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Let $X$ be a random $N \times n$ matrix with iid entries from $\mathcal N(0, 1)$ and with $n/N =: \lambda(N,n) \le \lambda_0$, for some $\lambda_0 \in (0, 1)$. That is, $X$ is genuinely rectangular (including the case where $X$ is an $N \times 1$ vector).

Question 1. Is it true that for every $C>0$, there exists $c,A>0$ (only depending on $C$ and $\lambda_0$) such that $P(s_\min(X) \le c\sqrt{N}) \le Ae^{-CN}$ ? What about the particular case when $C=1$ ?

In case the answer to the above is negative,

Question 2. Find $A,c,C>0$ such that $P(s_\min(X) \le c\sqrt{N}) \le Ae^{-CN}$.

Note that the case of Rademacher entries is solved in Theorem 2.7.1 this document (by T. Tao).

Update

Question 1 (and therefore, Question 2) has an affirmative answer with $A=2$ and without any constraint on $\lambda$ under than $\lambda < 1$. See https://mathoverflow.net/a/372119/78539

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The answer to your Question 2 is positive, in view of Theorem 3.1 and Fact 2.4 (for general subgaussian entries) in the paper by Litvak et al (referred to in the book by Tao that you linked).

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  • $\begingroup$ Thanks. Concerning the Tao et al. paper reference, I was referring to Q2. I read the paragraph just before 3.4 of this paper math.uci.edu/~rvershyn/papers/rv-ICM2010.pdf (Rudelson and Vershynin), and the authors say one can have $C=1$ in the subGaussian case, more precisely they claim: "if $n/N<1$, then there exists $c>0$ such that $ P(s_\min(A) \ge c\sqrt{N}) \ge 1 - 2e^{-N}$. Is this a typo ? $\endgroup$
    – dohmatob
    Commented Sep 18, 2020 at 12:53
  • $\begingroup$ OK, I see. I'm moving the discussion to a separate question mathoverflow.net/q/372015/78539. I really need the exponent to be $C$ (in order to absorb an entropy cost in the proof of something else I'm doing). Any input welcome! $\endgroup$
    – dohmatob
    Commented Sep 18, 2020 at 13:48
  • $\begingroup$ Actually, Question 1 has an affirmative answer, $A=2$, for arbitrary $C>0$, and without any restriction on $\lambda$ other than $\lambda < 1$. See mathoverflow.net/a/372119/78539 $\endgroup$
    – dohmatob
    Commented Sep 20, 2020 at 8:31

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