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Show that $p(\lambda)=\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots+a_{0}$ is the characteristic polynomial of the matrix
$A=\left(\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & & 0 \\ & \vdots & & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_{0} & -a_{1} & -a_{2} & \cdots & -a_{n-1} \end{array}\right),$
0 that is, $\operatorname{det}(\lambda I-A)=p(\lambda)$. In particular, every polynomial is the characteristic polynomial of a matrix.

If $\lambda$ is an eigenvalue of $A$ show that $\left(1, \lambda, \lambda^{2}, \ldots, \lambda^{n-1}\right)$ is a corresponding eigenvector. The above matrix is called the companion matrix of $p(\lambda)$. It arises naturally when writing a linear $\mathrm{n}^{\text {th }}$ order ordinary differential equation as a system of first order equations. Gershgorin's Theorem (Problem $134 \mathrm{~g}$ ) can be applied to this matrix to find simple, but useful, estimates of the roots of any polynomial.
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