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I have a given monic polynomial $P(s)=\sum\limits_{k=0}^N a_ks^k $ of degree $N$, and I construct this matrix which has $P(s)$ for a characteristic polynomial:

$$ M = \begin{bmatrix} -a_{N-1} & -a_{N-2} & -a_{N-3} & \cdots & -a_2 & -a_1 & -a_0 \\ 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 & 0\\ 0 & 0 & 0 & \cdots & 0 & 1 & 0\\ \end{bmatrix} $$

In other words, the top row is the negated coefficients except for the leading term, and the diagonal below the main diagonal contains all ones, and the remaining entries are zero.

Is there a name for this special matrix? (it comes up in the context of linear dynamic systems)

I know the eigenvalues are the roots of the characteristic polynomial, but is there a shortcut for finding the eigenvectors as a function of the coefficients?

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  • $\begingroup$ The polynomial $P$ is the minimal polynomial of $M$, not just the characteristic polynomial. $\endgroup$ Dec 17, 2015 at 21:43

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This matrix is very similar to the Comnpanion matrix, except you have the matrix transposed and flipped upside-down.

The eigenvectors of this matrix should be in the form $\begin{bmatrix}\lambda^{n-1}\\\lambda^{n-2}\\ \vdots \\ \lambda \\ 1\end{bmatrix}$ where $\lambda$ is a root of the characteristic polynomial.

EDIT: I'm assuming that the leading coefficient of the characteristic polynomial is $a_N = 1$. Let me know if that's not the case.

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  • $\begingroup$ correct, i had intended to include "monic" and somehow skipped it. Edited question. $\endgroup$
    – Jason S
    Dec 17, 2015 at 21:01

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