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.
Best Answer
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 $\lambda\neq1$.