Question

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.