= Solution
{c}
A <scalar> $\lambda$ is an <eigenvalue> of $A$ if and only if there is a nonzero <vector> $v$ with $Av=\lambda v$. By the preceding <kernel> criterion,
$$
Av=\lambda v\text{ for some }v\ne0
\iff (\lambda I_n-A)v=0\text{ for some }v\ne0
\iff\det(\lambda I_n-A)=0.
$$
The last condition is $\chi_A(\lambda)=0$, exactly the definition of a <polynomial root>. For a <real matrix>, this proves the assertion for real <eigenvalues>; the same proof over the <complex numbers> gives the usual complex version.
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