Bound of solutions of autonomous linear ODEs Given the linear system $\dot{z} = Az$. 
(a) Assume all the eigenvalues of $A$ have negative real part. Give a counterexample to this statement: every solution of $\dot{z} = Az$ satisfies $|z(t)|\leq |z(s)| \ \forall\ t> s $
(b) Assume A is symmetric and all the eigenvalues of $A$ have negative real part. Prove that every solution satisfies $|z(t)|\leq |z(s)|\ \forall\  t> s$.
My thought: I was trying many different types of matrices (the ones which has $2$ real eigenvalues with $2$ negative real parts or $2$ complex eigenvalues with $2$ negative real part), but they all satisfy the inequality. Can someone please help me with an example?
For part (b), I think of using $A$ as symmetric must have all real eigenvalues. Thus, by applying Lemma A, which is following: 
Consider $\dot{x} = Ax + f(t,x) + f_0(t)$ with $A,\  f$ and $f_0$ are continuous and $f(t,0) = 0 \ \forall\ t\in R$. Assume there exists constants $K\geq 1$, $M,L\geq 0$ and $\theta > \lambda + KL$ ($\lambda < 0$ is the negative real part). Then if we have:
(a) $||e^{At}||\leq Ke^{\lambda t}$ for $t\geq 0$
(b) $||f(t,x) - f(t,y)||\leq L||x-y||$ for all $L,x,y\in R$.
(c) $||f_0(t)||\leq Me^{\theta t}$ for $t\geq \tau$ ($\tau$ is some fixed constant).
Then if $u(t)$ is a solution, then for all $t\geq \tau$, we have: $||u(t)||e^{-\theta t}\leq K||u(\tau)||e^{-\theta \tau} + \frac{KM}{\theta - \lambda - KL}$.
Applying the result above into part (b) with sufficiently large $K\geq 1,\ M=L=\theta = 0$ (since $\lambda < 0$), we have:  $||u(t)||\leq K||u(\tau)||$. But how do we "cancel" the constant $K$ here?

The only way to get an initial increase is to have eigenvalues with multiplicity greater than 1. So try
$$
A=\begin{bmatrix}-1 & N\\ 0 & -1\end{bmatrix}
$$
and make $N$ large enough (positive or negative) to produce that initial increase.

One can probably also construct examples where the eigenvectors are not "orthogonal" for the vector norm used. Then in the initial point there could be some cancellation that is undone by the different decay velocities, resulting in an initial growth.

Both those principles are not true for symmetric matrices using the euclidean norm. (Why else use symmetric matrices.) Then one can use the Pythagorean theorem to separate the dimensions.
