# Find the characteristic and minimal polynomial of the given matrix

Let $a_1, ..., a_{n-1} \in \mathbb{C}$ and let

$$A = \begin{pmatrix} 0 & 0 & 0 & \cdots & a_0 \\ 1 & 0 & & \cdots & a_1 \\ 0 & 1 & \ddots & & \vdots \\ \vdots & \vdots & \ddots & 0 & a_{n-2} \\ 0 & 0 & \cdots & 1 & a_{n-1} \end{pmatrix}$$

Find the characteristic and minimal polynomial of $A$.

A general solution would suffice as long as it is clear how it would be accomplished.

Thanks everyone!

• What have you tried so far? Also, for the characteristic polynomial, look at your matrix and think about which 'sub' determinants will equal zero. – mattos Nov 23 '15 at 23:46
• I was able to calculate the characterisitic polynomial from the matrix, but did not see any way to deduce the minimal polynomial from it. that made me think that perhaps a straight-forward approach was not the correct way. – Yotam Alon Nov 24 '15 at 10:21

Proof by induction that $\chi_A=(-1)^nX^n-a_{n-1}X^{n-1}-...-a_1X-a_0$ :

For $n=2$ see that $$\chi_A=\begin{vmatrix}-X&a_0\\1&a_1-X\end{vmatrix}=X^2-a_1X-a_0.$$ Suppose it works for $(n-1)\in\mathbb{N}.$ Then : \begin{align*} \chi_A &= \begin{vmatrix} -X & 0 & 0 & \cdots & a_0 \\ 1 & -X & & \cdots & a_1 \\ 0 & 1 & \ddots & & \vdots \\ \vdots & \vdots & \ddots & -X & a_{n-2} \\ 0 & 0 & \cdots & 1 & a_{n-1}-X \end{vmatrix}\\ &=-X\times\begin{vmatrix} -X & & \cdots & a_1 \\ 1 & \ddots & & \vdots \\ \vdots & \ddots & -X & a_{n-2} \\ 0 & \cdots & 1 & a_{n-1}-X \end{vmatrix}-1\times\underset{=a_0}{\underbrace{\begin{vmatrix} 0 & 0 & \cdots & a_0 \\ 1 & \ddots & & \vdots \\ \vdots & \ddots & -X & a_{n-2} \\ 0 & \cdots & 1 & a_{n-1}-X \end{vmatrix}}}\\ &=-X\times\big((-1)^{n-1}X^{n-1}-a_{n-1}X^{n-2}-...-a_2X-a_1\big)-a_0\\ &=(-1)^nX^n-a_{n-1}X^{n-1}-...-a_1X-a_0.\end{align*} Then you get your result.

For the minimal polynomial $\mu_{C(P)}$, you can see that if you note $f$ the endomorphism associated at $C(P)$ in the canonical base of $\mathbb{C}^n,$ then $$\forall i\in\{0,...,n-1\}, f(e_{i+1})=f^{(i)}(e_1),$$ and $\big(f^{(i)}(e_1)\big)_{0\leq i\leq n-1}$ is an independant family of $n$ vectors, so the punctual minimal polynomial $\mu_{f,e_1}$ of $f$ about $e_1$ is of degree $n$ or more. As this polynomial divides $\mu_f,$ you know that $\mu_f$ is of degree $n$ or more and so $$\mu_f=(-1)^n\chi_f.$$

• I got that far myself, not by induction though induction is much simplar, but my problem was deducing the minimal polynomial from this charasteristic polynomial. – Yotam Alon Nov 24 '15 at 6:25
• I will edit that ! – Balloon Nov 24 '15 at 10:42
• Incredible! :D thanks! – Yotam Alon Nov 25 '15 at 5:39
• You're welcome ! – Balloon Nov 25 '15 at 6:30