Consider the $n$-th order ODE $$0=a_0x+\dots+a_{n-1}x^{(n-1)}+a_nx^{(n)}\quad (\star_1)$$ Any solution $u$ of $\dot x = Ax$ $(\star_2)$ with $$A = \begin{bmatrix} 0 & 1 & 0 & \dots & 0\\ \vdots & \ddots\ & \ddots & \ddots & \vdots\\ \vdots & & \ddots & \ddots & 0\\ 0 & \dots & \dots & 0 & 1\\ \frac{-a_0}{a_n}& \dots & \dots & \dots & \frac{-a_{n-1}}{a_n}\end{bmatrix}$$ gives a solution to the first ODE via $x= \operatorname{pr}_1\circ u$, and any solution $x$ of the first ODE gives a solution to the second via $u = (x, \dot x,\dots, x^{(n-1)})$.
Now, if $\lambda_1,\dots,\lambda_r\in\mathbb{C}$ are the Eigenvalues of $A$ with respective multiplicity $p_i$, $i=1,\dots, r$, how is $$e^{\lambda_1\cdot}, te^{\lambda_1\cdot},\dots, t^{p_1-1}e^{\lambda_1\cdot}, \dots, e^{\lambda_r\cdot}, te^{\lambda_r\cdot},\dots, t^{p_r-1}e^{\lambda_r\cdot}$$ a basis of the space of solutions of the first ODE? How do I show this using the Jordan normal form, that is, using $A = TJT^{-1}$ and the columns of $Te^JT^{-1}$ (that should be a basis)? (I can show it directly)
In this case, are the columns of $e^J$ already a basis of the space of solutions of the second ODE?
The Jordan form looks, in a block form, like $$ J=\mathrm{diag}(J_0,J_1,\ldots,J_k), $$ where $J_0,J_1,\ldots,J_k$ are submatrices. In particular, $J_0$ is diagonal $$ J_0=\mathrm{diag}(\lambda_1,\ldots,\lambda_s), $$ and $J_\ell$, $\ell=1,\ldots,k$, are Jordan blocks $$ J_\ell=\left(\begin{array}{ccccccc} \mu_\ell & 1 & 0 & 0 & \cdots &0& 0\\ 0 & \mu_\ell & 1 & 0 & \cdots & 0&0\\ \vdots & \vdots & \vdots & \vdots & &\vdots&\vdots \\ 0&0&0&0&&\mu_\ell&1 \\ 0&0&0&0&&0&\mu_\ell \end{array}\right)\in\mathbb C^{n_\ell\times n_\ell}. $$ Then $$ \exp(tJ)=\mathrm{diag}\big(\exp(tJ_0),\exp(tJ_1),\ldots,\exp(tJ_k)\big), $$ and $$ \exp(tJ_0)= \mathrm{diag}\big(\mathrm{e}^{t\lambda_1},\ldots,\mathrm{e}^{t\lambda_s}\big) $$ while $$ \exp(tJ_\ell)=\mathrm{e}^{t\mu_\ell}\left(\begin{array}{ccccccc} 1 & t & t^2/2! & t^3/3! & \cdots &t^{n_\ell-2}/(n_\ell-2)!& t^{n_\ell-1}/(n_\ell-1)! \\ 0 & 1 & t & t^2/2! & \cdots & t^{n_\ell-3}/(n_\ell-3)!&t^{n_\ell-2}/(n_\ell-2)!\\ \vdots & \vdots & \vdots & \vdots & &\vdots&\vdots\\ 0&0&0&0&&1&t\\ 0&0&0&0&&0&1 \end{array}\right). $$ The columns of $\mathrm{e}^{tJ}$ are NOT in general a basis of the solution space of $x'=Ax$.
For example $A=\left(\begin{array}{cc} 0&0\\ 1&0\end{array}\right)$ is similar to $J=\left(\begin{array}{cc} 0&1\\ 0&0\end{array}\right)$, and $$ \exp(tA)=\left(\begin{array}{cc} 1&0\\ t&1\end{array}\right) \quad\text{while}\quad \exp(tJ)=\left(\begin{array}{cc} 1&t\\ 0&1\end{array}\right), $$ and the columns of the first matrix are NOT linear combinations of the columns of the second matrix.