<s>[INST] Finding Jordan-normal form of a special $n\times n$ matrix Let $$A := \begin{pmatrix} 
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_n & -a_{n - 1} & -a_{n - 2} & \cdots & -a_1
\end{pmatrix}$$
I have to find $B$, $J$ such that $A = BJB^{-1}$ such that $J$ is a Jordan matrix. I already proved that $$\det(\lambda E - A) = \lambda^n + a_1\lambda^{n - 1} + ... + a_n$$ (just simple induction). I know that this gives the eigenvalues, and if $a_n, ..., a_1$ are known, I can find the eigenvalues and generalised eigenvectors, and write down the Jordan normal form, but I don't see how there is a general solution to this problem, as $(A - \lambda E)^n$ does not have a nice closed form. This is homework for a differential equation course, so maybe we can use some of that theory here? 
 [/INST] Let $\lambda$ be a root of $x^n+a_1x^{n-1}+\cdots a_n=0$ then with 
$$A=\begin{pmatrix}
0&1&0\\
0&0&1\\
-a_3&-a_2&-a_1
\end{pmatrix} $$
we have 
$$A-\lambda I=\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} $$
and 
$$\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} \begin{pmatrix}
1\\
\lambda\\
\lambda^2
\end{pmatrix}=
\begin{pmatrix}
0\\0\\-(a_3+a_2\lambda+a_1\lambda^2+\lambda^3)\end{pmatrix} 
=\begin{pmatrix}
0\\0\\0\end{pmatrix}  $$
Note further that 
$$\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} \begin{pmatrix}
0\\
1\\
2\lambda\\
\end{pmatrix}
=\begin{pmatrix}
1\\\lambda\\-a_2-2\lambda a_1-2\lambda^2\end{pmatrix}
=\begin{pmatrix}
1\\\lambda\\\lambda^2\end{pmatrix}  $$
provided that 
$$p^{\prime}(\lambda)=0$$
</s>