A question about the paper "The Condition Number of a Randomly Perturbed Matrix" My question pertains to this paper by Terence Tao and Van Vu, https://arxiv.org/abs/math/0703307
Both my questions pertain to the argument presented in this paper in its section 6 (page 5). We are looking at a $n-$dimensional square random matrix $M_n$ satisfying the conditions stated through Definitions 2.15, Definition 2.17 and Theorem 2.18 on page 3. 
Now we are saying that lets assume the existence of a unit vector $v \in \mathbb{R}^n$ such that for some $B >10$ we have, $\Vert M_n v\Vert < n^{-B}$. The vector $\tilde{v}$ is created by truncating each coordinate of $v$ to the nearest multiple of $n^{-B-2}$. So if I understand this correctly then we have that for each coordinate $i$, $\vert v_i - \tilde{v}_i\vert \leq \frac{1}{2n^{B+2}}$. This is now supposed to imply a number of things which arent clear to me, 


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*The paper claims, $0.9 \leq \Vert \tilde{v}\Vert \leq 1.1$. Why? 
If I just think directly then we have, $\Vert \tilde{v}\Vert \leq \Vert v + (\tilde{v}-v)\Vert \leq \Vert v\Vert + \sqrt{n} \frac{1}{2n^{B+2}} = 1 + \frac{n^{-B-\frac{3}{2}}}{2}$.For this to match the claim we need, $\frac{n^{-B-\frac{3}{2}}}{2} = 0.1$. But this is now incompatible with the initial statement that we need $B>10$. What am I missing? 

*The paper claims that the following is also true that, $\Vert M_n \tilde{v}\Vert \leq 2n^{-B}$. Why? 
Just as above if I again think just directly then we have, that $\Vert M_n \tilde{v}\Vert = \Vert M_n(v+(\tilde{v}-v))\Vert \leq \Vert M_n v \Vert + \Vert M_n (\tilde{v}-v)\Vert \leq n^{-B} + \Vert M_n \Vert \frac{n^{-B-\frac{3}{2}}}{2}$. One way this can be compatible with the claim is if we have, $\Vert M_n \Vert \leq 2n^{1.5}$. 
One might now go back to the ``boundedness" part of Definition 2.15 and the first bullet point of Theorem 2.18 on page 3 to see that with probability $1$ the entries of the matrix $M_n$ are integers bounded as $n^C$ for some constant $C>0$. This gives by Frobenius norms, $\Vert M_n \Vert \leq \Vert M_n \Vert_F \leq n^{1+C}$. So for compatibility with the previous bound we need, $n^{1+C} \leq 2n^{1.5}$. But then such an equation is now an upperbound on the constant $C$ and that is not something that Theorem 2.18 enforced. What am I missing? 
 A: *

*One does not need to have $n^{-B-3/2}/2$ to be equal to $0.1$, it is enough for it to be less than or equal to $0.1$, which is certainly the case for $n$ large enough.

*Thanks for pointing out this typo (or more precisely, set of typos) in this paper.  As you point out, the exponents here are adapted to the case of small $C$ (in particular $C=0$, which is the most frequent case in applications) and require some adjustment for large $C$.  [We do remark several times in the paper that the precise exponents are not to be taken too seriously.]  Basically, if one replaces all occurrences of $n^2$ in this argument with, say, $n^{2(1+C)}$ (and makes some similar adjustments to some other similar factors such as $n^{-4}$) then this issue will be avoided.
It is also worth mentioning that the results in this paper have been superseded by subsequent stronger results, e.g. 
Tao, Terence; Vu, Van, Smooth analysis of the condition number and the least singular value, Math. Comput. 79, No. 272, 2333-2352 (2010). ZBL1253.65067.
