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Context. Studing a problem in machine-learning, I'm led to consider the following problem in RMT...


Definition. Given positive integers $m$ and $n$ and positive real numbers $c_1$ and $c_2$, let's say an $m$-by-$n$ real matrix $X$ is $(c_1,c_2)$-incrompressible if every submatrix of $Z$ with $k \ge c_1 m$ rows and $n$ columns satisfies $c_2\mathbb B^k \subseteq Z\mathbb B^n$, where $c_2\mathbb B^k$ is the ball of radius $c_2$ in $\mathbb R^k$ and $Z\mathbb B^n := \{Zv \mid v \in \mathbb B^n\}$.

Question. Is the definition somehow linked to the classical "restricted isometry property" ?

An answer to this the above question would allow me to directly tap into the vast RMT literature, for the purposes of attacking my subsequent questions (see below).

Special case when $c_1=1$. In the particular case where $c_1=1$, we note that $X$ is $(1,c_2)$-incompressible iff its smallest singular value is $c_2$ or greater.

Now, fix $\delta \in (0, 1)$ once and for all.

Question. Do there exist universal constants $c_1 > 0$ and $c_2 > 0$ (depending only on $\delta$) such that the following phenomenon holds ?

The phenomenon. Let $m$ and $n$ be positive integers with $m \le \delta n$ and $n$ large, and let $k \ge c_1 m$. Let $X$ an $m$-by-$n$ random real matrix with iid $N(0,1)$ entries.

Question. With high probability, $X$ is $(c_1,c_2)$-incompressible!

Also, how large can this probability be as a function of $n$ and $\delta$ ?

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  • $\begingroup$ Is there no relation between $m$ and $n$ except that they are both large? If $n < k$, it's impossible to have $c_2 {\mathbb B}^k \subseteq Z \mathbb B^n$. $\endgroup$ Commented Jul 24, 2020 at 15:53
  • $\begingroup$ Indeed. Fixed. You may assume $m \le \delta n$ with $\delta \in (0, 1)$. Thus in the question, the universal constants $c_1$ and $c_2$ only depend on $\delta$. $\endgroup$
    – dohmatob
    Commented Jul 24, 2020 at 16:03
  • $\begingroup$ I mean, if we're talking probabilistically, there is a nonzero probability of arbitrarily big balls being in the range. Maybe it's unlikely, but asymptotically you could fit some truly ginormous balls in there. $\endgroup$ Commented Jul 25, 2020 at 1:42
  • $\begingroup$ Could you make this claim more formal ? $\endgroup$
    – dohmatob
    Commented Aug 6, 2020 at 10:10

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