Say, we have the scalar ODE $$ x^{(n)} = \sum_{j=0}^{n-1} a_jx^{(j)}. $$ We now that we can write this as a system $$ y' = Ay $$ where $$ A = \begin{pmatrix} 0 & 1 & \dots & 0\\ \vdots & 0 & \ddots & \vdots\\ 0 & & & 1 \\ a_1 & \dots &\dots &a_n \end{pmatrix}. $$ From the solution formula for the scalar ODE, we know that if $\lambda$ is a root of the characteristic polynomial od the scalar ODE (which coincides with $\pm$ the characteristic polynomial of $A$), then we have quasipolynomials $\exp(\lambda t) , \exp(\lambda t)t, \exp(\lambda t)t^2, ...$ up to the multiplicity of $\lambda$. So this means (looking at the exponential of the Jordan normal form of $A$) that $\lambda$ is an eigenvalue of geometric multiplicity $1$.
My question is whether there is any linear algebra (!) way to see that the eigenvalues of $A$ have geometric multiplicity $1$.
