I am using Lawrence Perko's book Differential Equations and Dynamical Systems, for my Differential Equations course.
At the moment we are going over Jordan Forms of a linear system $x^{'}(t) = Ax$, where A is an $n\times n$ matrix. The solution of the system is $x(t) = e^{At}x_0 = Pe^{Bt}P^{-1}x_0$.
For the purpose of this problem, we are considering two elementary Jordan blocks.
\begin{align} B_1 & = \begin{bmatrix} \lambda & 1 & 0 & \dots & 0\\ 0 & \lambda & 1 & \dots & 0\\ \vdots & & \ddots\\ 0 & \dots & & \lambda & 1\\ 0 & \dots & & 0 & \lambda \end{bmatrix} & B_2 & = \begin{bmatrix} D & I_2 & 0 & \dots & 0\\ 0 & D& I_2 & \dots & 0\\ \vdots & & \ddots\\ 0 & \dots & & D& I_2\\ 0 & \dots & & 0 & D \end{bmatrix} \end{align}
1st Elementary case is where we have $\lambda$ to be one of the real eigenvalues of some matrix $A$. \
2nd Elementary case is where consider $\lambda$ is a complex number $a+ib$ with
\begin{align} D & = \begin{bmatrix} a & -b\\ b & a \end{bmatrix} & I & = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & \text{and }& & 0 & = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{align}
Here is my question:
Suppose that the elementary blocks $B$ in the Jordan form of the matrix $A$, given by $B_1$ or $B_2$, have no ones or $I_2$ blocks off the diagonal respectively. Show that if all of the eigenvalues of $A$ have nonpositive real parts, then for each $x_0\in\mathbb{R}^n$ there is a positive constant $M$ such that $|x(t)|\leq M$ for all $t\geq 0$.
Now since we know for a fact that the real parts of all eigenvalues $\lambda_i$ are negative or zero, then every $y_i = e^{\lambda t}\rightarrow 0$ or $y_i = e^{\lambda t}\rightarrow 1$ as $t\rightarrow \infty$ (implying the system is stable or centered). Since we also don't have the ones and $I_2$ (for each respective case), we have matrix $A$ to be semisimple.
Focusing on the first case. Since we are looking to prove the norm of our solution is less than some constant $M$ for all $t\geq 0$. We have by the properties of norm operators on a linear transformation
\begin{align} |x(t)| = &|e^{At}x_0|\leq ||e^{At}||\cdot |x_{0}| &=& ||Pe^{Bt}P^{-1}||\cdot|x_{0}|\\ & & \leq & ||P||\cdot ||e^{Bt}||\cdot||P^{-1}||\cdot |x_{0}|\\ & & \leq & ||P||\cdot ||e^{Bt}||\cdot\frac{1}{||P||}\cdot |x_{0}|\\ & & \leq & ||e^{Bt}||\cdot|x_{0}|\\ & & \leq & e^{||B||\cdot|t|}\cdot|x_{0}|\\ \end{align}
Since the initial condition $x_0$ is a constant vector, bounded by some value $M$, and $||B||=\max_{|x|\leq 1}|T(x)| = 0$. Thus $e^{||B|||t|}x_0\leq e^{0}\cdot M = $ M.
Am I on the right track? I would assume The second case follows a similar procedure. Is there anything I may have incorrectly? Please let me know.
Thank You for you time. I appreciate any feedback, comments, or suggestions you may have. Thank you in advance and have a wonderful day.