I've been tasked with the following:
Show that the companion matrix $C(p)$ of $p(x) = x^2 + ax + b$ has characteristic polynomial $\lambda^2 + a\lambda + b$.
Show that if $\lambda$ is an eigenvalue of the companion matrix $C(p)$, then $\begin{pmatrix}\lambda\\1\end{pmatrix}$ is an eigenvector of $C( p)$ corresponding to $\lambda$.
Now the first part is rather simple, one just has to show
$$\text{det}(C(p)-\lambda I)$$
yields the desired equation.
$$C(p)-\lambda I=\left( \begin{array}{ccc}
-a-\lambda & -b \\
1 & 0-\lambda \end{array} \right)$$
$$\left| \begin{array}{ccc}
-a-\lambda & -b \\
1 & 0-\lambda \end{array} \right|=(-a-\lambda)(-\lambda)-(-1)(b)$$
$$=a\lambda+\lambda^2+b$$
$$=\lambda^2+a\lambda+b$$
What gets me is the second part, how does one prove that? From everything I've seen it shouldn't actually be true.
If $\lambda$ is an eigenvalue of $C(p)$ with eigenvector $\begin{pmatrix}\lambda\\1\end{pmatrix}$ then
$$C(p)\begin{pmatrix}\lambda\\1\end{pmatrix}=\lambda\begin{pmatrix}\lambda\\1\end{pmatrix}$$
Should be true, but
$$\left( \begin{array}{ccc}
-a & -b \\
1 & 0 \end{array} \right)
\begin{pmatrix}\lambda\\1\end{pmatrix}=\begin{pmatrix}-a\lambda-b\\\lambda\end{pmatrix}$$
Which does not equal
$$\lambda\begin{pmatrix}\lambda\\1\end{pmatrix}=\begin{pmatrix}\lambda^2\\\lambda\end{pmatrix}$$
What am I missing here?
Best Answer
$\lambda$ is a root of the characteristic polynomial $x^2 + ax + b$, hence $-a\lambda-b=\lambda^2$, and you are done.