So I'm trying to make an exercise in ode's.
Let $A \in M_5(\mathbb{R})$ be a matrix with 3 eigen-values $\lambda_1 = -1$ with multiplicity $3$ and $\dim(\text{Ker}(A+I))=1$, $\lambda_{2,3} = \pm 2i$. Let $J$ be the Jordan form of the matrix $A$.
Now the exercise asks to write down $J$, I simply get, $$ J_1 = \begin{pmatrix} -1 & 1 & 0\\\\ 0 & -1 & 1\\\\ 0 & 0 & -1 \end{pmatrix} $$ $$ J_2 = \begin{pmatrix} +2i\\\\ \end{pmatrix} $$ $$ J_3 = \begin{pmatrix} -2i\\\\ \end{pmatrix} $$ $$ J = \begin{pmatrix} J_1 & 0 & 0\\\\ 0 & J_2 & 0\\\\ 0 & 0 & J_3 \end{pmatrix} $$
I'm curious if this is correct. After that the exercise asks to prove that the supremum norm on $||e^{Jt}||\leq 1 \quad \forall t \geq 0$. With this fact the problem stated in the title is easily solved. I know that for the Jordan matrix we have $$ e^{Jt} = \begin{pmatrix} e^{J_1 t} & 0 & 0\\\\ 0 & e^{J_2 t} & 0\\\\ 0 & 0 & e^{J_3 t} \end{pmatrix} $$
Now if we verify for the different blocks it seems to work for every block except the $e^{J_2 t}$ block as is simply a scalar value equal to $e^{\lambda_2 t} = e^{2 i t}$ for which the norm obviously is greater than one for all $t \geq 0$. I don't think the exercise is wrong but that I'm missing something here, maybe I should use the $\dim(\text{Ker}(A+I))=1$ property as I haven't really used it?
After that the exercise asks to express the solution to a generic cauchy problem with the linear operator $A$ as a function of $J, t_0, x_0$ and ofcourse the transormation matrix to get the Jordan form.
Anyway if someone could help me here, I would very much appreciate it!