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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?

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  • $\begingroup$ It seems to me that $A=-C^T$, where $C$ is the companion matrix of the polynomial $x^n-a_{n-1}x^{n-1}-a_{n-2}x^{n-2}\ldots$ $\endgroup$ – Giuseppe Negro May 21 at 9:51
  • $\begingroup$ @GiuseppeNegro How to derive its corresponding characteristic polynomial via the polynomial mentioned in your comment? Could you elaborate a bit more? $\endgroup$ – mathdoge May 22 at 11:23
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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 $$ P_A(x)=P_{-C^T}(x)=(-1)^nP_C(x)=(-1)^np(x). $$

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