with $k$-dimensional range. Perhaps you mean $k$ rows from a random orthogonal matrix? $\endgroup$ May 13, 2015 at 18:24
  • $\begingroup$ @RobertIsrael Yes, that's exactly what I meant $\endgroup$
    – user58955
    May 13, 2015 at 19:46
  • $\begingroup$ youe want to first take the expectation value of the matrix between curly brackets and then study its eigenvalues? the eigenvalues of the matrix before taking the expectation value are much simpler to find.... $\endgroup$ May 13, 2015 at 20:58
  • $\begingroup$ @CarloBeenakker So what are the eigenvalues of the matrix inside the curly brackets? $\endgroup$
    – user58955
    May 13, 2015 at 23:01
  • $\begingroup$ well, since $\Pi\Pi^T=\mathbb{1}$ (exactly in your case (b) and approximately in case (a) for $k\ll n$), and since the matrix products $AB$ and $BA$ have the same set of nonzero eigenvalues, the nonzero eigenvalues of $M=\Pi^T (\Pi A^{-1}\Pi^T )^{-1}\Pi$ are those of $Y=(\Pi A^{-1}\Pi^T )^{-1}$, which has an inverse Wishart distribution. $\endgroup$ May 14, 2015 at 9:57

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