# Upper bound on bilinear form $|u^*Av|$ with $(1-\delta)I\preceq A \preceq (1+\delta)I$

I am interested in finding a better upper bound for $$|u^*Av|$$, with $$A$$ a random Hermitian matrix that satisfies $$(1-\delta)I\preceq A \preceq (1+\delta)I$$ with high probability- at least $$1-C\mathrm{e}^{-c(\delta)}$$, with $$C$$, $$c$$ absolute constants independent of $$A$$ or its size.

It's obvious that
\begin{align} |u^*Av|\leq\|A\|\|u\|\|v\| \end{align} but I believe that there are cases where the bound can be improved, in particular for the off-diagonal terms, i.e. $$|e_i^\mathsf{T}Ae_j|$$ with $$e_n$$ the $$n$$-th canonical vector and $$i\neq j$$.

I am aware of the Hanson-Wright inequality for the tail of these inner products (see this MO answer), but wasn't sure how to use this to bound the inner product in reference to $$\delta$$.

Since $$(1-\delta)I \preceq A \preceq (1+\delta)I$$ w.h.p., we have $$-\delta I \preceq A-I \preceq \delta I$$ w.h.p., i.e. $$\|A-I\| \le \delta$$ w.h.p.
Hence, we can trivially derive the bound \begin{align*} |u^*Av| &= \left|u^*v + u^*(A-I)v\right| \\ &\le |u^*v| + |u^*(A-I)v| \\ &\le |u^*v|+\|A-I\|\|u\|\|v\| \\ &\le |u^*v|+\delta\|u\|\|v\| \end{align*} with high probability. If $$u$$ and $$v$$ are nearly parallel (or anti-parallel), this will be similar to the bound you obtained. But if $$u$$ and $$v$$ are nearly orthogonal and $$\delta$$ is small, this bound will be much smaller than $$(1+\delta)\|u\|\|v\|$$.