2
$\begingroup$

Hi fellow mathematicians,

I've tried to find the characteristic polynomial of the following matrix

$$ A=\begin{pmatrix} 0 & 0 & \cdots & 0 & -a_n \\ 1 & \ddots & \ddots & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots& \ddots & 0 & -a_2 \\ 0 & \cdots & 0 & 1 & -a_1 \end{pmatrix}\in M_{n \times n}(\mathbb{K}) $$ for $\mathbb K$ being some element of a field and $n\in\mathbb N$.

I tried for several hours but I just couldn't succeed! Maybe someone can find an easy solution? Appreciated!

Edit: Apparently this matrix is called a companion matrix.

$\endgroup$
1
$\begingroup$

Take $A-\lambda I$. Now

  1. add $\lambda^{-1}$ times the first row to the second. By doing so you cancelled the $1$ in position $(2,1)$ (2nd row, 1st column) and the last element of the row is $-a_{n-1}-\lambda^{-1}a_n$;
  2. add $\lambda^{-1}$ times the second row to the third...

proceed like this, and in the end by doing these elementary matrix operations you end up with the following upper triangular matrix which has the same determinant of $A-\lambda I$: $$ \begin{pmatrix} -\lambda & & & -a_n \\ & -\lambda & & -a_{n-1}-\lambda^{-1}a_n \\ & & \ddots & \vdots \\ & & & -a_1-\lambda-\sum_{i=2}^na_i\lambda^{1-i} \\ \end{pmatrix} $$ The determinant is simply the product of the elements of the diagonal, hence yielding $$ {\rm det}(A-\lambda I) = (-\lambda)^{n-1} \left( -a_1-\lambda-\sum_{i=2}^na_i\lambda^{1-i} \right) = (-1)^n\sum_{i=1}^na_i\lambda^{n-i} $$

$\endgroup$
  • $\begingroup$ You are welcome :) $\endgroup$ – AndreasT Jan 27 '16 at 14:12
1
$\begingroup$

A proof without determinants:

Let $p(x)=x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$ and $(e_1, e_2, \dots, e_n)$ be the canonical basis of $\mathbb K^n$. We have \begin{align*} Ae_1&=e_2,& Ae_2&=A^2e_1=e_3, &&\dots&Ae_{n-1}&=e_n=A^{n-1}e_1, & Ae_n&= A^ne_1. \end{align*} On the other hand, the last columns says that \begin{align*} Ae_n&=-a_ne_1-a_{n-1}e_2-\dots-a_1e_n\\&=-a_nIe_1-a_{n-1}Ae_1-\dots-a_1A^{n-1}e_1, \end{align*} whence $\;A^ne_1+a_1A^{n-1}e_1+\dots+ a_{n-1}Ae_1+a_nIe_1=p(A)e_1=0$.

There results that $A^{k-1}p(A)e_1=0=p(A)A^{k-1}e_1=p(A)e_k\enspace(k=2,\dots,n)$. Since $p(A)=0$ on a basis, it is $0$.

Furthermore, $p(x)$ has degree $n$, so its characteristic polynomial is $$\chi_A(x)=(-1)^np(x).$$

Note: $p(x)$ is also the minimal polynomial of $A$.

Indeed, if there were a monic polynomial $q(x)=x^d+b_1x^{d-1}+\dots+b_{d-1}x+b_d$, of degree $d<n$ such that $q(A)=0$, we would have $\;A^d e_1++b_1A^{d-1}e_1+\dots+b_{d-1}Ae_1+b_d e_1=0$, i.e. a non-trivial relation: $$e_{d+1}+b_1e_d+\dots+b_{d-1}e_2+b_d e_1=0$$ between vectors of the basis.

$\endgroup$
  • $\begingroup$ I think one first has to show that $p(x)$ is the minimal polynomial, to conclude that $\chi_A(x) = (-1)^n p(x)$, because otherwise $p$ could be some multiple of the minimal polynomial which must not necessarily be $p$. $\endgroup$ – red_trumpet Nov 19 '18 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.