Modulus of eigenvalues of a companion matrix I have the companion matrix
$$\mathbf{M}:=\left(\begin{array}{ccccc}
1-p+pa_{1}&pa_{2}&pa_{3}&\cdots&pa_{m}\\
1&&&&\\
&1&&&\\
&&1&&\\
&&&1&\end{array}\right)$$
where $p\in(0,1)$ and $\sum_{i}a_{i}=1$, and the null entries are $0$.
So, obviously, $\lambda_{1}:=1$ is an eigenvalue and the corresponding right-eigenvector is $\mathbf{v}_{1}:=(1,1,\cdots,1)^{\mathsf{T}}$.
How can I show that the modulus of the other eigenvalues are strictly smaller than $1$?
I believe this is something well-known.
Thank you.
 A: Define the characteristic polynomial by $P(\lambda):=\det(\mathbf{M}-\lambda\mathbf{I})$, then
\begin{equation}
P(\lambda):=(\pm1)\bigl[\underbrace{\lambda^{m}-(1-p+pa_{1})\lambda^{m-1}-pa_{2}\lambda^{m-2}-\cdots-pa_{m}}_{Q(\lambda)}\bigr]=0.\nonumber
\end{equation}
If $P(\lambda)=0$ and $|\lambda|>1$, then
\begin{align}
1\leq{}&(1-p+pa_{1})\frac{1}{|\lambda|}+pa_{2}\frac{1}{|\lambda|^{2}}+\cdots+pa_{m}\frac{1}{|\lambda|^{m}}\nonumber\\
<{}&(1-p+pa_{1})+pa_{2}+\cdots+pa_{m}=1,\nonumber
\end{align}
which is a contradiction.
Thus, if $P(\lambda)=0$, then $|\lambda|\leq1$.
Clearly, $P(1)=0$ and
\begin{align}
Q^{\prime}(1)={}&1+(m-1)p-(m-1)pa_{1}-(m-2)pa_{2}-\cdots-pa_{m-1}\nonumber\\
\geq{}&1+(m-1)p-(m-1)pa_{1}-(m-1)pa_{2}-\cdots-(m-1)pa_{m-1}\nonumber\\
={}&1+(m-1)p[1-(a_{1}+a_{2}+\cdots+a_{m-1})]\geq1,\nonumber
\end{align}
which shows that $P^{\prime}(1)=\pm{}Q^{\prime}(1)\neq0$,
i.e., $1$ is simple.
Next, assume that $P(\lambda)=0$ and $|\lambda|=1$ such that $\lambda\neq1$, i.e, $\lambda=\mathrm{e}^{\mathrm{i}\theta}$, where $\theta\in(0,2\pi)$.
Since $\theta\in(0,2\pi)$, we have
\begin{align}
1={}&\bigl|(1-p+pa_{1})\mathrm{e}^{-\mathrm{i}\theta}+pa_{2}\mathrm{e}^{-2\mathrm{i}\theta}+\cdots+pa_{m}\mathrm{e}^{-m\mathrm{i}\theta}\bigr|\nonumber\\
<{}&(1-p+pa_{1})+pa_{2}+\cdots+pa_{m}=1,\nonumber
\end{align}
which is again a contradiction.
Thus, $|\lambda|<1$ if $P(\lambda)=0$ and $\lambda\neq1$.
A: Use Gershgorin theorem. The eigenvalues must be confined in the disks $\{λ\in\mathbb C\ :\ |\lambda|\le 1\}$ and $$\left\{\lambda\in\mathbb{C}:\ |\lambda- (1-p+p a_{1})|\le \sum_{j=2}^{m} a_{j}\right\}.$$ The second disk is included in the first. 
IMPORTANT. This shows that any eigenvalue $\lambda$ must satisfy $|\lambda|\le 1$, but it is not strong enough to prove the strict bound $|\lambda|<1$ except for $\lambda=1$. To obtain this stronger result I think there is no way around the computations with the characteristic polynomial of the other answer.
