= Solution
Expand the <vector field> about the reference trajectory:
$$
\dot{\delta\mathbf x}=\mathbf f(\mathbf x+\delta\mathbf x)-\mathbf f(\mathbf x)
=J(t)\delta\mathbf x+O(\|\delta\mathbf x\|^2),\qquad
J_{ij}=\frac{\partial f_i}{\partial x_j}.
$$
Over a sufficiently short interval, the <Linearization>[tangent-linear] <Fundamental matrix>[propagator] is
$$
\boxed{\delta\mathbf x(t+\delta t)=A\delta\mathbf x(t),\qquad
A=I+J\delta t+O(\delta t^2).}
$$
Consequently
$$
A^TA=I+(J+J^T)\delta t+O(\delta t^2).
$$
If $S=J+J^T$ has $S\mathbf e=\lambda\mathbf e$ and the perturbation is initially parallel to $\mathbf e$, its squared <Euclidean norm> obeys
$$
\boxed{\|\delta\mathbf x(t+\delta t)\|^2
=(1+\lambda\delta t)\|\delta\mathbf x(t)\|^2+O(\delta t^2\|\delta\mathbf x(t)\|^2).}
$$
Positive $\lambda$ means instantaneous norm-squared growth, negative $\lambda$ decay, and zero $\lambda$ no first-order change. The corresponding <norm> growth rate is $\lambda/2$, not $\lambda$. These are <Eigenvalue>[eigenvalues] of the <Symmetric matrix>[symmetric part] governing the chosen error metric, not generally the <Eigenvalue>[eigenvalues] of the nonsymmetric <Jacobian matrix> or the long-time <Lyapunov exponent>[Lyapunov exponents]. The direction and the reference <Jacobian matrix> evolve during a finite forecast.
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