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? 
 A: 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 
$$p^{\prime}(\lambda)=0$$
A: Your matrix is the transpose of the companion matrix. It can be shown that its characteristic polynomial and minimal polynomial are both equal to $$p(\lambda) = \lambda^n + a_1\lambda^{n - 1} + \cdots + a_n$$
Assume the minimal polynomial splits as $$p(\lambda) = (\lambda - \lambda_1)^{p_1}\cdots(\lambda - \lambda_k)^{p_k}$$ Then every Jordan block is of the maximal dimension $p_i$. 
$$\pmatrix{
\lambda_i & 1 & 0 &\cdots & 0 & 0\\
0 & \lambda_i & 1 &\cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots& \vdots & \vdots\\
0 & 0 &0& \cdots &\lambda_i & 1 \\
0 & 0 & 0&\cdots& 0 &\lambda_i \\
}$$
Hence the Jordan form is
$$J = \pmatrix{
\lambda_1 & 1 & & & & \\
 &\ddots & 1 & & & \\
& &\lambda_1 &  & & & \\
& & &\ddots &  & & & \\
& & & &\lambda_k & 1 \\
& & & & &\ddots & 1 \\
& & & & & &\lambda_k \\
}$$
Let $v_i \in \ker(A - \lambda_i I)^{p_i}$ be such that $(A - \lambda_i I)^{p_i-1}v_i \ne 0$. The corresponding Jordan basis can be obtained as
\begin{align}
&(A-\lambda_1 I)^{p_1-1}v_1, (A-\lambda_1 I)^{p_1-2}v_1, \ldots, (A-\lambda_1 I)v_1, v_1, \\
&(A-\lambda_2 I)^{p_2-1}v_2, (A-\lambda_2 I)^{p_2-2}v_2, \ldots, (A-\lambda_2 I)v_2, v_2,  \\
&\quad\quad\quad\quad\quad\quad\quad\quad\vdots\\
&(A-\lambda_k I)^{p_k-1}v_k, (A-\lambda_k I)^{p_k-2}v_k, \ldots, (A-\lambda_k I)v_k, v_k
\end{align}
The change-of-basis matrix $B$ has columns equal precisely to the above vectors.
A: 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
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
-c & 1 & 0 & 0 \\
c^2 & -2c & 1 & 0 \\
-c^3 & 3 c^2 & -3 c & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-c^4 & 4 c^3 & - 6 c^2 & 4 c \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
c & 1 & 0 & 0 \\
c^2 & 2c & 1 & 0 \\
c^3 & 3 c^2 & 3 c & 1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrr}
c & 1 & 0 & 0 \\
0 & c & 1 & 0 \\
0 & 0 & c & 1 \\
0 & 0 & 0 & c \\
\end{array}
\right)
$$
