Given $A$ is a $n\times n$ matrix, and $f: R\rightarrow R^n$ is continuous and bounded.
(a) If $A$ has no eigenvalues on the imaginary axis, prove that $\dot{x} = Ax + f(t)$ has a unique solution which is bounded on $R$.
(b) Show the counterexample to (a) when $A$ has eigenvalues on the imaginary axis.
My attempt: (a) Since $A$ has no eigenvalues on the imaginary axis, $A$ is infinitesimally hyperbolic. By a well-know theorem, we can rewrite a Jordan canonical form $J$ of $A$ such that $J = [A_s\ 0, 0\ A_u]^T$ where eigenvalues of matrix $A_s$ all have negative real parts and eigenvalues of $A_u$ all have positive real parts. Then the given ODE is equivalent to: $\dot{x} = PJP^{-1}x + f(t)$ where $P$ is matrix whose columns are eigenvectors associated with each eigenvalues of $A_s, A_u$ in the same order.
My question is: how to show that $PJP^{-1}$ transforms $A$ into 2 block matrices: one with all eigenvalues having positive real part, and one with all eigenvalues having negative real parts?
(b) I haven't been able to get this, despite spending several hours trying to find $A$ so that the solutions would have resonance effect. Can someone please give some thought about this?
The assertion in (a) is wrong. You need to have eigenvalues with negative real part for a bounded solution for all possible bounded $f(t)$.
As for the Jordan form you can write $$ A = \begin{bmatrix}P_s & P_u\end{bmatrix} \begin{bmatrix}A_s & 0 \\ 0 & A_u \end{bmatrix} \begin{bmatrix}P_s & P_u\end{bmatrix}^{-1} $$
where columns of $P_s$ and $P_u$ contains the (generalized) eigenvectors of the stable and unstable eigenvalues respectively. You can see this fact from the spectral decomposition of a matrix.
Now define $P = \begin{bmatrix}P_s & P_u\end{bmatrix}$ and $y = P^{-1} x$.
$$\dot{y} = P^{-1} \dot{x} = P^{-1} A x + P^{-1} f(t) = P^{-1} A P y + P^{-1} f(t)$$
$$\begin{bmatrix}\dot{y}_s \\ \dot{y}_u\end{bmatrix} = \begin{bmatrix}A_s & 0 \\ 0 & A_u \end{bmatrix} \begin{bmatrix}y_s \\ y_u\end{bmatrix} + P^{-1} \begin{bmatrix}f_s(t) \\ f_u(t) \end{bmatrix}$$