= Solution
Differentiate the <eigenvalue> equation
$$
A(t)v(t)=\lambda(t)v(t)
$$
at $t=0$, where $A'(0)=E$. Suppressing the argument $0$ gives
$$
Ev+Av'=\lambda'v+\lambda v'.
$$
Left-multiplication by the transpose of the unit <left eigenvector> $u$ and use of $u^TA=\lambda u^T$ cancel the terms involving $v'$:
$$
u^TEv=\lambda' u^Tv.
$$
Because the eigenvalue is <simple eigenvalue>[simple], $u^Tv\ne0$, and hence the <first-order perturbation of a simple eigenvalue> is
$$
\lambda'(0)=\frac{u(0)^TEv(0)}{u(0)^Tv(0)}.
$$
The definition of the <operator norm>, followed by the <Cauchy-Schwarz inequality>, gives
$$
|u^TEv|\leq\|u\|_2\|E\|_2\|v\|_2=\|E\|_2.
$$
Therefore
$$
\boxed{|\lambda'(0)|
\leq\frac{\|E\|_2}{|u(0)^Tv(0)|}}.
$$
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