I just read this blog entry, and it is unclear to me why the tail bound results are not trivial for all bounded random matrices.
For instance, Corollary 4 states that for an $n\times n$ random matrix $M$ with i.i.d. mean-zero entries absolutely bounded by $1$, $P(\|M\|_{op} > A \sqrt{n}) \leq C \exp(-c A n)$ for all $A \geq C$, where $\|\|_{op}$ is the operator norm and $C, c > 0 $ are constants.
Since $\|M\|_{op}$ is bounded due to the boundedness of $M$'s entries, we can simply choose $C$ high enough so that $A \sqrt{n}$ necessarily surpasses this bound. The probability $P(\|M\|_{op} > A \sqrt{n})$ is then $0$, and the tail bound holds trivially for all choices of $c$.
Given there are no further qualifications on the $c$ and $C$, why is the result not trivial?