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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?

mechanodroid
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user388557
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  • Do you have specific values for $a_i$ ? – Rene Schipperus Mar 03 '18 at 15:39
  • @ReneSchipperus no – user388557 Mar 03 '18 at 15:41
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    What kind of answer are you expecting ? You factor the polynomial and this gives you the Jordan form. I believe with this companion matrix the minimal and characteristic polynomials are equal. – Rene Schipperus Mar 03 '18 at 15:44
  • @ReneSchipperus I think there is a way to write down $B$ and find the length of the Jordan chains just by knowing the solutions to the charactersitic polynomial. – user388557 Mar 03 '18 at 15:47
  • In the comments and answers it is explained how there is only one choice for $J$. Finding $B$ seems to me to be (in general) a brute. But if the polynomial has distinct roots (as so often happens in DEs ;-) ) then I suggest that Lagrange interpolation and Vandermonde matrices may be your friends. – ancient mathematician Mar 03 '18 at 17:14

3 Answers3

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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

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

Rene Schipperus
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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.

mechanodroid
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  • But the question asks more than this: it asks for the matrix $B$ which transforms the Rational Form basis to a Jordan Form basis. – ancient mathematician Mar 03 '18 at 16:56
  • @ancientmathematician I have added how to obtain the matrix $B$. It is expressed in the roots $\lambda_1, \ldots, \lambda_k$ of the minimal polynomial (not only in its coefficients $a_i$). – mechanodroid Mar 03 '18 at 17:42
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    @ancientmathematician people ask for lots of things that they hope exist. – Will Jagy Mar 03 '18 at 17:55
  • Thanks @mechanodroid; of course this is the theoretical solution. But the other two answers give the change of basis matrix more explicitly in terms of the derivatives of the $(1,\lambda,\lambda^2, \dots)$ vector. For distinct roots it is essentially Lagrange interpolation/Vandermonde; for repeated roots it is a generalisation that requires the derivatives. – ancient mathematician Mar 04 '18 at 07:44
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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 $$ \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ -c & 1 & 0 & 0 \\ c^2 & -2c & 1 & 0 \\ -c^3 & 3 c^2 & -3 c & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -c^4 & 4 c^3 & - 6 c^2 & 4 c \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ c & 1 & 0 & 0 \\ c^2 & 2c & 1 & 0 \\ c^3 & 3 c^2 & 3 c & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} c & 1 & 0 & 0 \\ 0 & c & 1 & 0 \\ 0 & 0 & c & 1 \\ 0 & 0 & 0 & c \\ \end{array} \right) $$

Will Jagy
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