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Consider a random $n\times p$ matrix $X$ with $n\ll p$ and all entries of $X$ i.i.d. standard normal. For this $X$, the system of linear equations $y=Xw$ has infinitely many solutions, and the one with smallest $\ell^2$ norm is $\hat{w}=X^T(XX^T)^{-1}y.$ Based on this, I am interested in understanding $$\|X^T(XX^T)^{-1}\|_{\ell_1^n\to \ell_1^p}\leq \|X^T\|_{\ell_1^p\to \ell_1^n}\cdot \|(XX^T)^{-1}\|_{\ell_1^n\to \ell_1^n}.$$

The first term can be bounded using Chevet's inequality, but the closest thing I can find for the second term is this question, which deals with $\|\cdot \|_{\ell_2^n\to \ell_2^n}$.

So is anything known about arbitrary operator norms of the inverse Wishart distribution? Any help would be much appreciated!

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    $\begingroup$ I'm afraid I don't know about the particular question you ask, but my instinct is that the factorization you choose to get the inequality you state may not be such a good idea. The reason is that in the expression $X^\top (XX^\top)^{-1}$, eigenvectors of $XX^\top$ that have small eigenvalue get "blown up" by $(XX^\top)^{-1}$ and then "shrunk back down" by applying $X^\top$ again. When you split the product up and take norms of each component, you're not seeing this cancellation, so you could end up with an upper bound that is very bad $\endgroup$
    – Yemon Choi
    Commented Aug 6, 2019 at 21:25
  • $\begingroup$ Does this kind of argument apply even though the relevant quantities are not the eigenvalues/vectors? $\endgroup$
    – Samir K.
    Commented Aug 7, 2019 at 15:15
  • $\begingroup$ Well, roughly speaking, any reasonable norm on square matrices is going to be big when there are some large singular values (= eigenvalues for matrices of the form $XX^\top$). But I agree that what I outlined above is only a note of caution, not definite evidence that you will get a bad upper bound $\endgroup$
    – Yemon Choi
    Commented Aug 7, 2019 at 17:15

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