Finding Jordan-normal form of a special $n\times n$ matrix

Question

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?

Answer 1

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

$$p^{\prime}(\lambda)=0$$

Answer 2

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.

Answer 3

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 $$ \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ -c & 1 & 0 & 0 \\ c^2 & -2c & 1 & 0 \\ -c^3 & 3 c^2 & -3 c & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -c^4 & 4 c^3 & - 6 c^2 & 4 c \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ c & 1 & 0 & 0 \\ c^2 & 2c & 1 & 0 \\ c^3 & 3 c^2 & 3 c & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} c & 1 & 0 & 0 \\ 0 & c & 1 & 0 \\ 0 & 0 & c & 1 \\ 0 & 0 & 0 & c \\ \end{array} \right) $$