= Solution
At each fixed time, the maximum over nonzero initial conditions is the <optimal energy amplification of a linear system>:
$$
G(t)=\max_{x_0\ne0}\frac{\|A(t)x_0\|_2^2}{\|x_0\|_2^2}=\sigma_{\max}(A(t))^2.
$$
Write $a=e^{\lambda_1t}$, $d=e^{\lambda_2t}$ and $b=(a-d)/(\lambda_1-\lambda_2)$. The <symmetric matrix> $A^TA$ has <trace> $\tau=a^2+b^2+d^2$ and <determinant> $a^2d^2$. Its largest <eigenvalue> is
$$
\boxed{G(t)=\frac{\tau+\sqrt{\tau^2-4a^2d^2}}2.}
$$
An optimal initial condition is a <right singular vector> of $A(t)$ associated with its largest <singular value>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-331-transient-growth.png]
{title=Stable eigenmodes can combine to produce transient energy growth}
{description=For $\lambda_1=-1$ and $\lambda_2=-0.05$, the red directions on the unit circle have positive instantaneous energy derivative. The <eigenvectors> are <nonorthogonal>. The right panel compares the <optimal energy amplification of a linear system> with the monotonically decaying <energy> of each eigenmode.}
If $\lambda_2>\lambda_1$, let $\alpha=1/(\lambda_2-\lambda_1)$. Then $A(t)/e^{\lambda_2t}\to\begin{pmatrix}0&0\\\alpha&1\end{pmatrix}$, so the precise long-time asymptotic statement is
$$
\boxed{G(t)\sim(1+\alpha^2)e^{2\lambda_2t},\qquad \alpha^2=\frac1{(\lambda_2-\lambda_1)^2}.}
$$
The <Cauchy-Schwarz inequality> shows that the unit initial conditions achieving this leading factor are
$$
\boxed{x(0)=\pm\frac1{\sqrt{1+\alpha^2}}\begin{pmatrix}\alpha\\1\end{pmatrix}.}
$$
This is an <adjoint eigenvector>, since $L^T(\alpha,1)^T=\lambda_2(\alpha,1)^T$. It differs from the right <eigenvector> $v_2=(0,1)^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For $\lambda_2<0$, even the optimal <energy> ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.
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