<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] If you can get the companion matrix into a diagonal block decomposition of smaller companion matrices, one for each eigenvalue, you get a recipe for each block. For example $(x-c)^4 = x^4 - 4cx^3 + 6 c^2 x^2 - 4 c^3 x + c^4.$ We have
$$
A =
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-c^4 & 4 c^3 & - 6 c^2 & 4 c \\
\end{array}
\right)
$$
Take
$$
R =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
c & 1 & 0 & 0 \\
c^2 & 2c & 1 & 0 \\
c^3 & 3 c^2 & 3 c & 1 \\
\end{array}
\right)
$$
then
$$
R^{-1} =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
-c & 1 & 0 & 0 \\
c^2 & -2c & 1 & 0 \\
-c^3 & 3 c^2 & -3 c & 1 \\
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
-c & 1 & 0 & 0 \\
c^2 & -2c & 1 & 0 \\
-c^3 & 3 c^2 & -3 c & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-c^4 & 4 c^3 & - 6 c^2 & 4 c \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
c & 1 & 0 & 0 \\
c^2 & 2c & 1 & 0 \\
c^3 & 3 c^2 & 3 c & 1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrr}
c & 1 & 0 & 0 \\
0 & c & 1 & 0 \\
0 & 0 & c & 1 \\
0 & 0 & 0 & c \\
\end{array}
\right)
$$
</s>