Stable eigenvalues can amplify the state before every mode decays

Consider the planar linear system $\dot x=Ax$ with \[ A=\begin{pmatrix}-1&4\\0&-2\end{pmatrix}. \] Every coordinate of $x$ is a fully visible state: there is no output map, no cancelled factor, and no hidden mode. 1. Compute the eigenvalues of $A$ and prove that $A$ is Hurwitz and diagonalizable. Compute the matrix exponential \[ e^{At}=\begin{pmatrix}e^{-t}&4(e^{-t}-e^{-2t})\\0&e^{-2t}\end{pmatrix}. \] 2. Starting from $x(0)=(0,1)^{\mathsf T}$, obtain \[ x_1(t)=4(e^{-t}-e^{-2t}),\qquad x_2(t)=e^{-2t}. \] Prove that $x_1$ attains a unique maximum at $t=\ln 2$ with value $1$. Compute $\lVert x(\ln 2)\rVert_2$ and show that \[ \lVert x(\ln 2)\rVert_2=\frac{\sqrt{17}}4>1=\lVert x(0)\rVert_2. \] 3. Prove that $x(t)\to 0$ as $t\to\infty$. Explain the transient amplification: $A$ is stable but not normal. Verify $AA^{\mathsf T}\ne A^{\mathsf T}A$ by direct multiplication. 4. Audit these claims: (i) strictly negative eigenvalues force $\lVert x(t)\rVert_2$ to be monotone decreasing, so the state can never exceed its initial Euclidean norm; (ii) the observed growth is a hidden or cancelled mode, or else $A$ is a defective Jordan block rather than diagonalizable; (iii) the exercise is a Hankel-rank proof that a larger realization is nonminimal. Do not introduce an unobservable or uncontrollable mode, and do not replace the packet by a minimal-realization argument.

Solve $\dot x_2=-2x_2$ first, then integrate $\dot x_1+x_1=4x_2(t)$ with the factor $e^{t}$.