Jordan form of a companion matrix

Asked Mar 25 '25 at 07:50

Active Mar 25 '25 at 09:54

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Let $n \ge 2$ be an integer. For $i = 0, 1, \dots , n-1$, let $c_{i} = {n \choose i}$. Find the Jordan form of the following companion matrix.

$$ A = \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & 0 & \cdots & 0 & -c_1 \\ 0 & 1 & 0 & \cdots & 0 & -c_2 \\ 0 & 0 & 1 & \cdots & 0 & -c_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -c_{n-1} \\ \end{pmatrix}. $$

I know that this matrix happens to have characteristic polynomial $$ \text{ch}_A(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + x_0 = (x+1)^n, $$ so its Jordan canonical form must have all $-1$'s on the diagonal, but I don't know how to proceed from here.

edited Mar 25 '25 at 09:54

Rodrigo de Azevedo

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asked Mar 25 '25 at 07:50

Jonathan Huang

So here $A$ is the companion matrix of the polynomial $(x+1)^n$. Do you know the fact that the companion matrix has its characteristic polynomial equal to the minimal polynomial? Do you see how that solves the problem? – testaccount Mar 25 '25 at 07:55
Yes, the degree of the minimal polynomial is equal to the size of the largest Jordan block; this would imply $A$ is a Jordan block. But I do not know that the companion matrix has its characteristic polynomial equal to the minimal polynomial. – Jonathan Huang Mar 25 '25 at 08:01
Your question of comment can be answered by this – Myungheon Lee Mar 25 '25 at 08:05
See also this post – Ben Grossmann Mar 25 '25 at 13:58

Jordan form of a companion matrix

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