Differentiate the quadratic <energy>:
$$
\dot E=2\lambda_1x_1^2+2x_1x_2+2\lambda_2x_2^2=2x^THx,\qquad H=\frac{L+L^T}{2}=\begin{pmatrix}\lambda_1&1/2\\1/2&\lambda_2\end{pmatrix}.
$$
For $x_1\ne0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when $\lambda_2r^2+r+\lambda_1>0$. With both <eigenvalues> negative, such directions exist iff $1-4\lambda_1\lambda_2>0$. Writing $\delta=\sqrt{1-4\lambda_1\lambda_2}$, the growing initial conditions are the two opposite open wedges
$$
\boxed{x_1\ne0,\qquad \frac{-1+\delta}{2\lambda_2}<\frac{x_2}{x_1}<\frac{-1-\delta}{2\lambda_2}.}
$$
If $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly when
$$
\boxed{\lambda_1\lambda_2\geq\frac14\quad(\lambda_1,\lambda_2<0).}
$$
Indeed, this makes the <symmetric part of a matrix> $H$ a <negative semidefinite matrix>, so $\dot E\leq0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the <instantaneous energy-growth criterion for a linear system>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 3 / d / Solution

{title2=$\operatorname{sym}A=(A+A^T)/2$}

The symmetric part of a real square <matrix> is $H=(A+A^T)/2$. It controls the quadratic form because $x^TAx=x^THx$; the antisymmetric part contributes zero. In a <linear differential equation> $x'=Ax$, it determines the instantaneous derivative of the squared <Euclidean norm>.

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 Incoming links: Symmetric part of a matrix