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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?

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$\lambda$ is a root of the characteristic polynomial $x^2 + ax + b$, hence $-a\lambda-b=\lambda^2$, and you are done.

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  • $\begingroup$ Good god, I knew it was dead simple. This is what you get for doing math in the wee hours of the morning. $\endgroup$ – Nick Chapman Jul 27 '15 at 7:11
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Hint The argument so far does not use that $\lambda$ is an eigenvalue: Since $\lambda$ is an eigenvalue, it is a root of the characteristic polynomial.

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