Exercise 6.11: Unbounded Matrices and Bernstein Bounds
(a)
Following the hint, construct $Q_i$ from $A_i$ like in Exercise 6.10(a). We verify the Bernstein condition for $g_i Q_i$: \begin{align} \E[(g_i Q_i)^j] \le \frac{j!}{2} (b_1 b_2)^{j-2} \sigma^2 b_2^2 I. \end{align} Technically, we would need $\Var[g_i Q_i]$ on the r.h.s. instead of the upper bound $\sigma^2 b_2^2 I \ge \Var[g_i Q_i]$, but it can be seen that the Bernstein bound for matrices also works with $\Var[g_i Q_i]$ replaced by an upper bound. We now use a one-sided version of the Bernstein bound (one-sided to not accumulate an extra factor two) for matrices to deduce concentration for $\lambda_1(g_i Q_i)$. To conclude, note that $\lambda_1(g_i Q_i) = \sigma_1(g_i A_i) = \norm{g_i A_i}_2$, because the eigenvalues of $Q_i$ are symmetric (if $\lambda$ is an eigenvalue, then $-\lambda$ is also an eigenvalue), and use Exercise 6.10(a).
(b)
Follows immediately.