Consider the following $n \times n$ matrix (I believe this is similar to companion matrix):
$$ A = \begin{pmatrix}
0 & -1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 0 & -1 & \cdots & 0 & 0 & 0 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 0 & -1 & 0\\
0 & 0 & 0 & \cdots & 0 & 0 & -1\\
-a_{0} & -a_{1} & -a_{2} & \cdots & -a_{n-3} & -a_{n-2} & -a_{n-1}
\end{pmatrix}
$$
I tried to find its characteristic polynomial.
$$ \lambda I – A_n = \begin{pmatrix}
\lambda & -1 & 0 & \cdots & 0 & 0 & 0 \\
0 & \lambda & -1 & \cdots & 0 & 0 & 0 \\
0 & 0 & \lambda & \cdots & 0 & 0 & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \lambda & -1 & 0\\
0 & 0 & 0 & \cdots & 0 & \lambda & -1\\
a_{0} & a_{1} & a_{2} & \cdots & a_{n-3} & a_{n-2} & a_{n-1}+\lambda
\end{pmatrix}
$$
Expand it against the last column, we have that $$\det(\lambda I_n – A_n) = (a_{n-1}+\lambda)\lambda^{n-1} + \det(\lambda I_{n-1} – A_{n-1}),$$ where $I_n$ is the $n \times n$ identity matrix. So I think I will be able to find the characteristic polynomial recursively. So, $A_n$'s charateristic polynomial $$F_n(\lambda) = \lambda^n + (a_{n-1}+1)\lambda^{n-1} + (a_{n-2}+1) \lambda^{n-2} + \cdots + (a_2+1)\lambda^2 + a_1\lambda + a_0. (n \geq 3)$$
But when I tried to compute its eigenvector for given eigenvalue $\lambda$, I used $A_n(x_1, \cdots, x_n)^{T} = \lambda(x_1, \cdots, x_n)^{T}$, and found that $x_k = \lambda^{k-1}x_1$ and that $x_1(a_0 + a_1\lambda + \cdots + a_{n-1}\lambda^{n-1}) = 0$. If I could conclude that $a_0 + a_1\lambda + \cdots + a_{n-1}\lambda^{n-1} = 0$, then $x_1$ would be arbitrary, then the eigenvector is just $x_1(1, \lambda, \cdots, \lambda^{n-1})^{T}$. However, I don't think this is an immediate result from $F(\lambda) = 0$.
Could anyone give me some hint on these?
Best Answer
Let $C$ denote the companion matrix of $$ p(x)=x^n-a_{n-1}x^{n-1}-\ldots-a_0.$$ Then $A=-C^T$. Now use the notation $P_A$ to denote the characteristic polynomial of $A$; that is, $$ P_A(x):=\det(A-xI).$$ Then $$ P_A(x)=P_{-C^T}(x)=(-1)^nP_C(x)=(-1)^np(x). $$